CBSE Class 12 Mathematics Vector Algebra Assignment Set 01

Points to Remember

• Vector: A quantity that possesses both a numerical magnitude and a directional component is called a vector, which is represented geometrically by a directed line segment.
• Collinear Vectors: Two or more vectors are classified as collinear if they are parallel to the same straight line, regardless of their individual magnitudes.
• Position Vector: The position vector of any point \( P(a, b, c) \) relative to the coordinate origin \( O(0, 0, 0) \) is written as \( \vec{OP} = a\hat{i} + b\hat{j} + c\hat{k} \). The length or magnitude of this vector is given by:
  \[ |\vec{OP}| = \sqrt{a^2 + b^2 + c^2} \]
• Vector Joining Two Points: For any two points \( A(x_1, y_1, z_1) \) and \( B(x_2, y_2, z_2) \) in space, the vector starting at \( A \) and ending at \( B \) is:
  \[ \vec{AB} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k} \] The distance between the two points is the magnitude:
  \[ |\vec{AB}| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
• Triangle Law of Vector Addition: If two vectors are positioned to represent two successive sides of a triangle in order, their vector sum is represented in magnitude and direction by the third side of the triangle drawn in the reverse direction.
• Scalar Multiplication: Multiplying a vector \( \vec{a} \) by a real scalar \( \lambda \) produces a collinear vector \( \lambda\vec{a} \), whose magnitude is scaled by the absolute value of the scalar:
  \[ |\lambda\vec{a}| = |\lambda||\vec{a}| \]
• Collinearity Condition: Two vectors \( \vec{a} \) and \( \vec{b} \) are collinear if and only if one is a scalar multiple of the other:
  \[ \vec{a} = \lambda\vec{b} \]
• Unit Vector: Any non-zero vector \( \vec{a} \) can be normalized and expressed as \( \vec{a} = |\vec{a}|\hat{a} \), where \( \hat{a} = \frac{\vec{a}}{|\vec{a}|} \) is a unit vector pointing in the direction of \( \vec{a} \).
• Section Formula: If \( \vec{a} \) and \( \vec{b} \) are the position vectors of points \( A \) and \( B \), the position vector \( \vec{c} \) of a point \( C \) dividing the line segment \( AB \) in the ratio \( m : n \) is given by:
  • For internal division: \( \vec{c} = \frac{m\vec{b} + n\vec{a}}{m + n} \)
  • For external division: \( \vec{c} = \frac{m\vec{b} - n\vec{a}}{m - n} \)
• Direction Cosines and Angles: If a vector \( \vec{r} = a\hat{i} + b\hat{j} + c\hat{k} \) makes angles \( \alpha, \beta, \gamma \) with the positive x, y, z-axes respectively, then \( \cos\alpha, \cos\beta, \cos\gamma \) are its direction cosines, usually written as \( l, m, n \):
  \[ l = \frac{a}{|\vec{r}|}, \quad m = \frac{b}{|\vec{r}|}, \quad n = \frac{c}{|\vec{r}|} \] These satisfy the identity:
  \[ l^2 + m^2 + n^2 = 1 \]
• Direction Ratios: Any triplet of real numbers \( a, b, c \) that are directly proportional to the direction cosines \( l, m, n \) are known as direction ratios.
• Scalar (Dot) Product: The scalar product of two vectors is a real number defined by:
  \[ \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta \] where \( \theta \) is the angle between them (\( 0 \le \theta \le \pi \)).
  • Commutative Property: \( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} \)
  • Perpendicularity: \( \vec{a} \cdot \vec{b} = 0 \iff \vec{a} \perp \vec{b} \) (for non-zero vectors)
  • Self-product: \( \vec{a} \cdot \vec{a} = |\vec{a}|^2 \)
  • In orthonormal components: \( \vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \)
• Vector Projection:
  • The scalar projection of \( \vec{a} \) on \( \vec{b} \) is: \( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \)
  • The vector projection of \( \vec{a} \) along \( \vec{b} \) is: \( \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \right) \hat{b} \)
• Vector (Cross) Product: The cross product of two vectors is a vector defined by:
  \[ \vec{a} \times \vec{b} = |\vec{a}||\vec{b}|\sin\theta \hat{n} \] where \( \hat{n} \) is a unit vector perpendicular to the plane of both \( \vec{a} \) and \( \vec{b} \), obeying the right-hand rule.
  • Anti-commutative: \( \vec{a} \times \vec{b} = - (\vec{b} \times \vec{a}) \)
  • Parallelism: \( \vec{a} \times \vec{b} = \vec{0} \iff \vec{a} \parallel \vec{b} \) (for non-zero vectors)
  • Component Form:
    \[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} \]
  • Unit vector perpendicular to both \( \vec{a} \) and \( \vec{b} \): \( \pm \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} \)
• Geometric Areas using Vectors:
  • Area of a parallelogram with adjacent sides \( \vec{a} \) and \( \vec{b} \): \( \text{Area} = |\vec{a} \times \vec{b}| \)
  • Area of a parallelogram with diagonals \( \vec{d}_1 \) and \( \vec{d}_2 \): \( \text{Area} = \frac{1}{2}|\vec{d}_1 \times \vec{d}_2| \)
  • Area of a triangle with adjacent sides \( \vec{a} \) and \( \vec{b} \): \( \text{Area} = \frac{1}{2}|\vec{a} \times \vec{b}| \)
• Scalar Triple Product: The dot product of a vector with the cross product of two other vectors is defined as:
  \[ [\vec{a}\ \vec{b}\ \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c}) \]
  • It represents the volume of a parallelepiped with coterminous edges \( \vec{a}, \vec{b}, \vec{c} \).
  • Coplanarity: \( \vec{a}, \vec{b}, \vec{c} \) are coplanar if and only if \( [\vec{a}\ \vec{b}\ \vec{c}] = 0 \).
  • Determinant Form:
    \[ [\vec{a}\ \vec{b}\ \vec{c}] = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} \]

Very Short Answer Type Questions (1 Mark)

Question 1. What are the horizontal and vertical components of a vector \( \vec{a} \) of magnitude 5 making an angle of 150° with the direction of x-axis.
Answer: Let \( \theta = 150^\circ \) be the angle the vector \( \vec{a} \) makes with the positive x-axis. The horizontal component \( a_x \) is calculated as:
\( a_x = |\vec{a}| \cos(150^\circ) = 5 \cos(180^\circ - 30^\circ) = 5 (-\cos 30^\circ) = 5 \left( -\frac{\sqrt{3}}{2} \right) = -\frac{5\sqrt{3}}{2} \)
The vertical component \( a_y \) is calculated as:
\( a_y = |\vec{a}| \sin(150^\circ) = 5 \sin(180^\circ - 30^\circ) = 5 \sin 30^\circ = 5 \left( \frac{1}{2} \right) = \frac{5}{2} \)
Therefore, the horizontal and vertical components are \( -\frac{5\sqrt{3}}{2} \) and \( \frac{5}{2} \) respectively.
In simple words: We find how much the vector stretches along the flat ground and how much it goes straight up by multiplying its length by the cosine and sine of its angle.

Exam Tip: Remember that \( 150^\circ \) lies in the second quadrant, where the horizontal (cos) component is negative and the vertical (sin) component is positive.

Question 2. What is \( a \in \mathbb{R} \) such that \( |a\vec{x}| = 1 \), where \( \vec{x} = \hat{i} - 2\hat{j} + 2\hat{k} \)?
Answer: First, let us compute the magnitude of the given vector \( \vec{x} \):
\( |\vec{x}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \)
Using the properties of vector magnitude, we have:
\( |a\vec{x}| = |a| |\vec{x}| = 1 \)
Substituting the value of \( |\vec{x}| \):
\( |a| (3) = 1 \implies |a| = \frac{1}{3} \implies a = \pm \frac{1}{3} \)
In simple words: The length of the original vector is 3. To scale it so that its new length is exactly 1, we must multiply it by either positive or negative one-third.

Exam Tip: Do not miss the negative sign; since \( |a| \) is an absolute value, \( a \) can be both positive and negative \( \frac{1}{3} \).

Question 3. When is \( |\vec{x} + \vec{y}| = |\vec{x}| + |\vec{y}| \)?
Answer: According to the triangle inequality for vectors, \( |\vec{x} + \vec{y}| \le |\vec{x}| + |\vec{y}| \). The equality \( |\vec{x} + \vec{y}| = |\vec{x}| + |\vec{y}| \) holds true if and only if the vectors \( \vec{x} \) and \( \vec{y} \) point in the exact same direction. In other words, \( \vec{x} \) and \( \vec{y} \) must be like parallel (collinear) vectors, meaning the angle \( \theta \) between them is \( 0^\circ \).
In simple words: The combined length of two vectors is only equal to their individual lengths added together if they are lined up straight in the same direction.

Exam Tip: Use the term "like parallel vectors" or state that the angle between them is zero to secure full marks.

Question 4. What is the area of a parallelogram whose sides are given by \( 2\hat{i} - \hat{j} \) and \( \hat{i} + 5\hat{k} \)?
Answer: Let \( \vec{a} = 2\hat{i} - \hat{j} \) and \( \vec{b} = \hat{i} + 5\hat{k} \) represent the adjacent sides of the parallelogram. The area of the parallelogram is given by the magnitude of their cross product, \( |\vec{a} \times \vec{b}| \). First, find the cross product using a determinant:
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 0 \\ 1 & 0 & 5 \end{vmatrix} \)
\( \implies \vec{a} \times \vec{b} = \hat{i}((-1)(5) - 0) - \hat{j}((2)(5) - 0) + \hat{k}(0 - (-1)(1)) \)
\( \implies \vec{a} \times \vec{b} = -5\hat{i} - 10\hat{j} + \hat{k} \bk \)
Now, calculate its magnitude:
\( \text{Area} = |\vec{a} \times \vec{b}| = \sqrt{(-5)^2 + (-10)^2 + 1^2} = \sqrt{25 + 100 + 1} = \sqrt{126} \text{ sq. units} \)
In simple words: To find the area of a parallelogram when we know its side vectors, we calculate their cross product and then find the length of the resulting vector.

Exam Tip: Ensure your determinant calculation is correct; a common slip is forgetting the negative sign of the \( \hat{j} \) term.

Question 5. What is the angle between \( \vec{a} \) and \( \vec{b} \), If \( \vec{a} \cdot \vec{b} = 3 \) and \( |\vec{a} \times \vec{b}| = 3\sqrt{3} \).
Answer: Let \( \theta \) be the angle between the vectors. We know that:
\( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta = 3 \)
\( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin\theta = 3\sqrt{3} \)
Dividing the second equation by the first:
\( \frac{|\vec{a}| |\vec{b}| \sin\theta}{|\vec{a}| |\vec{b}| \cos\theta} = \frac{3\sqrt{3}}{3} \implies \tan\theta = \sqrt{3} \)
Since \( 0 \le \theta \le \pi \), we have:
\( \theta = \frac{\pi}{3} \text{ or } 60^\circ \)
In simple words: By dividing the cross product's magnitude by the dot product, we get the tangent of the angle. Since the tangent value is \( \sqrt{3} \), the angle is 60 degrees.

Exam Tip: Writing the ratio \( \tan\theta = \frac{|\vec{a} \times \vec{b}|}{\vec{a} \cdot \vec{b}} \) directly is a quick and clean way to solve this type of problem.

Question 6. Write a unit vector which makes an angle of \( \frac{\pi}{4} \) with x-axis and \( \frac{\pi}{3} \) with z-axis and an acute angle with y-axis.
Answer: Let the required unit vector be \( \hat{u} = l\hat{i} + m\hat{j} + n\hat{k} \), where \( l, m, n \) are its direction cosines. We are given:
\( l = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \)
\( n = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \)
Using the identity \( l^2 + m^2 + n^2 = 1 \):
\( \left(\frac{1}{\sqrt{2}}\right)^2 + m^2 + \left(\frac{1}{2}\right)^2 = 1 \implies \frac{1}{2} + m^2 + \frac{1}{4} = 1 \)
\( \implies m^2 = 1 - \frac{3}{4} = \frac{1}{4} \implies m = \pm \frac{1}{2} \bk \)
Since \( \hat{u} \) makes an acute angle with the y-axis, the direction cosine \( m = \cos\beta \) must be positive. Thus, \( m = \frac{1}{2} \). The unit vector is:
\( \hat{u} = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{2}\hat{j} + \frac{1}{2}\hat{k} \)
In simple words: The directions of a unit vector are related by a special sum of squares that equals 1. Using the given angles, we find the missing coordinate and choose the positive value for the acute angle.

Exam Tip: Always justify why you selected the positive value of \( m \) by referencing the "acute angle with y-axis" condition.

Question 7. If A is the point (4, 5) and vector \( \vec{AB} \) has components 2 and 6 along x-axis and y-axis respectively then write point B.
Answer: Let the coordinates of point B be \( (x, y) \). The vector \( \vec{AB} \) is given by the difference of coordinates:
\( \vec{AB} = (x - 4)\hat{i} + (y - 5)\hat{j} \)
We are given that the components of \( \vec{AB} \) along the x and y axes are 2 and 6, respectively. This means:
\( \vec{AB} = 2\hat{i} + 6\hat{j} \)
By comparing components:
\( x - 4 = 2 \implies x = 6 \)
\( y - 5 = 6 \implies y = 11 \bk \)
Therefore, the coordinates of point B are \( (6, 11) \).
In simple words: To find the ending point B, we simply add the x and y shift values of the vector to the coordinates of the starting point A.

Exam Tip: Be sure to write the final answer as a coordinate pair \( (6, 11) \) rather than just leaving the values of \( x \) and \( y \) separate.

Question 8. What is the point of trisection of PQ nearer to P if positions of P and Q are \( 3\hat{i} + 3\hat{j} - 4\hat{k} \) and \( 9\hat{i} + 8\hat{j} - 10\hat{k} \) respectively?
Answer: Let \( \vec{p} = 3\hat{i} + 3\hat{j} - 4\hat{k} \) and \( \vec{q} = 9\hat{i} + 8\hat{j} - 10\hat{k} \) be the position vectors of P and Q. The point of trisection of PQ nearer to P divides the segment PQ internally in the ratio \( 1 : 2 \). Using the section formula, the position vector \( \vec{r} \) of this point is:
\( \vec{r} = \frac{1\vec{q} + 2\vec{p}}{1 + 2} = \frac{(9\hat{i} + 8\hat{j} - 10\hat{k}) + 2(3\hat{i} + 3\hat{j} - 4\hat{k})}{3} \)
\( \implies \vec{r} = \frac{(9 + 6)\hat{i} + (8 + 6)\hat{j} + (-10 - 8)\hat{k}}{3} \)
\( \implies \vec{r} = \frac{15\hat{i} + 14\hat{j} - 18\hat{k}}{3} = 5\hat{i} + \frac{14}{3}\hat{j} - 6\hat{k} \)
The coordinates of the point are \( \left(5, \frac{14}{3}, -6\right) \).
In simple words: The point that sits one-third of the way along the line PQ divides it in a 1 to 2 ratio. We use a weighted average formula to calculate its coordinates.

Exam Tip: Pay close attention to the phrasing "nearer to P" vs "nearer to Q" as this determines whether the ratio is \( 1:2 \) or \( 2:1 \).

Question 9. Write the vector in the direction of \( 2\hat{i} + 3\hat{j} + 2\sqrt{3}\hat{k} \), whose magnitude is 10 units.
Answer: Let \( \vec{v} = 2\hat{i} + 3\hat{j} + 2\sqrt{3}\hat{k} \). The magnitude of \( \vec{v} \) is:
\( |\vec{v}| = \sqrt{2^2 + 3^2 + (2\sqrt{3})^2} = \sqrt{4 + 9 + 12} = \sqrt{25} = 5 \)
To find a unit vector \( \hat{v} \) in the direction of \( \vec{v} \):
\( \hat{v} = \frac{\vec{v}}{|\vec{v}|} = \frac{2\hat{i} + 3\hat{j} + 2\sqrt{3}\hat{k}}{5} \bk \)
The required vector of magnitude 10 is:
\( 10\hat{v} = 10 \left( \frac{2\hat{i} + 3\hat{j} + 2\sqrt{3}\hat{k}}{5} \right) = 4\hat{i} + 6\hat{j} + 4\sqrt{3}\hat{k} \)
In simple words: First we find the unit-length version of the given vector by dividing it by its magnitude, 5. Then we multiply it by 10 to scale it up to the required size.

Exam Tip: Scaling a vector is always done by finding the unit vector first and then multiplying by the desired magnitude.

Question 10. What are the direction cosines of a vector equiangular with co-ordinate axes?
Answer: Let \( l, m, n \) be the direction cosines of the vector. Since the vector is equiangular with the coordinate axes, we have \( \alpha = \beta = \gamma \implies l = m = n \). We know that:
\( l^2 + m^2 + n^2 = 1 \implies 3l^2 = 1 \implies l = \pm \frac{1}{\sqrt{3}} \)
Thus, the direction cosines of the equiangular vector are:
\( \left( \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}} \right) \)
In simple words: Since the vector is tilted at the same angle to all three axes, all of its direction coordinates are equal. Squaring them and adding them up must equal 1, which gives us these fractional values.

Exam Tip: Write the sign \( \pm \) for all components since a line or vector can point in either direction of the octant.

Question 11. What is the angle which the vector \( 3\hat{i} - 6\hat{j} + 2\hat{k} \) makes with the x-axis?
Answer: Let \( \vec{a} = 3\hat{i} - 6\hat{j} + 2\hat{k} \). The magnitude of \( \vec{a} \) is computed as:
\( |\vec{a}| = \sqrt{3^2 + (-6)^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7 \)
If \( \alpha \) is the angle that the vector makes with the positive x-axis, then:
\( \cos\alpha = \frac{a_x}{|\vec{a}|} = \frac{3}{7} \implies \alpha = \cos^{-1}\left(\frac{3}{7}\right) \)
In simple words: The cosine of the angle with the x-axis is the x-coordinate of the vector divided by its overall length. We then apply the inverse cosine to find the angle.

Exam Tip: State the magnitude of the vector clearly as part of your step-by-step presentation to ensure full marks.

Question 12. Write a unit vector perpendicular to both the vectors \( 3\hat{i} - 2\hat{j} + \hat{k} \) and \( -2\hat{i} + \hat{j} - 2\hat{k} \).
Answer: Let \( \vec{a} = 3\hat{i} - 2\hat{j} + \hat{k} \) and \( \vec{b} = -2\hat{i} + \hat{j} - 2\hat{k} \). A vector perpendicular to both \( \vec{a} \) and \( \vec{b} \) is given by their cross product, \( \vec{a} \times \vec{b} \):
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -2 & 1 \\ -2 & 1 & -2 \end{vmatrix} \)
\( \implies \vec{a} \times \vec{b} = \hat{i}(4 - 1) - \hat{j}(-6 - (-2)) + \hat{k}(3 - 4) = 3\hat{i} + 4\hat{j} - \hat{k} \)
The magnitude of this perpendicular vector is:
\( |\vec{a} \times \vec{b}| = \sqrt{3^2 + 4^2 + (-1)^2} = \sqrt{9 + 16 + 1} = \sqrt{26} \)
The required unit vector is:
\( \hat{n} = \pm \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \pm \frac{3\hat{i} + 4\hat{j} - \hat{k}}{\sqrt{26}} \)
In simple words: To find a vector that is perpendicular to two other vectors, we use the cross product. We then divide the resulting vector by its length to make it a unit vector.

Exam Tip: Including the \( \pm \) sign is good practice as there are always two unit vectors pointing in opposite directions that are perpendicular to a plane.

Question 13. What is the projection of the vector \( \hat{i} - \hat{j} \) on the vector \( \hat{i} + \hat{j} \)?
Answer: Let \( \vec{a} = \hat{i} - \hat{j} \) and \( \vec{b} = \hat{i} + \hat{j} \). The scalar projection of \( \vec{a} \) on \( \vec{b} \) is given by:
\( \text{Projection} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \)
Calculate the dot product:
\( \vec{a} \cdot \vec{b} = (1)(1) + (-1)(1) = 1 - 1 = 0 \)
Since the dot product is zero, the projection is also zero:
\( \text{Projection} = \frac{0}{\sqrt{1^2 + 1^2}} = 0 \)
In simple words: Since these two vectors are perpendicular, one casts no shadow onto the other, meaning the projection is exactly zero.

Exam Tip: A projection of zero indicates that the vectors are orthogonal. Mention this connection to show a deep conceptual understanding.

Question 14. If \( |\vec{a}| = 2 \), \( |\vec{b}| = 2\sqrt{3} \) and \( \vec{a} \perp \vec{b} \), what is the value of \( |\vec{a} + \vec{b}| \)?
Answer: Since \( \vec{a} \perp \vec{b} \), their dot product \( \vec{a} \cdot \vec{b} = 0 \). We know the identity:
\( |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) \)
Substitute the given values:
\( |\vec{a} + \vec{b}|^2 = (2)^2 + (2\sqrt{3})^2 + 2(0) = 4 + 12 = 16 \)
Taking the square root:
\( |\vec{a} + \vec{b}| = \sqrt{16} = 4 \)
In simple words: Because the vectors are perpendicular, they form the legs of a right-angled triangle. We can use the Pythagorean theorem to find the length of their sum.

Exam Tip: For orthogonal vectors, you can directly use the simplified relation \( |\vec{a} + \vec{b}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2} \).

Question 15. For what value of \( \lambda \), \( \vec{a} = \lambda\hat{i} + \hat{j} + 4\hat{k} \) is perpendicular to \( \vec{b} = 2\hat{i} + 6\hat{j} + 3\hat{k} \)?
Answer: Two vectors are perpendicular if and only if their dot product is zero:
\( \vec{a} \cdot \vec{b} = 0 \)
Substitute the components of \( \vec{a} \) and \( \vec{b} \):
\( (\lambda)(2) + (1)(6) + (4)(3) = 0 \)
\( \implies 2\lambda + 6 + 12 = 0 \)
\( \implies 2\lambda + 18 = 0 \implies 2\lambda = -18 \implies \lambda = -9 \)
In simple words: For the vectors to meet at a right angle, their coordinate product sum must equal zero. Solving this simple equation gives \( \lambda = -9 \).

Exam Tip: Always state the essential condition \( \vec{a} \cdot \vec{b} = 0 \) clearly before solving the linear equation.

Question 16. What is \( |\vec{a}| \), if \( (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 3 \) and \( 2|\vec{b}| = |\vec{a}| \)?
Answer: Expanding the given dot product:
\( (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = |\vec{a}|^2 - |\vec{b}|^2 = 3 \)
We are also given that \( 2|\vec{b}| = |\vec{a}| \implies |\vec{b}| = \frac{|\vec{a}|}{2} \). Substitute this relation into the expanded equation:
\( |\vec{a}|^2 - \left(\frac{|\vec{a}|}{2}\right)^2 = 3 \implies |\vec{a}|^2 - \frac{|\vec{a}|^2}{4} = 3 \)
\( \implies \frac{3|\vec{a}|^2}{4} = 3 \implies |\vec{a}|^2 = 4 \implies |\vec{a}| = 2 \)
In simple words: The dot product expands like a difference of squares. Substituting one magnitude in terms of the other lets us solve directly for the length of the vector, which is 2.

Exam Tip: Since vector magnitude represents a physical length, discard the negative option \( |\vec{a}| = -2 \) and output only the positive value.

Question 17. What is the angle between \( \vec{a} \) and \( \vec{b} \), if \( |\vec{a} - \vec{b}| = |\vec{a} + \vec{b}| \)?
Answer: Squaring both sides of the given equation:
\( |\vec{a} - \vec{b}|^2 = |\vec{a} + \vec{b}|^2 \)
\( \implies |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b}) = |\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) \)
Subtracting identical terms from both sides:
\( -2(\vec{a} \cdot \vec{b}) = 2(\vec{a} \cdot \vec{b}) \implies 4(\vec{a} \cdot \vec{b}) = 0 \implies \vec{a} \cdot \vec{b} = 0 \)
Since the dot product of the non-zero vectors is zero, they are perpendicular. Thus, the angle between them is:
\( \theta = \frac{\pi}{2} \text{ or } 90^\circ \)
In simple words: If adding two vectors gives a result of the same length as subtracting them, they must be perpendicular to each other, meaning the angle between them is 90 degrees.

Exam Tip: Geometrically, this equation states that the diagonals of a parallelogram are equal in length, which only occurs when the parallelogram is a rectangle.

Question 18. In a parallelogram ABCD, \( \vec{AB} = 2\hat{i} - \hat{j} + 4\hat{k} \) and \( \vec{AC} = \hat{i} + \hat{j} + 4\hat{k} \). What is the length of side BC?
Answer: In triangle ABC, by the triangle law of vector addition:
\( \vec{AB} + \vec{BC} = \vec{AC} \implies \vec{BC} = \vec{AC} - \vec{AB} \)
Substitute the given vectors:
\( \vec{BC} = (\hat{i} + \hat{j} + 4\hat{k}) - (2\hat{i} - \hat{j} + 4\hat{k}) = -\hat{i} + 2\hat{j} \)
The length of side BC is the magnitude of vector \( \vec{BC} \):
\( |\vec{BC}| = \sqrt{(-1)^2 + 2^2 + 0^2} = \sqrt{1 + 4} = \sqrt{5} \text{ units} \)
In simple words: Using the triangle rule, the vector for side BC is the difference between vectors AC and AB. Finding the length of this difference vector gives us \( \sqrt{5} \).

Exam Tip: Be sure to write out the vector subtraction step clearly before calculating the magnitude.

Question 19. What is the area of a parallelogram whose diagonals are given by vectors \( 2\hat{i} + \hat{j} - 2\hat{k} \) and \( -\hat{i} + 2\hat{k} \)?
Answer: Let the diagonal vectors be \( \vec{d}_1 = 2\hat{i} + \hat{j} - 2\hat{k} \) and \( \vec{d}_2 = -\hat{i} + 2\hat{k} \). The area of a parallelogram in terms of its diagonals is given by:
\( \text{Area} = \frac{1}{2} |\vec{d}_1 \times \vec{d}_2| \)
First, find the cross product of the diagonals:
\( \vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ -1 & 0 & 2 \end{vmatrix} \)
\( \implies \vec{d}_1 \times \vec{d}_2 = \hat{i}(2 - 0) - \hat{j}(4 - 2) + \hat{k}(0 - (-1)) = 2\hat{i} - 2\hat{j} + \hat{k} \)
Now, calculate the magnitude of this cross product:
\( |\vec{d}_1 \times \vec{d}_2| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \)
Thus, the area is:
\( \text{Area} = \frac{1}{2} (3) = 1.5 \text{ sq. units} \)
In simple words: When we are given the diagonals instead of the sides of a parallelogram, we compute their cross product, find its length, and then take exactly half of that value.

Exam Tip: Always remember the \( \frac{1}{2} \) factor when using diagonals; leaving it out is a common mistake that costs marks.

Question 20. Find \( |\vec{x}| \) if for a unit vector \( \hat{a} \), \( (\vec{x} - \hat{a}) \cdot (\vec{x} + \hat{a}) = 12 \).
Answer: Since \( \hat{a} \) is a unit vector, its magnitude is \( |\hat{a}| = 1 \). Expanding the given dot product expression:
\( (\vec{x} - \hat{a}) \cdot (\vec{x} + \hat{a}) = |\vec{x}|^2 - |\hat{a}|^2 = 12 \)
Substituting \( |\hat{a}|^2 = 1 \):
\( |\vec{x}|^2 - 1 = 12 \implies |\vec{x}|^2 = 13 \implies |\vec{x}| = \sqrt{13} \)
In simple words: The expression simplifies to the square of the unknown vector's length minus 1. Solving this gives the length of the vector as \( \sqrt{13} \).

Exam Tip: Always state the unit vector condition \( |\hat{a}| = 1 \) clearly in your steps to show the examiner how you simplified the equation.

Question 21. If \( \vec{a} \) and \( \vec{b} \) are two unit vectors and \( \vec{a} + \vec{b} \) is also a unit vector then what is the angle between \( \vec{a} \) and \( \vec{b} \)?
Answer: We are given that \( |\vec{a}| = 1 \), \( |\vec{b}| = 1 \), and \( |\vec{a} + \vec{b}| = 1 \). Squaring the magnitude equation:
\( |\vec{a} + \vec{b}|^2 = 1 \)
Using the vector identity:
\( |\vec{a}|^2 + |\vec{b}|^2 + 2 |\vec{a}| |\vec{b}| \cos\theta = 1 \)
Substitute the magnitudes of the unit vectors:
\( 1^2 + 1^2 + 2(1)(1)\cos\theta = 1 \implies 2 + 2\cos\theta = 1 \)
\( \implies 2\cos\theta = -1 \implies \cos\theta = -\frac{1}{2} \)
Since the angle \( \theta \) between vectors lies in \( [0, \pi] \):
\( \theta = \frac{2\pi}{3} \text{ or } 120^\circ \)
In simple words: If two vectors of unit length add up to a third vector of unit length, they must form an equilateral-like triangle structure where the angle between them is 120 degrees.

Exam Tip: Write down the angle in both radians (\( \frac{2\pi}{3} \)) and degrees (\( 120^\circ \)) to be safe and clear.

Question 22. If \( \hat{i}, \hat{j}, \hat{k} \) are the usual three mutually perpendicular unit vectors then what is the value of \( \hat{i} \cdot (\hat{j} \times \hat{k}) + \hat{j} \cdot (\hat{i} \times \hat{k}) + \hat{k} \cdot (\hat{j} \times \hat{i}) \)?
Answer: Recall the standard cross products for the unit vectors in a right-handed system:
\( \hat{j} \times \hat{k} = \hat{i} \)
\( \hat{i} \times \hat{k} = -\hat{j} \)
\( \hat{j} \times \hat{i} = -\hat{k} \)
Substitute these back into the original expression:
\( \hat{i} \cdot \hat{i} + \hat{j} \cdot (-\hat{j}) + \hat{k} \cdot (-\hat{k}) \)
Since \( \hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1 \):
\( 1 - 1 - 1 = -1 \bk \)
In simple words: We find the individual cross-product vectors first and then calculate their dot products. Summing up these values yields a final answer of \( -1 \).

Exam Tip: Take extra care with the direction of cyclic cross products; reversing the cycle changes the sign of the resulting vector.

Question 23. What is the angle between \( \vec{x} \) and \( \vec{y} \) if \( \vec{x} \cdot \vec{y} = |\vec{x} \times \vec{y}| \)?
Answer: Let \( \theta \) be the angle between \( \vec{x} \) and \( \vec{y} \). Using the definitions of dot product and cross product magnitude:
\( |\vec{x}| |\vec{y}| \cos\theta = |\vec{x}| |\vec{y}| \sin\theta \)
Assuming non-zero vectors, we can divide both sides by \( |\vec{x}| |\vec{y}| \):
\( \cos\theta = \sin\theta \implies \tan\theta = 1 \)
Since \( \theta \in [0, \pi] \):
\( \theta = \frac{\pi}{4} \text{ or } 45^\circ \)
In simple words: The dot and cross products are equal when the sine and cosine of their angle are equal. This occurs at exactly 45 degrees.

Exam Tip: Be sure to write the final angle in radians as it is the standard unit of measurement in senior math exams.

Question 24. Write a unit vector in xy-plane, making an angle of 30° with the +ve direction of x–axis.
Answer: Any unit vector \( \hat{u} \) lying in the xy-plane and making an angle \( \theta \) with the positive x-axis can be written as:
\( \hat{u} = \cos\theta\hat{i} + \sin\theta\hat{j} \)
We are given \( \theta = 30^\circ \). Substitute this value:
\( \hat{u} = \cos(30^\circ)\hat{i} + \sin(30^\circ)\hat{j} \)
\( \hat{u} = \frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j} \)
In simple words: A unit vector on a flat plane has coordinates given by the cosine and sine of its angle with the horizontal axis. For 30 degrees, this gives us these exact fractions.

Exam Tip: Always double check that the magnitude of your final vector is indeed 1 using \( \sqrt{x^2+y^2} \).

Question 25. If \( \vec{a}, \vec{b} \) and \( \vec{c} \) are unit vectors with \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \), then what is the value of \( \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} \)?
Answer: We are given that \( |\vec{a}| = |\vec{b}| = |\vec{c}| = 1 \) and \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \). Squaring both sides of the vector sum equation:
\( |\vec{a} + \vec{b} + \vec{c}|^2 = 0 \)
Using the standard identity:
\( |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \)
Substitute the magnitudes of the unit vectors:
\( 1^2 + 1^2 + 1^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \)
\( \implies 3 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \)
\( \implies \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{3}{2} \)
In simple words: By squaring the sum of three vectors that add to zero, we can relate their lengths to the sum of their dot products, giving us a final value of negative 1.5.

Exam Tip: This classic question is highly favored by examiners; mastering this algebraic expansion can save you a lot of time.

Question 26. If \( \vec{a} \) and \( \vec{b} \) are unit vectors such that \( (\vec{a} + 2\vec{b}) \) is perpendicular to \( (5\vec{a} - 4\vec{b}) \), then what is the angle between \( \vec{a} \) and \( \vec{b} \)?
Answer: Since the two vectors are perpendicular, their dot product must be zero:
\( (\vec{a} + 2\vec{b}) \cdot (5\vec{a} - 4\vec{b}) = 0 \)
Expand the expression:
\( 5|\vec{a}|^2 - 4(\vec{a} \cdot \vec{b}) + 10(\vec{a} \cdot \vec{b}) - 8|\vec{b}|^2 = 0 \)
Since \( \vec{a} \) and \( \vec{b} \) are unit vectors, substitute \( |\vec{a}| = 1 \) and \( |\vec{b}| = 1 \):
\( 5(1) + 6(\vec{a} \cdot \vec{b}) - 8(1) = 0 \)
\( \implies 6(\vec{a} \cdot \vec{b}) - 3 = 0 \implies \vec{a} \cdot \vec{b} = \frac{3}{6} = \frac{1}{2} \)
Now, let \( \theta \) be the angle between them:
\( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta = \frac{1}{2} \implies \cos\theta = \frac{1}{2} \)
This gives:
\( \theta = \frac{\pi}{3} \text{ or } 60^\circ \)
In simple words: Expanding the dot product and substituting the unit lengths allows us to find the dot product of the two vectors, which leads directly to an angle of 60 degrees.

Exam Tip: Be methodical during the expansion process to ensure you correctly distribute the terms and keep the signs accurate.

Short Answer Type Questions (4 Marks)

Question 27. If ABCDEF is a regular hexagon then using triangle law of addition prove that : \( \vec{AB} + \vec{AC} + \vec{AD} + \vec{AE} + \vec{AF} = 3\vec{AD} = 6\vec{AO} \), O being the centre of hexagon.
Answer: Let O be the center of the regular hexagon ABCDEF. By the properties of a regular hexagon, we know:
\( \vec{AD} = 2\vec{AO} \)
We can pair the remaining vectors based on geometric symmetry:
\( \vec{AB} + \vec{AE} \quad \text{and} \quad \vec{AC} + \vec{AF} \)
Since \( \vec{AF} \) is parallel and equal to \( \vec{CD} \), and \( \vec{AB} \) is parallel and equal to \( \vec{ED} \):
\( \vec{AC} + \vec{AF} = \vec{AC} + \vec{CD} = \vec{AD} \) (using the triangle law in \( \Delta ACD \))
\( \vec{AE} + \vec{AB} = \vec{AE} + \vec{ED} = \vec{AD} \) (using the triangle law in \( \Delta AED \))
Now, substitute these grouped terms into the main expression:
\( \vec{AB} + \vec{AC} + \vec{AD} + \vec{AE} + \vec{AF} = (\vec{AC} + \vec{AF}) + (\vec{AE} + \vec{AB}) + \vec{AD} \)
\( \implies \vec{AD} + \vec{AD} + \vec{AD} = 3\vec{AD} \)
Since \( \vec{AD} = 2\vec{AO} \):
\( 3\vec{AD} = 3(2\vec{AO}) = 6\vec{AO} \)
Thus, \( \vec{AB} + \vec{AC} + \vec{AD} + \vec{AE} + \vec{AF} = 3\vec{AD} = 6\vec{AO} \). Hence proved.
In simple words: By grouping opposite sides of the hexagon together, we can simplify the vector sum into three copies of the main diagonal vector AD, which is also six times the vector to the center O.

Exam Tip: Draw a neat, labeled sketch of the regular hexagon to clearly illustrate the parallel vectors and triangle paths to the examiner.

Question 28. Points L, M, N divides the sides BC, CA, AB of a \( \Delta ABC \) in the ratios 1 : 4, 3 : 2, 3 : 7 respectively. Prove that \( \vec{AL} + \vec{BM} + \vec{CN} \) is a vector parallel to \( \vec{CK} \) where K divides AB in ratio 1 : 3.
Answer: Let \( \vec{a}, \vec{b}, \vec{c} \) be the position vectors of the vertices A, B, C of the triangle with respect to an origin. Using the section formula, we can find the position vectors of points L, M, N, and K:
\( \vec{l} = \frac{1\vec{c} + 4\vec{b}}{5} \)
\( \vec{m} = \frac{3\vec{a} + 2\vec{c}}{5} \)
\( \vec{n} = \frac{3\vec{b} + 7\vec{a}}{10} \)
\( \vec{k} = \frac{1\vec{b} + 3\vec{a}}{4} \)
Now, write the vectors \( \vec{AL}, \vec{BM}, \vec{CN} \) and \( \vec{CK} \) in terms of these position vectors:
\( \vec{AL} = \vec{l} - \vec{a} = \frac{4\vec{b} + \vec{c} - 5\vec{a}}{5} \)
\( \vec{BM} = \vec{m} - \vec{b} = \frac{3\vec{a} + 2\vec{c} - 5\vec{b}}{5} \)
\( \vec{CN} = \vec{n} - \vec{c} = \frac{7\vec{a} + 3\vec{b} - 10\vec{c}}{10} \)
Adding these three vectors together:
\( \vec{AL} + \vec{BM} + \vec{CN} = \frac{2(4\vec{b} + \vec{c} - 5\vec{a}) + 2(3\vec{a} + 2\vec{c} - 5\vec{b}) + (7\vec{a} + 3\vec{b} - 10\vec{c})}{10} \)
\( \implies \vec{AL} + \vec{BM} + \vec{CN} = \frac{8\vec{b} + 2\vec{c} - 10\vec{a} + 6\vec{a} + 4\vec{c} - 10\vec{b} + 7\vec{a} + 3\vec{b} - 10\vec{c}}{10} \)
\( \implies \vec{AL} + \vec{BM} + \vec{CN} = \frac{3\vec{a} + \vec{b} - 4\vec{c}}{10} \)
Now, write the vector \( \vec{CK} \):
\( \vec{CK} = \vec{k} - \vec{c} = \frac{3\vec{a} + \vec{b} - 4\vec{c}}{4} \)
By comparing the two final vector expressions:
\( \vec{AL} + \vec{BM} + \vec{CN} = \frac{4}{10} \vec{CK} = \frac{2}{5} \vec{CK} \)
Since one vector is a scalar multiple of the other, they are parallel. Hence proved.
In simple words: By writing each divider point as a fraction of the corner coordinates, we sum the three path vectors and show that the total sum vector is just a scaled-down version of vector CK.

Exam Tip: Be very careful when finding a common denominator for the fractions to ensure the coefficients of \( \vec{a}, \vec{b}, \vec{c} \) are summed correctly.

Question 29. The scalar product of vector \( \hat{i} + \hat{j} + \hat{k} \) with a unit vector along the sum of the vectors \( 2\hat{i} + 4\hat{j} - 5\hat{k} \) and \( \lambda\hat{i} + 2\hat{j} + 3\hat{k} \) is equal to 1. Find the value of \( \lambda \).
Answer: Let \( \vec{a} = 2\hat{i} + 4\hat{j} - 5\hat{k} \) and \( \vec{b} = \lambda\hat{i} + 2\hat{j} + 3\hat{k} \). The sum of these two vectors is:
\( \vec{s} = \vec{a} + \vec{b} = (2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k} \)
The magnitude of this sum vector is:
\( |\vec{s}| = \sqrt{(2 + \lambda)^2 + 6^2 + (-2)^2} = \sqrt{\lambda^2 + 4\lambda + 4 + 36 + 4} = \sqrt{\lambda^2 + 4\lambda + 44} \)
The unit vector \( \hat{u} \) along the direction of the sum vector is:
\( \hat{u} = \frac{\vec{s}}{|\vec{s}|} = \frac{(2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{\lambda^2 + 4\lambda + 44}} \)
We are given that the dot product of \( \hat{i} + \hat{j} + \hat{k} \) with this unit vector is 1:
\( (\hat{i} + \hat{j} + \hat{k}) \cdot \hat{u} = 1 \)
\( \implies \frac{(2 + \lambda)(1) + (6)(1) + (-2)(1)}{\sqrt{\lambda^2 + 4\lambda + 44}} = 1 \)
\( \implies \frac{\lambda + 6}{\sqrt{\lambda^2 + 4\lambda + 44}} = 1 \implies \lambda + 6 = \sqrt{\lambda^2 + 4\lambda + 44} \)
Squaring both sides (noting that \( \lambda + 6 \ge 0 \implies \lambda \ge -6 \)):
\( (\lambda + 6)^2 = \lambda^2 + 4\lambda + 44 \implies \lambda^2 + 12\lambda + 36 = \lambda^2 + 4\lambda + 44 \bk \)
\( \implies 8\lambda = 8 \implies \lambda = 1 \)
Since \( \lambda = 1 \ge -6 \), this solution is valid.
In simple words: We find the sum vector, divide it by its length to make it a unit vector, and then calculate its dot product with the test vector. Solving the resulting equation gives \( \lambda = 1 \).

Exam Tip: Always double check that your final value of \( \lambda \) satisfies the original equation, especially after squaring both sides, to avoid extraneous solutions.

Question 30. \( \vec{a}, \vec{b} \) and \( \vec{c} \) are three mutually perpendicular vectors of equal magnitude. Show that \( \vec{a} + \vec{b} + \vec{c} \) makes equal angles with \( \vec{a}, \vec{b} \) and \( \vec{c} \) with each angle as \( \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \).
Answer: Let the equal magnitude of the mutually perpendicular vectors be \( |\vec{a}| = |\vec{b}| = |\vec{c}| = k \). Since they are mutually perpendicular:
\( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 0 \)
Now, let us find the magnitude of the sum vector \( \vec{s} = \vec{a} + \vec{b} + \vec{c} \):
\( |\vec{s}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \)
\( \implies |\vec{s}|^2 = k^2 + k^2 + k^2 + 2(0) = 3k^2 \implies |\vec{s}| = \sqrt{3}k \)
Let \( \theta_1, \theta_2, \theta_3 \) be the angles that \( \vec{s} \) makes with \( \vec{a}, \vec{b}, \vec{c} \), respectively. For the angle with \( \vec{a} \):
\( \cos\theta_1 = \frac{\vec{s} \cdot \vec{a}}{|\vec{s}| |\vec{a}|} = \frac{(\vec{a} + \vec{b} + \vec{c}) \cdot \vec{a}}{(\sqrt{3}k)(k)} = \frac{|\vec{a}|^2 + \vec{b}\cdot\vec{a} + \vec{c}\cdot\vec{a}}{\sqrt{3}k^2} \)
\( \implies \cos\theta_1 = \frac{k^2 + 0 + 0}{\sqrt{3}k^2} = \frac{1}{\sqrt{3}} \implies \theta_1 = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \)
Similarly, by symmetry:
\( \cos\theta_2 = \frac{1}{\sqrt{3}} \implies \theta_2 = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \)
\( \cos\theta_3 = \frac{1}{\sqrt{3}} \implies \theta_3 = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \bk \)
Thus, \( \vec{a} + \vec{b} + \vec{c} \) makes equal angles with each of the three vectors. Hence proved.
In simple words: Because the three vectors are of equal length and are perpendicular to each other, their sum vector points directly out of the corner of the cube they form, making a matching angle with each leg.

Exam Tip: Be sure to write out the expansion for the magnitude of the sum of three vectors as it is a critical step in the proof.

Question 31. If \( \vec{\alpha} = 3\hat{i} - \hat{j} \) and \( \vec{\beta} = 2\hat{i} + \hat{j} - 3\hat{k} \), then express \( \vec{\beta} \) in the form of \( \vec{\beta} = \vec{\beta}_1 + \vec{\beta}_2 \), where \( \vec{\beta}_1 \) is parallel to \( \vec{\alpha} \) and \( \vec{\beta}_2 \) is perpendicular to \( \vec{\alpha} \).
Answer: Since \( \vec{\beta}_1 \) is parallel to \( \vec{\alpha} \), we can write:
\( \vec{\beta}_1 = k\vec{\alpha} = k(3\hat{i} - \hat{j}) = 3k\hat{i} - k\hat{j} \)
Since \( \vec{\beta} = \vec{\beta}_1 + \vec{\beta}_2 \), we can write the perpendicular vector as:
\( \vec{\beta}_2 = \vec{\beta} - \vec{\beta}_1 = (2\hat{i} + \hat{j} - 3\hat{k}) - (3k\hat{i} - k\hat{j}) = (2 - 3k)\hat{i} + (1 + k)\hat{j} - 3\hat{k} \)
Given that \( \vec{\beta}_2 \) is perpendicular to \( \vec{\alpha} \), their dot product is zero:
\( \vec{\beta}_2 \cdot \vec{\alpha} = 0 \)
\( \implies (2 - 3k)(3) + (1 + k)(-1) + (-3)(0) = 0 \)
\( \implies 6 - 9k - 1 - k = 0 \implies 5 - 10k = 0 \implies k = \frac{1}{2} \)
Now, substitute the value of \( k \) back:
\( \vec{\beta}_1 = \frac{3}{2}\hat{i} - \frac{1}{2}\hat{j} \)
\( \vec{\beta}_2 = \left(2 - \frac{3}{2}\right)\hat{i} + \left(1 + \frac{1}{2}\right)\hat{j} - 3\hat{k} = \frac{1}{2}\hat{i} + \frac{3}{2}\hat{j} - 3\hat{k} \)
Thus, the vector is decomposed as:
\( \vec{\beta} = \left(\frac{3}{2}\hat{i} - \frac{1}{2}\hat{j}\right) + \left(\frac{1}{2}\hat{i} + \frac{3}{2}\hat{j} - 3\hat{k}\right) \)
In simple words: We split the vector into two parts. The first part is a multiplier of vector alpha. The second part is whatever remains, which must have a dot product of zero with alpha.

Exam Tip: You can quickly check your final answer by adding \( \vec{\beta}_1 \) and \( \vec{\beta}_2 \) together to verify that they sum to \( \vec{\beta} \).

Question 32. If \( \vec{a}, \vec{b}, \vec{c} \) are three vectors such that \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \) then prove that \( \vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a} \).
Answer: We are given that:
\( \vec{a} + \vec{b} + \vec{c} = \vec{0} \)
Take the cross product of both sides with vector \( \vec{b} \):
\( (\vec{a} + \vec{b} + \vec{c}) \times \vec{b} = \vec{0} \times \vec{b} \)
\( \implies \vec{a} \times \vec{b} + \vec{b} \times \vec{b} + \vec{c} \times \vec{b} = \vec{0} \)
Since the cross product of any vector with itself is zero (\( \vec{b} \times \vec{b} = \vec{0} \)):
\( \vec{a} \times \vec{b} + \vec{c} \times \vec{b} = \vec{0} \implies \vec{a} \times \vec{b} = - (\vec{c} \times \vec{b}) \)
Using the anti-commutative property of cross product (\( - (\vec{c} \times \vec{b}) = \vec{b} \times \vec{c} \)):
\( \vec{a} \times \vec{b} = \vec{b} \times \vec{c} \quad \text{--- (Equation 1)} \)
Similarly, take the cross product of the original sum equation with vector \( \vec{c} \):
\( (\vec{a} + \vec{b} + \vec{c}) \times \vec{c} = \vec{0} \)
\( \implies \vec{a} \times \vec{c} + \vec{b} \times \vec{c} + \vec{c} \times \vec{c} = \vec{0} \)
Since \( \vec{c} \times \vec{c} = \vec{0} \):
\( \vec{a} \times \vec{c} + \vec{b} \times \vec{c} = \vec{0} \implies \vec{b} \times \vec{c} = - (\vec{a} \times \vec{c}) \)
Using \( - (\vec{a} \times \vec{c}) = \vec{c} \times \vec{a} \):
\( \vec{b} \times \vec{c} = \vec{c} \times \vec{a} \quad \text{--- (Equation 2)} \bk \)
Combining Equations 1 and 2, we obtain:
\( \vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a} \). Hence proved.
In simple words: Since the three vectors add up to zero, they form a closed triangle. The cross product of any two sides represents the area of this triangle, which must be equal regardless of which two sides we choose.

Exam Tip: Clearly show the use of anti-commutative property as it is a crucial step in transforming the signs of the cross products.

Question 33. If \( |\vec{a}| = 3, |\vec{b}| = 5, |\vec{c}| = 7 \) and \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \), find the angle between \( \vec{a} \) and \( \vec{b} \).
Answer: We are given the relation:
\( \vec{a} + \vec{b} + \vec{c} = \vec{0} \implies \vec{a} + \vec{b} = -\vec{c} \)
Squaring both sides of the equation:
\( |\vec{a} + \vec{b}|^2 = |-\vec{c}|^2 \)
\( \implies |\vec{a}|^2 + |\vec{b}|^2 + 2|\vec{a}||\vec{b}|\cos\theta = |\vec{c}|^2 \)
where \( \theta \) is the angle between \( \vec{a} \) and \( \vec{b} \). Substitute the given magnitudes:
\( 3^2 + 5^2 + 2(3)(5)\cos\theta = 7^2 \)
\( \implies 9 + 25 + 30\cos\theta = 49 \implies 34 + 30\cos\theta = 49 \)
\( \implies 30\cos\theta = 15 \implies \cos\theta = \frac{15}{30} = \frac{1}{2} \)
Since \( \theta \in [0, \pi] \):
\( \theta = \frac{\pi}{3} \text{ or } 60^\circ \)
In simple words: By moving vector c to the other side and squaring, we create a triangle relationship. Substituting the lengths allows us to find the cosine of the angle, which is 60 degrees.

Exam Tip: Be sure to isolate the vector term whose angle is requested on one side before squaring.

Question 34. Let \( \vec{a} = \hat{i} - \hat{j} \), \( \vec{b} = 3\hat{j} - \hat{k} \) and \( \vec{c} = 7\hat{i} - \hat{k} \), find a vector \( \vec{d} \) which is perpendicular to \( \vec{a} \) and \( \vec{b} \) and \( \vec{c} \cdot \vec{d} = 1 \).
Answer: Since the vector \( \vec{d} \) must be perpendicular to both \( \vec{a} \) and \( \vec{b} \), it must point along the direction of their cross product, \( \vec{a} \times \vec{b} \):
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 0 \\ 0 & 3 & -1 \end{vmatrix} \)
\( \implies \vec{a} \times \vec{b} = \hat{i}(1 - 0) - \hat{j}(-1 - 0) + \hat{k}(3 - 0) = \hat{i} + \hat{j} + 3\hat{k} \)
Thus, we can write \( \vec{d} \) as a scalar multiple of this cross product:
\( \vec{d} = k(\hat{i} + \hat{j} + 3\hat{k}) \)
We are given that \( \vec{c} \cdot \vec{d} = 1 \). Substitute the vectors:
\( (7\hat{i} - \hat{k}) \cdot k(\hat{i} + \hat{j} + 3\hat{k}) = 1 \)
\( \implies k [ (7)(1) + (0)(1) + (-1)(3) ] = 1 \)
\( \implies k(7 - 3) = 1 \implies 4k = 1 \implies k = \frac{1}{4} \)
Substituting \( k = \frac{1}{4} \) back into the expression for \( \vec{d} \):
\( \vec{d} = \frac{1}{4}(\hat{i} + \hat{j} + 3\hat{k}) \)
In simple words: A vector perpendicular to two others is parallel to their cross product. We find this cross product, then scale it using a multiplier so that its dot product with vector c is exactly 1.

Exam Tip: Working with the parametric constant \( k \) is the most reliable way to find specific scaled perpendicular vectors.

Question 35. If \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \), \( \vec{c} = \hat{j} - \hat{k} \) are the given vectors then find a vector \( \vec{b} \) satisfying the equation \( \vec{a} \times \vec{b} = \vec{c} \), \( \vec{a} \cdot \vec{b} = 3 \).
Answer: Let the required vector be \( \vec{b} = x\hat{i} + y\hat{j} + z\hat{k} \). Using the cross product equation \( \vec{a} \times \vec{b} = \vec{c} \):
\( \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix} = \hat{j} - \hat{k} \)
\( \implies (z - y)\hat{i} - (z - x)\hat{j} + (y - x)\hat{k} = \hat{j} - \hat{k} \)
Equating the individual vector components:
\( z - y = 0 \implies y = z \quad \text{--- (Equation 1)} \)
\( x - z = 1 \implies x = z + 1 \quad \text{--- (Equation 2)} \)
\( y - x = -1 \implies x = y + 1 \quad \text{--- (Equation 3, which matches Eq 2 since } y=z) \)
Using the dot product equation \( \vec{a} \cdot \vec{b} = 3 \):
\( (1)(x) + (1)(y) + (1)(z) = 3 \implies x + y + z = 3 \)
Substitute \( x = z + 1 \) and \( y = z \) into this sum:
\( (z + 1) + z + z = 3 \implies 3z + 1 = 3 \implies 3z = 2 \implies z = \frac{2}{3} \)
Using the relations:
\( y = \frac{2}{3} \)
\( x = \frac{2}{3} + 1 = \frac{5}{3} \)
Thus, the vector \( \vec{b} \) is:
\( \vec{b} = \frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k} \)
In simple words: By writing vector b with unknown components, we create a system of simple equations from its cross and dot products. Solving these gives us the unique coordinates.

Exam Tip: Be sure to write out the individual component comparisons from the determinant expansion clearly to show a rigorous method.

Question 36. Find a unit vector perpendicular to plane ABC, when position vectors of A, B, C are \( 3\hat{i} - \hat{j} + 2\hat{k} \), \( \hat{i} - \hat{j} - 3\hat{k} \) and \( 4\hat{i} - 3\hat{j} + \hat{k} \) respectively.
Answer: First, find two vectors lying in the plane of the triangle, namely \( \vec{AB} \) and \( \vec{AC} \):
\( \vec{AB} = (\hat{i} - \hat{j} - 3\hat{k}) - (3\hat{i} - \hat{j} + 2\hat{k}) = -2\hat{i} - 5\hat{k} \)
\( \vec{AC} = (4\hat{i} - 3\hat{j} + \hat{k}) - (3\hat{i} - \hat{j} + 2\hat{k}) = \hat{i} - 2\hat{j} - \hat{k} \)
A vector perpendicular to the plane ABC is given by the cross product \( \vec{AB} \times \vec{AC} \):
\( \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 0 & -5 \\ 1 & -2 & -1 \end{vmatrix} \)
\( \implies \vec{AB} \times \vec{AC} = \hat{i}(0 - 10) - \hat{j}(2 - (-5)) + \hat{k}(4 - 0) = -10\hat{i} - 7\hat{j} + 4\hat{k} \bk \)
The magnitude of this perpendicular vector is:
\( |\vec{AB} \times \vec{AC}| = \sqrt{(-10)^2 + (-7)^2 + 4^2} = \sqrt{100 + 49 + 16} = \sqrt{165} \)
The required unit vector is:
\( \hat{n} = \pm \frac{-10\hat{i} - 7\hat{j} + 4\hat{k}}{\sqrt{165}} \)
In simple words: We find two vectors along the sides of the triangle. Their cross product points perpendicular to the plane. Dividing this by its magnitude gives the unit vector.

Exam Tip: Choose point A as a common vertex for both side vectors to keep your subtraction steps clean and easy to verify.

Question 37. For any two vector, show that \( |\vec{a} + \vec{b}| \le |\vec{a}| + |\vec{b}| \).
Answer: Let us consider the square of the magnitude of the sum of two vectors:
\( |\vec{a} + \vec{b}|^2 = (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = |\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) \)
Using the definition of dot product, \( \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta \). Since \( \cos\theta \le 1 \) for all real angles:
\( \vec{a} \cdot \vec{b} \le |\vec{a}||\vec{b}| \)
Substitute this inequality back into the squared equation:
\( |\vec{a} + \vec{b}|^2 \le |\vec{a}|^2 + |\vec{b}|^2 + 2|\vec{a}||\vec{b}| \)
Recognizing the right-hand side as a perfect square:
\( |\vec{a} + \vec{b}|^2 \le (|\vec{a}| + |\vec{b}|)^2 \bk \)
Taking the square root on both sides (since magnitudes are non-negative):
\( |\vec{a} + \vec{b}| \le |\vec{a}| + |\vec{b}| \). Hence proved.
In simple words: This is the classic triangle inequality. It states that the direct path between two points is always shorter than or equal to going through a third detour point.

Exam Tip: Be sure to explicitly state the condition \( \cos\theta \le 1 \) to justify the transition from equality to inequality.

Question 38. Evaluate \( |\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 \).
Answer: Let \( \vec{a} = x\hat{i} + y\hat{j} + z\hat{k} \). Compute each cross product term individually:
\( \vec{a} \times \hat{i} = (x\hat{i} + y\hat{j} + z\hat{k}) \times \hat{i} = -y\hat{k} + z\hat{j} \implies |\vec{a} \times \hat{i}|^2 = y^2 + z^2 \)
\( \vec{a} \times \hat{j} = (x\hat{i} + y\hat{j} + z\hat{k}) \times \hat{j} = x\hat{k} - z\hat{i} \implies |\vec{a} \times \hat{j}|^2 = x^2 + z^2 \)
\( \vec{a} \times \hat{k} = (x\hat{i} + y\hat{j} + z\hat{k}) \times \hat{k} = -x\hat{j} + y\hat{i} \implies |\vec{a} \times \hat{k}|^2 = x^2 + y^2 \)
Summing these three squared magnitudes together:
\( |\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = (y^2 + z^2) + (x^2 + z^2) + (x^2 + y^2) \)
\( = 2(x^2 + y^2 + z^2) = 2|\vec{a}|^2 \)
Therefore, the evaluated sum is \( 2|\vec{a}|^2 \).
In simple words: Each cross product with a unit coordinate axis eliminates that specific component's contribution. Adding all three together counts each component twice, giving twice the squared length of the vector.

Exam Tip: Writing \( \vec{a} \) in component form makes this algebraic proof straightforward and highly convincing to the grader.

Question 39. If \( \hat{a} \) and \( \hat{b} \) are unit vector inclined at an angle \( \theta \) than prove that : (i) \( \sin\frac{\theta}{2} = \frac{1}{2}|\hat{a} - \hat{b}| \) (ii) \( \tan\frac{\theta}{2} = \frac{|\hat{a} - \hat{b}|}{|\hat{a} + \hat{b}|} \).
Answer:
(i) Consider the square of the difference vector magnitude:
\( |\hat{a} - \hat{b}|^2 = |\hat{a}|^2 + |\hat{b}|^2 - 2(\hat{a} \cdot \hat{b}) \)
Since \( \hat{a} \) and \( \hat{b} \) are unit vectors, \( |\hat{a}| = 1 \) and \( |\hat{b}| = 1 \):
\( |\hat{a} - \hat{b}|^2 = 1 + 1 - 2\cos\theta = 2(1 - \cos\theta) \)
Using the half-angle trigonometric identity \( 1 - \cos\theta = 2\sin^2\frac{\theta}{2} \):
\( |\hat{a} - \hat{b}|^2 = 2 \left( 2\sin^2\frac{\theta}{2} \right) = 4\sin^2\frac{\theta}{2} \)
Taking the square root:
\( |\hat{a} - \hat{b}| = 2\sin\frac{\theta}{2} \implies \sin\frac{\theta}{2} = \frac{1}{2}|\hat{a} - \hat{b}| \). Hence proved.

(ii) Similarly, consider the square of the sum vector magnitude:
\( |\hat{a} + \hat{b}|^2 = |\hat{a}|^2 + |\hat{b}|^2 + 2(\hat{a} \cdot \hat{b}) = 1 + 1 + 2\cos\theta = 2(1 + \cos\theta) \)
Using the half-angle identity \( 1 + \cos\theta = 2\cos^2\frac{\theta}{2} \):
\( |\hat{a} + \hat{b}|^2 = 2 \left( 2\cos^2\frac{\theta}{2} \right) = 4\cos^2\frac{\theta}{2} \)
Taking the square root:
\( |\hat{a} + \hat{b}| = 2\cos\frac{\theta}{2} \)
Dividing the equation from part (i) by this equation:
\( \frac{|\hat{a} - \hat{b}|}{|\hat{a} + \hat{b}|} = \frac{2\sin(\theta/2)}{2\cos(\theta/2)} = \tan\frac{\theta}{2} \). Hence proved.
In simple words: We expand the vector magnitudes using dot products, which leads to trigonometric half-angle forms. Dividing these sine and cosine forms gives the tangent relationship.

Exam Tip: Be sure to write out the basic trigonometric double angle/half angle identities as they are critical links in this proof.

Question 40. For any two vectors, show that \( |\vec{a} \times \vec{b}|^2 = a^2b^2 - (\vec{a} \cdot \vec{b})^2 \).
Answer: Let \( \theta \) be the angle between the two vectors \( \vec{a} \) and \( \vec{b} \). Using the definitions of cross and dot products:
\( |\vec{a} \times \vec{b}| = a b \sin\theta \implies |\vec{a} \times \vec{b}|^2 = a^2 b^2 \sin^2\theta \)
\( \vec{a} \cdot \vec{b} = a b \cos\theta \implies (\vec{a} \cdot \vec{b})^2 = a^2 b^2 \cos^2\theta \bk \)
Starting with the right-hand side of the identity:
\( a^2 b^2 - (\vec{a} \cdot \vec{b})^2 = a^2 b^2 - a^2 b^2 \cos^2\theta \)
\( = a^2 b^2 (1 - \cos^2\theta) \)
Using the fundamental identity \( 1 - \cos^2\theta = \sin^2\theta \):
\( a^2 b^2 \sin^2\theta = |\vec{a} \times \vec{b}|^2 \)
LHS = RHS. Hence proved.
In simple words: This is Lagrange's Identity. It shows that the squared sizes of the cross product and dot product always sum up to the squared product of the vector lengths.

Exam Tip: This identity is exceptionally useful for solving vector systems where you need to switch between dot and cross products.

Question 41. \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \), \( \vec{b} = \hat{i} - \hat{j} + 2\hat{k} \) and \( \vec{c} = x\hat{i} + (x - 2)\hat{j} - \hat{k} \). If \( \vec{c} \) lies in the plane of \( \vec{a} \) and \( \vec{b} \), then find the value of x.
Answer: Since \( \vec{c} \) lies in the plane containing \( \vec{a} \) and \( \vec{b} \), the three vectors are coplanar. The scalar triple product of coplanar vectors is zero:
\( [\vec{a}\ \vec{b}\ \vec{c}] = \begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 2 \\ x & x - 2 & -1 \end{vmatrix} = 0 \)
Evaluating the determinant along the first row:
\( 1 [ (-1)(-1) - 2(x - 2) ] - 1 [ (1)(-1) - 2x ] + 1 [ 1(x - 2) - (-1)x ] = 0 \)
\( \implies 1 [ 1 - 2x + 4 ] - 1 [ -1 - 2x ] + 1 [ x - 2 + x ] = 0 \)
\( \implies (5 - 2x) + (1 + 2x) + (2x - 2) = 0 \)
\( \implies 2x + 4 = 0 \implies 2x = -4 \implies x = -2 \)
Therefore, the value of \( x \) is \( -2 \).
In simple words: When three vectors lie on the same flat surface, they cannot enclose any 3D volume, so their scalar triple product (determinant) is zero. Solving this gives us the value of \( x \).

Exam Tip: Be highly methodical when expanding the determinant; sign mistakes are the primary cause of lost points in this type of question.

Question 42. Prove that angle between any two diagonals of a cube is \( \cos^{-1}\left(\frac{1}{3}\right) \).
Answer: Let a cube of side length \( a \) be placed in a Cartesian coordinate system with one vertex at the origin \( (0,0,0) \). The coordinate vectors of the vertices can be written in terms of side \( a \). Two main diagonals of this cube are:
\( \vec{d}_1 = a\hat{i} + a\hat{j} + a\hat{k} \)
\( \vec{d}_2 = -a\hat{i} + a\hat{j} + a\hat{k} \)
The magnitudes of these diagonal vectors are:
\( |\vec{d}_1| = \sqrt{a^2 + a^2 + a^2} = \sqrt{3}a \)
\( |\vec{d}_2| = \sqrt{(-a)^2 + a^2 + a^2} = \sqrt{3}a \)
The dot product of the two diagonals is:
\( \vec{d}_1 \cdot \vec{d}_2 = (a)(-a) + (a)(a) + (a)(a) = -a^2 + a^2 + a^2 = a^2 \)
Let \( \theta \) be the angle between the two diagonals:
\( \cos\theta = \frac{\vec{d}_1 \cdot \vec{d}_2}{|\vec{d}_1| |\vec{d}_2|} = \frac{a^2}{(\sqrt{3}a)(\sqrt{3}a)} = \frac{a^2}{3a^2} = \frac{1}{3} \)
\( \implies \theta = \cos^{-1}\left(\frac{1}{3}\right) \). Hence proved.
In simple words: By placing a cube on a grid and writing out coordinate directions for its internal diagonals, we calculate their dot product divided by their lengths to find the angle.

Exam Tip: Standardizing the side of the cube as \( a \) allows all dimensions to cancel out cleanly during the final cosine calculation.

Question 43. Let \( \hat{a}, \hat{b} \) and \( \hat{c} \) are unit vectors such that \( \hat{a} \cdot \hat{b} = \hat{a} \cdot \hat{c} = 0 \) and the angle between \( \hat{b} \) and \( \hat{c} \) is \( \frac{\pi}{6} \), then prove that \( \hat{a} = \pm 2 (\hat{b} \times \hat{c}) \).
Answer: Since \( \hat{a} \cdot \hat{b} = 0 \) and \( \hat{a} \cdot \hat{c} = 0 \), the unit vector \( \hat{a} \) is perpendicular to both \( \hat{b} \) and \( \hat{c} \). This means \( \hat{a} \) is parallel to the cross product vector \( \hat{b} \times \hat{c} \). We can express this relationship as:
\( \hat{a} = \pm \frac{\hat{b} \times \hat{c}}{|\hat{b} \times \hat{c}|} \)
Now, let us calculate the magnitude of the cross product of the unit vectors \( \hat{b} \) and \( \hat{c} \):
\( |\hat{b} \times \hat{c}| = |\hat{b}| |\hat{c}| \sin\left(\frac{\pi}{6}\right) \)
Substitute \( |\hat{b}| = 1 \), \( |\hat{c}| = 1 \), and \( \sin(30^\circ) = \frac{1}{2} \):
\( |\hat{b} \times \hat{c}| = (1)(1)\left(\frac{1}{2}\right) = \frac{1}{2} \)
Substitute this magnitude back into the vector equation:
\( \hat{a} = \pm \frac{\hat{b} \times \hat{c}}{1/2} = \pm 2 (\hat{b} \times \hat{c}) \). Hence proved.
In simple words: Since vector a is perpendicular to both b and c, it must lie along their cross product direction. We calculate the size of this cross product as 0.5, meaning we must multiply by 2 to keep a at unit length.

Exam Tip: Clearly write down that \( \hat{a} \) being perpendicular to both vectors means it is collinear with their cross product, to justify your starting step.

Question 44. Prove that the normal vector to the plane containing three points with position vectors \( \vec{a}, \vec{b} \) and \( \vec{c} \) lies in the direction of vector \( \vec{b} \times \vec{c} + \vec{c} \times \vec{a} + \vec{a} \times \vec{b} \).
Answer: Let A, B, C be the three points with position vectors \( \vec{a}, \vec{b}, \vec{c} \). Two vectors in the plane of these three points are:
\( \vec{AB} = \vec{b} - \vec{a} \)
\( \vec{AC} = \vec{c} - \vec{a} \)
The normal vector to the plane containing A, B, C must be perpendicular to both \( \vec{AB} \) and \( \vec{AC} \). Thus, it lies along their cross product:
\( \vec{n} = \vec{AB} \times \vec{AC} = (\vec{b} - \vec{a}) \times (\vec{c} - \vec{a}) \)
Expand the cross product using distributive properties:
\( \vec{n} = \vec{b} \times \vec{c} - \vec{b} \times \vec{a} - \vec{a} \times \vec{c} + \vec{a} \times \vec{a} \)
Using the properties of cross products, \( \vec{a} \times \vec{a} = \vec{0} \), \( -\vec{b} \times \vec{a} = \vec{a} \times \vec{b} \), and \( -\vec{a} \times \vec{c} = \vec{c} \times \vec{a} \):
\( \vec{n} = \vec{b} \times \vec{c} + \vec{a} \times \vec{b} + \vec{c} \times \vec{a} = \vec{b} \times \vec{c} + \vec{c} \times \vec{a} + \vec{a} \times \vec{b} \)
Therefore, the normal vector indeed lies in the direction of \( \vec{b} \times \vec{c} + \vec{c} \times \vec{a} + \vec{a} \times \vec{b} \). Hence proved.
In simple words: The normal direction is found by crossing two sides of the triangle. Expanding this cross product algebraically leaves us with three cyclic cross products of the position vectors.

Exam Tip: Be mindful of signs during the algebraic distribution; a common slip is incorrect ordering of the cross product terms.

Question 45. If \( \vec{a}, \vec{b}, \vec{c} \) are position vectors of the vertices A, B, C of a triangle ABC then show that the area of \( \Delta ABC \) is \( \frac{1}{2} |\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}| \).
Answer: The area of triangle ABC is defined as:
\( \text{Area} = \frac{1}{2} |\vec{AB} \times \vec{AC}| \)
The vectors representing the sides can be written as:
\( \vec{AB} = \vec{b} - \vec{a} \)
\( \vec{AC} = \vec{c} - \vec{a} \)
Now, let us calculate the cross product of these side vectors:
\( \vec{AB} \times \vec{AC} = (\vec{b} - \vec{a}) \times (\vec{c} - \vec{a}) \)
Expanding this expression as in the previous proof:
\( \vec{AB} \times \vec{AC} = \vec{b} \times \vec{c} - \vec{b} \times \vec{a} - \vec{a} \times \vec{c} + \vec{a} \times \vec{a} \)
Since \( \vec{a} \times \vec{a} = \vec{0} \), \( -\vec{b} \times \vec{a} = \vec{a} \times \vec{b} \), and \( -\vec{a} \times \vec{c} = \vec{c} \times \vec{a} \):
\( \vec{AB} \times \vec{AC} = \vec{b} \times \vec{c} + \vec{a} \times \vec{b} + \vec{c} \times \vec{a} = \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} \)
Substituting this back into the area equation:
\( \text{Area} = \frac{1}{2} |\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}| \). Hence proved.
In simple words: We find the area of a triangle using its side vectors. When we write these sides in terms of the corner coordinates, we get a beautiful symmetric sum of coordinate cross products.

Exam Tip: This formula is highly elegant because it does not depend on selecting any specific corner vertex as the origin.

Question 46. If \( \vec{a} \times \vec{b} = \vec{c} \times \vec{d} \) and \( \vec{a} \times \vec{c} = \vec{b} \times \vec{d} \), then prove that \( \vec{a} - \vec{d} \) is parallel to \( \vec{b} - \vec{c} \) provided \( \vec{a} \ne \vec{d} \) and \( \vec{b} \ne \vec{c} \).
Answer: Two non-zero vectors are parallel if and only if their cross product is zero. Let us compute the cross product of the two given difference vectors:
\( (\vec{a} - \vec{d}) \times (\vec{b} - \vec{c}) \)
Expand this expression using distributive properties:
\( = \vec{a} \times \vec{b} - \vec{a} \times \vec{c} - \vec{d} \times \vec{b} + \vec{d} \times \vec{c} \)
Rearrange terms and use anti-commutative properties where appropriate:
\( = \vec{a} \times \vec{b} - \vec{a} \times \vec{c} + \vec{b} \times \vec{d} - \vec{c} \times \vec{d} \)
Now, substitute the given equations \( \vec{a} \times \vec{b} = \vec{c} \times \vec{d} \) and \( \vec{a} \times \vec{c} = \vec{b} \times \vec{d} \):
\( = (\vec{c} \times \vec{d}) - (\vec{b} \times \vec{d}) + (\vec{b} \times \vec{d}) - (\vec{c} \times \vec{d}) \)
\( = \vec{0} \)
Since the cross product of the two non-zero vectors is zero, the vector \( \vec{a} - \vec{d} \) is indeed parallel to \( \vec{b} - \vec{c} \). Hence proved.
In simple words: To prove that two vectors are parallel, we cross them together. Substituting the given relationships makes all terms cancel out, proving they are parallel.

Exam Tip: State the zero cross-product condition of parallel vectors clearly at the beginning of your steps to set up the proof properly.

Question 47. Dot product of a vector with vectors \( \hat{i} + \hat{j} - 3\hat{k} \), \( \hat{i} + 3\hat{j} - 2\hat{k} \) and \( 2\hat{i} + \hat{j} + 4\hat{k} \) is 0, 5 and 8 respectively. Find the vectors.
Answer: Let the required vector be \( \vec{v} = x\hat{i} + y\hat{j} + z\hat{k} \). Using the three given dot product equations, we set up a system of three linear equations:
1) \( \vec{v} \cdot (\hat{i} + \hat{j} - 3\hat{k}) = x + y - 3z = 0 \)
2) \( \vec{v} \cdot (\hat{i} + 3\hat{j} - 2\hat{k}) = x + 3y - 2z = 5 \bk \)
3) \( \vec{v} \cdot (2\hat{i} + \hat{j} + 4\hat{k}) = 2x + y + 4z = 8 \)
Subtract equation (1) from equation (2) to eliminate \( x \):
\( (x + 3y - 2z) - (x + y - 3z) = 5 - 0 \implies 2y + z = 5 \implies z = 5 - 2y \)
From equation (1), write \( x \) in terms of \( y \):
\( x = 3z - y = 3(5 - 2y) - y = 15 - 6y - y = 15 - 7y \)
Substitute \( x = 15 - 7y \) and \( z = 5 - 2y \) into equation (3):
\( 2(15 - 7y) + y + 4(5 - 2y) = 8 \)
\( \implies 30 - 14y + y + 20 - 8y = 8 \)
\( \implies 50 - 21y = 8 \implies 21y = 42 \implies y = 2 \)
Now, find the values of \( x \) and \( z \):
\( z = 5 - 2(2) = 1 \)
\( x = 15 - 7(2) = 1 \)
Thus, the required vector is:
\( \vec{v} = \hat{i} + 2\hat{j} + \hat{k} \)
In simple words: We write the vector with unknown components and create a system of three equations. Solving this system gives us the unique coordinate components.

Exam Tip: Be sure to verify your final vector by checking its dot product with all three given vectors to ensure no calculation mistakes were made.

Question 48. If \( \vec{a} = 5\hat{i} - \hat{j} + 7\hat{k} \), \( \vec{b} = \hat{i} - \hat{j} - \lambda\hat{k} \), find \( \lambda \) such that \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \) are orthogonal.
Answer: Since the vectors \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \) are orthogonal, their dot product is zero:
\( (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 0 \)
This simplifies directly to:
\( |\vec{a}|^2 - |\vec{b}|^2 = 0 \implies |\vec{a}|^2 = |\vec{b}|^2 \)
Calculate the squared magnitudes of \( \vec{a} \) and \( \vec{b} \):
\( |\vec{a}|^2 = 5^2 + (-1)^2 + 7^2 = 25 + 1 + 49 = 75 \)
\( |\vec{b}|^2 = 1^2 + (-1)^2 + (-\lambda)^2 = 1 + 1 + \lambda^2 = 2 + \lambda^2 \)
Equate these two values:
\( 75 = 2 + \lambda^2 \implies \lambda^2 = 73 \implies \lambda = \pm \sqrt{73} \)
In simple words: The orthogonality condition simplifies to the requirement that both vectors must be of equal length. Solving this squared length equation yields \( \lambda = \pm\sqrt{73} \).

Exam Tip: Always include both the positive and negative roots in the final answer unless a specific quadrant or sign constraint is mentioned.

Question 49. Let \( \vec{a} \) and \( \vec{b} \) be vectors such that \( |\vec{a}| = |\vec{b}| = |\vec{a} - \vec{b}| = 1 \), then find \( |\vec{a} + \vec{b}| \).
Answer: We are given \( |\vec{a}| = 1 \), \( |\vec{b}| = 1 \), and \( |\vec{a} - \vec{b}| = 1 \). Square the difference equation:
\( |\vec{a} - \vec{b}|^2 = 1 \)
\( \implies |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b}) = 1 \)
Substitute the known values:
\( 1^2 + 1^2 - 2(\vec{a} \cdot \vec{b}) = 1 \implies 2 - 2(\vec{a} \cdot \vec{b}) = 1 \implies 2(\vec{a} \cdot \vec{b}) = 1 \)
Now, let us evaluate the squared magnitude of the sum vector:
\( |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) \)
Substitute the values of the components:
\( |\vec{a} + \vec{b}|^2 = 1^2 + 1^2 + 1 = 3 \)
Taking the square root:
\( |\vec{a} + \vec{b}| = \sqrt{3} \)
In simple words: We find the value of the shared dot product term from the subtraction equation and plug it into the addition equation, giving a final length of \( \sqrt{3} \).

Exam Tip: This algebraic trick of substituting the whole term \( 2(\vec{a} \cdot \vec{b}) \) avoids having to calculate intermediate angles.

Question 50. If \( |\vec{a}| = 2 \), \( |\vec{b}| = 5 \) and \( \vec{a} \times \vec{b} = 2\hat{i} + \hat{j} - 2\hat{k} \), find the value of \( \vec{a} \cdot \vec{b} \).
Answer: First, calculate the squared magnitude of the cross product vector:
\( |\vec{a} \times \vec{b}|^2 = 2^2 + 1^2 + (-2)^2 = 4 + 1 + 4 = 9 \)
Using Lagrange's Identity, which relates dot and cross products:
\( |\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 \)
Substitute the given values into this identity:
\( 9 + (\vec{a} \cdot \vec{b})^2 = (2)^2 (5)^2 \)
\( \implies 9 + (\vec{a} \cdot \vec{b})^2 = 4 \times 25 = 100 \)
\( \implies (\vec{a} \cdot \vec{b})^2 = 91 \implies \vec{a} \cdot \vec{b} = \pm \sqrt{91} \bk \)
In simple words: Using Lagrange's identity, we can find the missing dot product directly from the lengths of the vectors and their cross product size.

Exam Tip: Be sure to write the \( \pm \) sign in your final answer since the dot product can be positive or negative depending on whether the angle is acute or obtuse.

Question 51. \( \vec{a}, \vec{b}, \vec{c} \) are three vectors such that \( \vec{b} \times \vec{c} = \vec{a} \) and \( \vec{a} \times \vec{b} = \vec{c} \). Prove that \( \vec{a}, \vec{b} \) and \( \vec{c} \) are mutually perpendicular to each other and \( |\vec{b}| = 1 \), \( |\vec{c}| = |\vec{a}| \).
Answer: From the equation \( \vec{b} \times \vec{c} = \vec{a} \), we know that the cross product result is perpendicular to both input vectors:
\( \vec{a} \perp \vec{b} \quad \text{and} \quad \vec{a} \perp \vec{c} \)
From the equation \( \vec{a} \times \vec{b} = \vec{c} \), we have:
\( \vec{c} \perp \vec{a} \quad \text{and} \quad \vec{c} \perp \vec{b} \)
Combining these conditions shows that the vectors \( \vec{a}, \vec{b}, \vec{c} \) are mutually perpendicular.
Now, let us analyze the magnitudes of the cross products. Since they are mutually perpendicular, the angle between them is \( 90^\circ \):
\( |\vec{c}| = |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin(90^\circ) = |\vec{a}| |\vec{b}| \quad \text{--- (Equation 1)} \)
\( |\vec{a}| = |\vec{b} \times \vec{c}| = |\vec{b}| |\vec{c}| \sin(90^\circ) = |\vec{b}| |\vec{c}| \quad \text{--- (Equation 2)} \)
Multiply Equation 1 and Equation 2 together:
\( |\vec{a}| |\vec{c}| = |\vec{b}|^2 |\vec{a}| |\vec{c}| \)
Assuming non-zero vectors, we can divide both sides by \( |\vec{a}| |\vec{c}| \):
\( |\vec{b}|^2 = 1 \implies |\vec{b}| = 1 \)
Substitute \( |\vec{b}| = 1 \) back into Equation 1:
\( |\vec{c}| = |\vec{a}| (1) \implies |\vec{c}| = |\vec{a}| \). Hence proved.
In simple words: Since the cross product of any two gives the third vector, they must all be at right angles to each other. Comparing their scaling factors shows that vector b has unit length and vectors a and c are of equal size.

Exam Tip: Be sure to justify the transition to magnitudes by stating that the angle between mutually perpendicular vectors is \( 90^\circ \) so \( \sin(90^\circ) = 1 \).

Question 52. If \( \vec{a} = 2\hat{i} - 3\hat{j} \), \( \vec{b} = \hat{i} + \hat{j} - \hat{k} \) and \( \vec{c} = 3\hat{i} - \hat{k} \) find \( [\vec{a}\ \vec{b}\ \vec{c}] \).
Answer: The scalar triple product is computed using the determinant of the components of the three vectors:
\( [\vec{a}\ \vec{b}\ \vec{c}] = \begin{vmatrix} 2 & -3 & 0 \\ 1 & 1 & -1 \\ 3 & 0 & -1 \end{vmatrix} \)
Expand the determinant along the first row:
\( [\vec{a}\ \vec{b}\ \vec{c}] = 2 [ (1)(-1) - (0)(-1) ] - (-3) [ (1)(-1) - (3)(-1) ] + 0 \)
\( = 2 [ -1 ] + 3 [ -1 + 3 ] \)
\( = -2 + 3(2) = -2 + 6 = 4 \)
Thus, the value of the scalar triple product is 4.
In simple words: We arrange the components of the three vectors in a grid and calculate its determinant. Expanding this grid gives us the final answer of 4.

Exam Tip: Writing the missing components as \( 0 \) (such as the \( \hat{k} \) term in \( \vec{a} \)) helps keep the determinant structure correct and easy to solve.

Question 53. Find volume of parallelepiped whose coterminous edges are given by vectors \( \vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k} \), \( \vec{b} = \hat{i} + 2\hat{j} - \hat{k} \), and \( \vec{c} = 3\hat{i} - \hat{j} + 2\hat{k} \).
Answer: The volume of a parallelepiped is equal to the absolute value of the scalar triple product of its coterminous edge vectors:
\( \text{Volume} = |[\vec{a}\ \vec{b}\ \vec{c}]| \)
First, find the scalar triple product using the determinant:
\( [\vec{a}\ \vec{b}\ \vec{c}] = \begin{vmatrix} 2 & 3 & 4 \\ 1 & 2 & -1 \\ 3 & -1 & 2 \end{vmatrix} \bk \)
Expand along the first row:
\( = 2 [ (2)(2) - (-1)(-1) ] - 3 [ (1)(2) - (3)(-1) ] + 4 [ (1)(-1) - (3)(2) ] \)
\( = 2 [ 4 - 1 ] - 3 [ 2 + 3 ] + 4 [ -1 - 6 ] \)
\( = 2(3) - 3(5) + 4(-7) = 6 - 15 - 28 = -37 \)
The volume is the absolute value of this determinant:
\( \text{Volume} = |-37| = 37 \text{ cubic units} \)
In simple words: The volume is found by calculating the determinant of the edge coordinates. Since a physical volume cannot be negative, we convert the negative determinant into a positive value.

Exam Tip: Always state "cubic units" as the unit of measurement to make sure you get full marks for presentation.

Question 54. Find the value of \( \lambda \) such that \( \vec{a} = \hat{i} - \hat{j} + \hat{k} \), \( \vec{b} = 2\hat{i} + \hat{j} - \hat{k} \) and \( \vec{c} = \lambda\hat{i} - \hat{j} + \lambda\hat{k} \) are coplanar.
Answer: Since the three vectors are coplanar, their scalar triple product is zero:
\( [\vec{a}\ \vec{b}\ \vec{c}] = \begin{vmatrix} 1 & -1 & 1 \\ 2 & 1 & -1 \\ \lambda & -1 & \lambda \end{vmatrix} = 0 \)
Expanding the determinant along the first row:
\( 1 [ (1)(\lambda) - (-1)(-1) ] - (-1) [ (2)(\lambda) - (\lambda)(-1) ] + 1 [ (2)(-1) - (\lambda)(1) ] = 0 \)
\( \implies 1 [ \lambda - 1 ] + 1 [ 2\lambda + \lambda ] + 1 [ -2 - \lambda ] = 0 \)
\( \implies (\lambda - 1) + 3\lambda - (2 + \lambda) = 0 \)
\( \implies \lambda - 1 + 3\lambda - 2 - \lambda = 0 \)
\( \implies 3\lambda - 3 = 0 \implies 3\lambda = 3 \implies \lambda = 1 \)
Therefore, the value of \( \lambda \) is 1.
In simple words: For the vectors to lie flat on the same plane, their determinant must equal zero. Solving this linear equation gives us \( \lambda = 1 \).

Exam Tip: Write down the coplanarity condition \( [\vec{a}\ \vec{b}\ \vec{c}] = 0 \) as your first step to establish a clear mathematical basis.

Question 55. Show that the four points (–1, 4, –3), (3, 2, –5) (–3, 8, –5) and (–3, 2, 1) are coplanar.
Answer: Let the points be \( A(-1, 4, -3) \), \( B(3, 2, -5) \), \( C(-3, 8, -5) \), and \( D(-3, 2, 1) \). First, find three vectors from point A to the other points:
\( \vec{AB} = (3 - (-1))\hat{i} + (2 - 4)\hat{j} + (-5 - (-3))\hat{k} = 4\hat{i} - 2\hat{j} - 2\hat{k} \)
\( \vec{AC} = (-3 - (-1))\hat{i} + (8 - 4)\hat{j} + (-5 - (-3))\hat{k} = -2\hat{i} + 4\hat{j} - 2\hat{k} \bk \)
\( \vec{AD} = (-3 - (-1))\hat{i} + (2 - 4)\hat{j} + (1 - (-3))\hat{k} = -2\hat{i} - 2\hat{j} + 4\hat{k} \)
The four points are coplanar if these three vectors are coplanar, meaning their scalar triple product is zero:
\( [\vec{AB}\ \vec{AC}\ \vec{AD}] = \begin{vmatrix} 4 & -2 & -2 \\ -2 & 4 & -2 \\ -2 & -2 & 4 \end{vmatrix} \)
Expand the determinant along the first row:
\( = 4 [ (4)(4) - (-2)(-2) ] - (-2) [ (-2)(4) - (-2)(-2) ] + (-2) [ (-2)(-2) - (-2)(4) ] \)
\( = 4 [ 16 - 4 ] + 2 [ -8 - 4 ] - 2 [ 4 + 8 ] \)
\( = 4(12) + 2(-12) - 2(12) = 48 - 24 - 24 = 0 \)
Since the scalar triple product is zero, the vectors are coplanar, which proves that the four points are coplanar.
In simple words: We create three vectors connecting one point to the other three. Since their coordinate determinant equals zero, they all lie on the same flat surface.

Exam Tip: Using the first point as a common reference point for all three vectors makes subtraction easier and reduces calculation mistakes.

Question 56. For any three vectors \( \vec{a}, \vec{b} \) and \( \vec{c} \), prove that \( [\vec{a} + \vec{b}\ \ \vec{b} + \vec{c}\ \ \vec{c} + \vec{a}] = 2[\vec{a}\ \vec{b}\ \vec{c}] \).
Answer: Let us expand the scalar triple product on the left-hand side:
\( [\vec{a} + \vec{b}\ \ \vec{b} + \vec{c}\ \ \vec{c} + \vec{a}] = (\vec{a} + \vec{b}) \cdot [(\vec{b} + \vec{c}) \times (\vec{c} + \vec{a})] \)
First, expand the cross product inside the bracket:
\( (\vec{b} + \vec{c}) \times (\vec{c} + \vec{a}) = \vec{b} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{c} + \vec{c} \times \vec{a} \)
Since \( \vec{c} \times \vec{c} = \vec{0} \), the expression simplifies to:
\( = \vec{b} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{a} \)
Now, take the dot product of this with \( (\vec{a} + \vec{b}) \):
\( = (\vec{a} + \vec{b}) \cdot (\vec{b} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{a}) \)
\( = \vec{a} \cdot (\vec{b} \times \vec{c}) + \vec{a} \cdot (\vec{b} \times \vec{a}) + \vec{a} \cdot (\vec{c} \times \vec{a}) + \vec{b} \cdot (\vec{b} \times \vec{c}) + \vec{b} \cdot (\vec{b} \times \vec{a}) + \vec{b} \cdot (\vec{c} \times \vec{a}) \)
Recall that any scalar triple product with repeating vectors is zero:
\( \vec{a} \cdot (\vec{b} \times \vec{a}) = 0 \), \( \vec{a} \cdot (\vec{c} \times \vec{a}) = 0 \), \( \vec{b} \cdot (\vec{b} \times \vec{c}) = 0 \), and \( \vec{b} \cdot (\vec{b} \times \vec{a}) = 0 \). This leaves only two terms:
\( = \vec{a} \cdot (\vec{b} \times \vec{c}) + \vec{b} \cdot (\vec{c} \times \vec{a}) \)
Using the cyclic property of scalar triple product, \( \vec{b} \cdot (\vec{c} \times \vec{a}) = \vec{a} \cdot (\vec{b} \times \vec{c}) \):
\( = [\vec{a}\ \vec{b}\ \vec{c}] + [\vec{a}\ \vec{b}\ \vec{c}] = 2[\vec{a}\ \vec{b}\ \vec{c}] \). Hence proved.
In simple words: Expanding the cross and dot products yields several terms. All terms with repeated vectors cancel out, leaving exactly two matching copies of the scalar triple product.

Exam Tip: Clearly state that scalar triple products containing identical vectors are zero to justify your simplification steps.

Question 57. For any three vectors \( \vec{a}, \vec{b} \) and \( \vec{c} \), prove that \( \vec{a} - \vec{b} \), \( \vec{b} - \vec{c} \) and \( \vec{c} - \vec{a} \) are coplanar.
Answer: To prove the three vectors are coplanar, we must show that their scalar triple product is zero:
\( [(\vec{a} - \vec{b})\ \ (\vec{b} - \vec{c})\ \ (\vec{c} - \vec{a})] = (\vec{a} - \vec{b}) \cdot [(\vec{b} - \vec{c}) \times (\vec{c} - \vec{a})] \bk \)
First, expand the cross product term:
\( (\vec{b} - \vec{c}) \times (\vec{c} - \vec{a}) = \vec{b} \times \vec{c} - \vec{b} \times \vec{a} - \vec{c} \times \vec{c} + \vec{c} \times \vec{a} \)
Since \( \vec{c} \times \vec{c} = \vec{0} \) and \( - (\vec{b} \times \vec{a}) = \vec{a} \times \vec{b} \):
\( = \vec{b} \times \vec{c} + \vec{a} \times \vec{b} + \vec{c} \times \vec{a} \)
Now, take the dot product with \( (\vec{a} - \vec{b}) \):
\( = (\vec{a} - \vec{b}) \cdot (\vec{b} \times \vec{c} + \vec{a} \times \vec{b} + \vec{c} \times \vec{a}) \)
\( = \vec{a} \cdot (\vec{b} \times \vec{c}) + \vec{a} \cdot (\vec{a} \times \vec{b}) + \vec{a} \cdot (\vec{c} \times \vec{a}) - \vec{b} \cdot (\vec{b} \times \vec{c}) - \vec{b} \cdot (\vec{a} \times \vec{b}) - \vec{b} \cdot (\vec{c} \times \vec{a}) \)
Any triple product term containing duplicate vectors is zero:
\( = [\vec{a}\ \vec{b}\ \vec{c}] + 0 + 0 - 0 - 0 - [\vec{b}\ \vec{c}\ \vec{a}] \)
Using the cyclic property, \( [\vec{b}\ \vec{c}\ \vec{a}] = [\vec{a}\ \vec{b}\ \vec{c}] \):
\( = [\vec{a}\ \vec{b}\ \vec{c}] - [\vec{a}\ \vec{b}\ \vec{c}] = 0 \)
Since the scalar triple product is zero, the three vectors are coplanar. Hence proved.
In simple words: Expanding the triple product of these differences yields several terms. All repeating terms cancel, and the two remaining cyclic products subtract to zero, proving they are coplanar.

Exam Tip: Be sure to keep track of the minus signs when distributing the subtraction vector term across the cross product terms.

Click on link below to download CBSE Class 12 Mathematics Vector Algebra Assignment Set A