CBSE Class 12 Mathematics Vector Algebra Worksheet Set 01

CBSE Class 12 Mathematics Worksheet - Vector Algebra. CBSE issues sample papers every year for students for class 12 board exams. Students should solve the CBSE issued sample papers to understand the pattern of the question paper which will come in class 12 board exams this year. The sample papers have been provided with marking scheme. It’s always recommended to practice as many CBSE sample papers as possible before the board examinations. Sample papers should be always practiced in examination condition at home or school and the student should show the answers to teachers for checking or compare with the answers provided. Students can download the sample papers in pdf format free and score better marks in examinations. Refer to other links too for latest sample papers.

Some Important Results/Concepts

• Position vector of point \(A(x, y, z) = \vec{OA} = x\hat{i} + y\hat{j} + z\hat{k}\)
• If \(A(x_1, y_1, z_1)\) and point \(B(x_2, y_2, z_2)\), then \(\vec{AB} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}\)
• If \(\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}\), then \(|\vec{a}| = \sqrt{x^2 + y^2 + z^2}\)
• Unit vector parallel to \(\vec{a} = \hat{a} = \frac{\vec{a}}{|\vec{a}|}\)
• Scalar Product (dot product) between two vectors: \(\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta\), where \(\theta\) is the angle between the vectors.
• \(\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\)
• If \(\vec{a} = a_1\hat{i} + b_1\hat{j} + c_1\hat{k}\) and \(\vec{b} = a_2\hat{i} + b_2\hat{j} + c_2\hat{k}\), then \(\vec{a} \cdot \vec{b} = a_1 a_2 + b_1 b_2 + c_1 c_2\)
• If \(\vec{a}\) is perpendicular to \(\vec{b}\), then \(\vec{a} \cdot \vec{b} = 0\)
• \(\vec{a} \cdot \vec{a} = |\vec{a}|^2\)
• Projection of \(\vec{a}\) on \(\vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\)
• Vector product between two vectors: \(\vec{a} \times \vec{b} = |\vec{a}||\vec{b}|\sin\theta \, \hat{n}\), where \(\hat{n}\) is the normal unit vector which is perpendicular to both \(\vec{a}\) and \(\vec{b}\).
• \(\hat{n} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}\)
• If \(\vec{a}\) is parallel to \(\vec{b}\), then \(\vec{a} \times \vec{b} = \vec{0}\)
• Area of triangle (whose sides are given by \(\vec{a}\) and \(\vec{b}\)) \(= \frac{1}{2} |\vec{a} \times \vec{b}|\)
• Area of parallelogram (whose adjacent sides are given by \(\vec{a}\) and \(\vec{b}\)) \(= |\vec{a} \times \vec{b}|\)
• Area of parallelogram (whose diagonals are given by \(\vec{a}\) and \(\vec{b}\)) \(= \frac{1}{2} |\vec{a} \times \vec{b}|\)

ASSIGNMENTS

(i) Vector and scalars, Direction ratio and direction cosines & Unit vector

LEVEL I

Question 1. If \(\vec{a} = \hat{i} + \hat{j} - 5\hat{k}\) and \(\vec{b} = \hat{i} - 4\hat{j} + 3\hat{k}\) find a unit vector parallel to \(\vec{a} + \vec{b}\).
Answer: First, calculate the sum of the two given vectors:
\(\vec{a} + \vec{b} = (\hat{i} + \hat{j} - 5\hat{k}) + (\hat{i} - 4\hat{j} + 3\hat{k}) = 2\hat{i} - 3\hat{j} - 2\hat{k}\)
Now, find the magnitude of this resultant vector:
\(|\vec{a} + \vec{b}| = \sqrt{2^2 + (-3)^2 + (-2)^2} = \sqrt{4 + 9 + 4} = \sqrt{17}\)
The required unit vector is parallel to this sum, which is given by dividing the vector by its magnitude:
\(\hat{n} = \frac{\vec{a} + \vec{b}}{|\vec{a} + \vec{b}|} = \frac{2\hat{i} - 3\hat{j} - 2\hat{k}}{\sqrt{17}}\)
In simple words: Add the two vectors together by combining their parts. Then, divide this new vector by its total length to make it a unit vector.

Exam Tip: A unit vector parallel to any given vector is found by dividing that vector by its absolute length. Ensure you show the steps for addition and magnitude calculation separately.

Question 2. Write a vector of magnitude 15 units in the direction of vector \(\hat{i} - 2\hat{j} + 2\hat{k}\).
Answer: Let \(\vec{r} = \hat{i} - 2\hat{j} + 2\hat{k}\).
First, find the unit vector pointing in the direction of \(\vec{r}\):
\(\hat{r} = \frac{\vec{r}}{|\vec{r}|} = \frac{\hat{i} - 2\hat{j} + 2\hat{k}}{\sqrt{1^2 + (-2)^2 + 2^2}} = \frac{\hat{i} - 2\hat{j} + 2\hat{k}}{\sqrt{9}} = \frac{\hat{i} - 2\hat{j} + 2\hat{k}}{3}\)
To get a vector with a magnitude of 15 units, multiply this unit vector by 15:
\(\vec{v} = 15 \hat{r} = 15 \left( \frac{\hat{i} - 2\hat{j} + 2\hat{k}}{3} \right) = 5(\hat{i} - 2\hat{j} + 2\hat{k}) = 5\hat{i} - 10\hat{j} + 10\hat{k}\)
In simple words: Find the direction vector of length 1 first. Then multiply each component of that direction by 15 to scale up its length.

Exam Tip: Always divide by the magnitude first to establish a baseline of 1 unit before scaling to the final magnitude requested by the question.

Question 3. If \(\vec{a} = \hat{i} + \hat{j} - \hat{k}\); \(\vec{b} = \hat{i} - \hat{j} + \hat{k}\); \(\vec{c} = -\hat{i} + \hat{j} + \hat{k}\) find a unit vector in the direction of \(\vec{a} + \vec{b} + \vec{c}\).
Answer: First, calculate the sum of the three given vectors:
\(\vec{s} = \vec{a} + \vec{b} + \vec{c} = (\hat{i} + \hat{j} - \hat{k}) + (\hat{i} - \hat{j} + \hat{k}) + (-\hat{i} + \hat{j} + \hat{k}) = \hat{i} + \hat{j} + \hat{k}\)
Now, find the magnitude of this sum:
\(|\vec{s}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}\)
The unit vector is obtained by dividing the vector sum by its magnitude:
\(\hat{s} = \frac{\vec{s}}{|\vec{s}|} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}\)
In simple words: Combine the three vectors into one single vector, then divide its components by its length, which is \(\sqrt{3}\).

Exam Tip: Simplify the vector sum carefully by grouping the corresponding \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) components before carrying out magnitude calculations.

Question 4. Find a unit vector in the direction of the vector \(\vec{a} = 2\hat{i} + \hat{j} + 2\hat{k}\).
Answer: The magnitude of the vector \(\vec{a}\) is:
\(|\vec{a}| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3\)
The unit vector \(\hat{a}\) in the same direction is:
\(\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{2\hat{i} + \hat{j} + 2\hat{k}}{3}\)
In simple words: Work out the total length of the vector, which is 3, and then divide each component of the vector by 3.

Exam Tip: This is a standard board exam question. Ensure you write the formula \(\hat{a} = \frac{\vec{a}}{|\vec{a}|}\) to score step-wise marks.

Question 5. Find a vector in the direction of vector \(\vec{a} = \hat{i} - 2\hat{j}\), whose magnitude is 7.
Answer: First, determine the unit vector in the direction of \(\vec{a}\):
\(\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{\hat{i} - 2\hat{j}}{\sqrt{1^2 + (-2)^2}} = \frac{\hat{i} - 2\hat{j}}{\sqrt{5}}\)
Now, scale this unit vector to a magnitude of 7:
\(\vec{v} = 7\hat{a} = \frac{7}{\sqrt{5}}(\hat{i} - 2\hat{j}) = \frac{7}{\sqrt{5}}\hat{i} - \frac{14}{\sqrt{5}}\hat{j}\)
In simple words: Find the unit-length direction of the vector by dividing it by \(\sqrt{5}\), and then multiply the result by 7.

Exam Tip: Do not worry if the components contain irrational numbers in the denominator; leaving the answer in fractional form with \(\sqrt{5}\) is mathematically precise and acceptable.

LEVEL II

Question 1. Find a vector of magnitude 5 units, perpendicular to each of the vectors \((\vec{a} + \vec{b})\), \((\vec{a} - \vec{b})\) where \(\vec{a} = \hat{i} + \hat{j} + \hat{k}\) and \(\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}\).
Answer: First, find the vectors \((\vec{a} + \vec{b})\) and \((\vec{a} - \vec{b})\):
\(\vec{a} + \vec{b} = (\hat{i} + \hat{j} + \hat{k}) + (\hat{i} + 2\hat{j} + 3\hat{k}) = 2\hat{i} + 3\hat{j} + 4\hat{k}\)
\(\vec{a} - \vec{b} = (\hat{i} + \hat{j} + \hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k}) = -\hat{j} - 2\hat{k}\)
A vector perpendicular to both is given by their cross product:
\(\vec{p} = (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})\)
\[ \vec{p} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 0 & -1 & -2 \end{vmatrix} = \hat{i}(3(-2) - 4(-1)) - \hat{j}(2(-2) - 0) + \hat{k}(2(-1) - 0) \] \(\vec{p} = -2\hat{i} + 4\hat{j} - 2\hat{k}\)
The magnitude of this perpendicular vector is:
\(|\vec{p}| = \sqrt{(-2)^2 + 4^2 + (-2)^2} = \sqrt{4 + 16 + 4} = \sqrt{24} = 2\sqrt{6}\)
The unit vector perpendicular to both is:
\(\hat{p} = \frac{\vec{p}}{|\vec{p}|} = \frac{-2\hat{i} + 4\hat{j} - 2\hat{k}}{2\sqrt{6}} = \frac{-\hat{i} + 2\hat{j} - \hat{k}}{\sqrt{6}}\)
Now, the required vector of magnitude 5 is:
\(\pm 5\hat{p} = \pm \frac{5}{\sqrt{6}}(-\hat{i} + 2\hat{j} - \hat{k})\)
In simple words: Add and subtract the vectors first. Use cross product to find a direction perpendicular to both. Finally, turn it into a unit vector and scale it by 5.

Exam Tip: A vector perpendicular to two given vectors is always parallel to their cross product. Remember to write both positive and negative directions (\(\pm\)) as both are perpendicular.

Question 2. If the sum of two unit vectors is a unit vector, show that the magnitude of their difference is \(\sqrt{3}\).
Answer: Let \(\hat{u}\) and \(\hat{v}\) be two unit vectors. Thus, \(|\hat{u}| = 1\) and \(|\hat{v}| = 1\).
Given that their sum is also a unit vector:
\(|\hat{u} + \hat{v}| = 1\)
Squaring both sides:
\(|\hat{u} + \hat{v}|^2 = 1^2 \implies |\hat{u}|^2 + |\hat{v}|^2 + 2(\hat{u} \cdot \hat{v}) = 1\)
Substituting \(1\) for their magnitudes:
\(1 + 1 + 2(\hat{u} \cdot \hat{v}) = 1 \implies 2(\hat{u} \cdot \hat{v}) = -1 \implies \hat{u} \cdot \hat{v} = -\frac{1}{2}\)
Now, we find the magnitude of their difference:
\(|\hat{u} - \hat{v}|^2 = |\hat{u}|^2 + |\hat{v}|^2 - 2(\hat{u} \cdot \hat{v})\)
\(|\hat{u} - \hat{v}|^2 = 1 + 1 - 2\left(-\frac{1}{2}\right) = 2 + 1 = 3\)
Taking the square root of both sides:
\(|\hat{u} - \hat{v}| = \sqrt{3}\)
Hence proved.
In simple words: Use the dot product of the sum of the unit vectors to find the angle relationship between them. Then, use this same relationship to calculate the length of their difference.

Exam Tip: Expanding vector square identities like \(|\vec{a} \pm \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 \pm 2\vec{a}\cdot\vec{b}\) is a vital tool for solving algebraic vector proofs.

Question 3. If \(\vec{a} = \hat{i} + \hat{j} + \hat{k}\), \(\vec{b} = 4\hat{i} - 2\hat{j} + 3\hat{k}\) and \(\vec{c} = \hat{i} - 2\hat{j} + \hat{k}\), find a vector of magnitude 6 units which is parallel to the vector \(2\vec{a} - \vec{b} + 3\vec{c}\).
Answer: Let \(\vec{r} = 2\vec{a} - \vec{b} + 3\vec{c}\).
Calculate the combined vector \(\vec{r}\):
\(\vec{r} = 2(\hat{i} + \hat{j} + \hat{k}) - (4\hat{i} - 2\hat{j} + 3\hat{k}) + 3(\hat{i} - 2\hat{j} + \hat{k})\)
\(\vec{r} = (2\hat{i} + 2\hat{j} + 2\hat{k}) - (4\hat{i} - 2\hat{j} + 3\hat{k}) + (3\hat{i} - 6\hat{j} + 3\hat{k})\)
\(\vec{r} = (2 - 4 + 3)\hat{i} + (2 + 2 - 6)\hat{j} + (2 - 3 + 3)\hat{k}\)
\(\vec{r} = \hat{i} - 2\hat{j} + 2\hat{k}\)
The magnitude of \(\vec{r}\) is:
\(|\vec{r}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{9} = 3\)
The unit vector parallel to \(\vec{r}\) is:
\(\hat{r} = \frac{\hat{i} - 2\hat{j} + 2\hat{k}}{3}\)
The required vector of magnitude 6 is:
\(6\hat{r} = 6 \left( \frac{\hat{i} - 2\hat{j} + 2\hat{k}}{3} \right) = 2(\hat{i} - 2\hat{j} + 2\hat{k}) = 2\hat{i} - 4\hat{j} + 4\hat{k}\)
In simple words: Combine the three vectors in the given proportion to get a single vector. Scale its length down to 1 unit, then scale it up to length 6.

Exam Tip: Be meticulous with sign changes when distributing negative signs across parentheses, especially on the components of \(\vec{b}\).

LEVEL III

Question 1. If a line make \(\alpha, \beta, \gamma\) with the X - axis , Y– axis and Z – axis respectively, then find the value of \(\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma\).
Answer: We know that the direction cosines of a line satisfy the fundamental identity:
\(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1\)
Using the trigonometric identity \(\cos^2 \theta = 1 - \sin^2 \theta\):
\((1 - \sin^2 \alpha) + (1 - \sin^2 \beta) + (1 - \sin^2 \gamma) = 1\)
\(3 - (\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma) = 1\)
\(\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 3 - 1 = 2\)
Therefore, the value is 2.
In simple words: Use the rule that the squared cosines of the angles add up to 1. Since each squared cosine is \(1\) minus the squared sine, the squared sines must add up to 2.

Exam Tip: This is a key conceptual question in both vector algebra and three-dimensional geometry. Remembering the sum is 2 will help verify your steps quickly.

Question 2. For what value of p, is \( (\hat{i} + \hat{j} + \hat{k})p \) a unit vector?
Answer: Let \(\vec{u} = p(\hat{i} + \hat{j} + \hat{k}) = p\hat{i} + p\hat{j} + p\hat{k}\).
For \(\vec{u}\) to be a unit vector, its magnitude must equal 1:
\(|\vec{u}| = 1 \implies \sqrt{p^2 + p^2 + p^2} = 1\)
\(\sqrt{3p^2} = 1 \implies |p|\sqrt{3} = 1\)
\(p = \pm \frac{1}{\sqrt{3}}\)
In simple words: The length of the basic vector is \(\sqrt{3}\). To make the final length equal to 1, the multiplier \(p\) must be plus or minus \(1/\sqrt{3}\).

Exam Tip: Do not forget the \(\pm\) sign, as the square root of \(p^2\) is \(|p|\), yielding both positive and negative solutions.

Question 3. What is the cosine of the angle which the vector \(\sqrt{2}\hat{i} + \hat{j} + \hat{k}\) makes with Y-axis?
Answer: Let \(\vec{a} = \sqrt{2}\hat{i} + \hat{j} + \hat{k}\).
The magnitude of \(\vec{a}\) is:
\(|\vec{a}| = \sqrt{(\sqrt{2})^2 + 1^2 + 1^2} = \sqrt{2 + 1 + 1} = \sqrt{4} = 2\)
The Y-component of the vector (the coefficient of \(\hat{j}\)) is \(a_y = 1\).
The direction cosine with the Y-axis is given by:
\(\cos \beta = \frac{a_y}{|\vec{a}|} = \frac{1}{2}\)
Therefore, the cosine of the angle is \(\frac{1}{2}\).
In simple words: Find the length of the vector, which is 2. The component along the Y-axis is 1. The cosine is just the Y-component divided by the total length, which is \(1/2\).

Exam Tip: The direction cosines of a vector are simply its components divided by the vector's total magnitude.

Question 4. Write the value of p for which \(\vec{a} = 3\hat{i} + 2\hat{j} + 9\hat{k}\) and \(\vec{b} = \hat{i} + p\hat{j} + 3\hat{k}\) are parallel vectors.
Answer: For two vectors \(\vec{a}\) and \(\vec{b}\) to be parallel, their respective directional components must be proportional:
\(\frac{a_x}{b_x} = \frac{a_y}{b_y} = \frac{a_z}{b_z}\)
\(\frac{3}{1} = \frac{2}{p} = \frac{9}{3}\)
From the relation \(\frac{3}{1} = \frac{2}{p}\):
\(3p = 2 \implies p = \frac{2}{3}\)
In simple words: Since the vectors are parallel, one is a multiple of the other. The first vector's components are exactly 3 times larger than the second vector's components, so \(3p = 2\), which means \(p = 2/3\).

Exam Tip: Equating component ratios is the fastest and most reliable way to solve for unknown parameters in parallel vector problems.

(ii) Position vector of a point and collinear vectors

LEVEL I

Question 1. Find the position vector of the midpoint of the line segment joining the points A(\(5\hat{i} + 3\hat{j}\)) and B(\(3\hat{i} - \hat{j}\)).
Answer: Let \(\vec{a} = 5\hat{i} + 3\hat{j}\) and \(\vec{b} = 3\hat{i} - \hat{j}\) be the position vectors of A and B respectively.
The position vector of the midpoint is given by averaging the position vectors of the endpoints:
\(\vec{m} = \frac{\vec{a} + \vec{b}}{2} = \frac{(5\hat{i} + 3\hat{j}) + (3\hat{i} - \hat{j})}{2} = \frac{8\hat{i} + 2\hat{j}}{2} = 4\hat{i} + \hat{j}\)
In simple words: Add the coordinates of the two points together and divide the result by 2 to find the middle position.

Exam Tip: The midpoint formula in vectors is identical to the midpoint formula in coordinate geometry. Just perform component-wise addition and divide by 2.

Question 2. In a triangle ABC, the sides AB and BC are represents by vectors \(2\hat{i} - \hat{j} + 2\hat{k}\) , \(\hat{i} + 3\hat{j} + 5\hat{k}\) respectively. Find the vector representing CA.
Answer: According to the triangle law of vector addition, the sum of the vectors along the sides taken in order is:
\(\vec{AB} + \vec{BC} = \vec{AC}\)
Substitute the given vectors into this relation:
\(\vec{AC} = (2\hat{i} - \hat{j} + 2\hat{k}) + (\hat{i} + 3\hat{j} + 5\hat{k}) = 3\hat{i} + 2\hat{j} + 7\hat{k}\)
Since the question asks for the vector representing \(\vec{CA}\), we reverse the direction of \(\vec{AC}\):
\(\vec{CA} = -\vec{AC} = -(3\hat{i} + 2\hat{j} + 7\hat{k}) = -3\hat{i} - 2\hat{j} - 7\hat{k}\)
In simple words: Use the triangle rule to find the path from A to C by adding the two paths together. Then, flip the signs to find the reverse path from C to A.

Exam Tip: Be careful with the directional sense of side vectors; \(\vec{CA}\) is the exact negative of \(\vec{AC}\).

Question 3. Show that the points (1,0), (6,0) ,(0,0) are collinear.
Answer: Let the points be \(P(1, 0)\), \(Q(6, 0)\), and \(R(0, 0)\).
Their corresponding position vectors are \(\vec{p} = \hat{i}\), \(\vec{q} = 6\hat{i}\), and \(\vec{r} = \vec{0}\).
Let us find the vectors representing segments PQ and PR:
\(\vec{PQ} = \vec{q} - \vec{p} = 6\hat{i} - \hat{i} = 5\hat{i}\)
\(\vec{PR} = \vec{r} - \vec{p} = \vec{0} - \hat{i} = -\hat{i}\)
Clearly, we can write \(\vec{PQ} = -5\vec{PR}\).
Since the vector \(\vec{PQ}\) is a scalar multiple of \(\vec{PR}\), these vectors are parallel and share a common point \(P\).
Therefore, the points \(P\), \(Q\), and \(R\) are collinear.
In simple words: All three points have a Y-coordinate of 0, meaning they all lie flat on the X-axis. Since they lie on the same straight line, they are collinear.

Exam Tip: To prove three points are collinear using vectors, show that the vector formed by two of the points is a scalar multiple of the vector formed by another pair.

LEVEL II

Question 1. Write the position vector of a point R which divides the line joining the points P and Q whose position vectors are \(\hat{i} + 2\hat{j} - \hat{k}\) and \(-\hat{i} + \hat{j} + \hat{k}\) respectively in the ratio 2 : 1 externally.
Answer: Let \(\vec{p} = \hat{i} + 2\hat{j} - \hat{k}\) and \(\vec{q} = -\hat{i} + \hat{j} + \hat{k}\).
The section formula for external division in the ratio \(m : n\) is:
\(\vec{r} = \frac{m\vec{q} - n\vec{p}}{m - n}\)
For the given ratio \(2 : 1\):
\(\vec{r} = \frac{2(-\hat{i} + \hat{j} + \hat{k}) - 1(\hat{i} + 2\hat{j} - \hat{k})}{2 - 1}\)
\(\vec{r} = \frac{(-2\hat{i} + 2\hat{j} + 2\hat{k}) - (\hat{i} + 2\hat{j} - \hat{k})}{1} = -3\hat{i} + 3\hat{k}\)
Thus, the position vector of R is \(-3\hat{i} + 3\hat{k}\).
In simple words: Use the external division formula with a ratio of 2 to 1. This formula subtracts the weighted position vectors to find the new point R.

Exam Tip: Double check that you use a minus sign in the numerator and denominator for external division, as opposed to a plus sign for internal division.

Question 2. Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are \((2\vec{a} + \vec{b})\) and \((\vec{a} - 3\vec{b})\) respectively, externally in the ratio 1:2. Also, show that P is the mid-point of the line segment RQ.
Answer: Let \(\vec{p} = 2\vec{a} + \vec{b}\) and \(\vec{q} = \vec{a} - 3\vec{b}\).
By the section formula for external division in the ratio \(1 : 2\):
\(\vec{r} = \frac{1\vec{q} - 2\vec{p}}{1 - 2} = \frac{(\vec{a} - 3\vec{b}) - 2(2\vec{a} + \vec{b})}{-1}\)
\(\vec{r} = \frac{\vec{a} - 3\vec{b} - 4\vec{a} - 2\vec{b}}{-1} = \frac{-3\vec{a} - 5\vec{b}}{-1} = 3\vec{a} + 5\vec{b}\)
Now, let us find the midpoint of the line segment RQ:
\(\text{Midpoint} = \frac{\vec{r} + \vec{q}}{2} = \frac{(3\vec{a} + 5\vec{b}) + (\vec{a} - 3\vec{b})}{2} = \frac{4\vec{a} + 2\vec{b}}{2} = 2\vec{a} + \vec{b}\)
Since the midpoint of RQ is \(2\vec{a} + \vec{b}\), which is exactly the position vector of P, P is the mid-point of segment RQ.
In simple words: Apply the external division formula to find the coordinates of point R. Then, find the average of Q and R's positions to show it matches P.

Exam Tip: Splitting this question into two logical steps (finding R, then checking the midpoint) makes it easier to verify your vector algebra.

(iii) Dot product of two vectors

LEVEL I

Question 1. Find \(\vec{a} \cdot \vec{b}\) if \(\vec{a} = 3\hat{i} - \hat{j} + 2\hat{k}\) and \(\vec{b} = 2\hat{i} + 3\hat{j} + 3\hat{k}\).
Answer: The dot product is the sum of the products of the corresponding components:
\(\vec{a} \cdot \vec{b} = a_x b_x + a_y b_y + a_z b_z\)
\(\vec{a} \cdot \vec{b} = (3)(2) + (-1)(3) + (2)(3)\)
\(\vec{a} \cdot \vec{b} = 6 - 3 + 6 = 9\)
In simple words: Multiply the X-parts, the Y-parts, and the Z-parts together, then add these three numbers to get the final answer.

Exam Tip: Keep careful track of negative signs during multiplication (e.g., \(-1 \times 3 = -3\)).

Question 2. If \(|\vec{a}| = \sqrt{3}\), \(|\vec{b}| = 2\) and \(\vec{a} \cdot \vec{b} = \sqrt{6}\). Then find the angle between \(\vec{a}\) and \(\vec{b}\).
Answer: Let \(\theta\) be the angle between the two vectors. The dot product relation is:
\(\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos \theta\)
Substitute the given values into the formula:
\(\sqrt{6} = \sqrt{3} \times 2 \times \cos \theta\)
\(\cos \theta = \frac{\sqrt{6}}{2\sqrt{3}} = \frac{\sqrt{2}\sqrt{3}}{2\sqrt{3}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}\)
Since \(\cos \theta = \frac{1}{\sqrt{2}}\), the angle between the vectors is:
\(\theta = \frac{\pi}{4}\) (or \(45^\circ\))
In simple words: Divide the dot product value by the product of the two vector lengths to find the cosine value. This cosine value of \(1/\sqrt{2}\) corresponds to a \(45^\circ\) angle.

Exam Tip: Rationalize fractions like \(\frac{\sqrt{6}}{2\sqrt{3}}\) step-by-step to avoid errors in identifying standard trigonometric values.

Question 3. Write the angle between two vectors \(\vec{a}\) and \(\vec{b}\) with magnitudes \(\sqrt{3}\) and 2 respectively having \(\vec{a} \cdot \vec{b} = \sqrt{6}\).
Answer: Using the angle formula for two vectors:
\(\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\)
Substitute the given magnitudes and dot product:
\(\cos \theta = \frac{\sqrt{6}}{\sqrt{3} \cdot 2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}\)
This gives the angle:
\(\theta = \frac{\pi}{4}\) (or \(45^\circ\))
In simple words: Calculate the cosine of the angle using the dot product divided by the lengths. The angle whose cosine is \(1/\sqrt{2}\) is \(45\) degrees.

Exam Tip: Clearly state whether you are writing the angle in degrees or radians; both are acceptable but showing both displays thoroughness.

LEVEL II

Question 1. The dot products of a vector with the vectors \(\hat{i} - 3\hat{j}\), \(\hat{i} - 2\hat{j}\) and \(\hat{i} + \hat{j} + 4\hat{k}\) are 0 , 5 and 8 respectively. Find the vectors.
Answer: Let the required vector be \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\).
Using the three given dot product equations:
1) \(\vec{r} \cdot (\hat{i} - 3\hat{j}) = 0 \implies x - 3y = 0 \implies x = 3y\)
2) \(\vec{r} \cdot (\hat{i} - 2\hat{j}) = 5 \implies x - 2y = 5\)
Substitute \(x = 3y\) into the second equation:
\(3y - 2y = 5 \implies y = 5\)
Thus, \(x = 3(5) = 15\).
3) \(\vec{r} \cdot (\hat{i} + \hat{j} + 4\hat{k}) = 8 \implies x + y + 4z = 8\)
Substitute \(x = 15\) and \(y = 5\) into the third equation:
\(15 + 5 + 4z = 8 \implies 20 + 4z = 8 \implies 4z = -12 \implies z = -3\).
Therefore, the required vector is:
\(\vec{r} = 15\hat{i} + 5\hat{j} - 3\hat{k}\)
In simple words: Write out the three rules as equations using \(x\), \(y\), and \(z\). Solve these equations step-by-step to find that \(x=15\), \(y=5\), and \(z=-3\).

Exam Tip: Substituting the first equation directly into the second simplifies this system of equations immediately to find \(y\).

Question 2. If \(\vec{a}\) and \(\vec{b}\) are two vectors such that \(|\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}|\), then what is the angle between \(\vec{a}\) and \(\vec{b}\).
Answer: Let \(\theta\) be the angle between the vectors.
We know the scalar formulas:
\(|\vec{a} \cdot \vec{b}| = |\vec{a}||\vec{b}|\cos\theta\)
\(|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta\)
Equating the two magnitudes:
\(|\vec{a}||\vec{b}|\cos\theta = |\vec{a}||\vec{b}|\sin\theta\)
Assuming \(\vec{a}\) and \(\vec{b}\) are non-zero vectors, we divide both sides by \(|\vec{a}||\vec{b}|\):
\(\cos\theta = \sin\theta \implies \tan\theta = 1\)
The angle is:
\(\theta = \frac{\pi}{4}\) (or \(45^\circ\))
In simple words: The magnitude of the dot product equals the magnitude of the cross product when the sine and cosine of their angle are equal. This happens at \(45\) degrees.

Exam Tip: State clearly that \(|\vec{a}| \neq 0\) and \(|\vec{b}| \neq 0\) to justify dividing both sides during the proof.

Question 3. If \(\vec{a} = 2\hat{i} + 2\hat{j} + 3\hat{k}\) , \(\vec{b} = -\hat{i} + 2\hat{j} + \hat{k}\) and \(\vec{c} = 3\hat{i} + \hat{j}\) are such that \(\vec{a} + \lambda\vec{b}\) is perpendicular to \(\vec{c}\) , find the value of \(\lambda\).
Answer: First, write the expression for the combined vector \(\vec{a} + \lambda\vec{b}\):
\(\vec{a} + \lambda\vec{b} = (2\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(-\hat{i} + 2\hat{j} + \hat{k}) = (2 - \lambda)\hat{i} + (2 + 2\lambda)\hat{j} + (3 + \lambda)\hat{k}\)
Since this vector is perpendicular to \(\vec{c} = 3\hat{i} + \hat{j}\), their dot product must equal zero:
\((\vec{a} + \lambda\vec{b}) \cdot \vec{c} = 0\)
\((2 - \lambda)(3) + (2 + 2\lambda)(1) + (3 + \lambda)(0) = 0\)
\(6 - 3\lambda + 2 + 2\lambda = 0\)
\(8 - \lambda = 0 \implies \lambda = 8\)
Therefore, the value of \(\lambda\) is 8.
In simple words: Create a new vector by combining \(\vec{a}\) and \(\lambda\vec{b}\). Since this new vector is perpendicular to \(\vec{c}\), their dot product must be 0, which lets us solve for \(\lambda\).

Exam Tip: Always set the dot product of perpendicular vectors to zero. This is one of the most frequently tested concepts in vector geometry.

LEVEL III

Question 1. If \(\vec{a}\) & \(\vec{b}\) are unit vectors inclined at an angle \(\theta\) , prove that \(\sin \frac{\theta}{2} = \frac{1}{2}|\vec{a} - \vec{b}|\).
Answer: Consider the square of the magnitude of their difference:
\(|\vec{a} - \vec{b}|^2 = (\vec{a} - \vec{b})\cdot(\vec{a} - \vec{b}) = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a}\cdot\vec{b})\)
Since \(\vec{a}\) and \(\vec{b}\) are unit vectors, \(|\vec{a}| = 1\), \(|\vec{b}| = 1\), and \(\vec{a}\cdot\vec{b} = \cos\theta\):
\(|\vec{a} - \vec{b}|^2 = 1 + 1 - 2\cos\theta = 2(1 - \cos\theta)\)
Using the trigonometric identity \(1 - \cos\theta = 2\sin^2\frac{\theta}{2}\):
\(|\vec{a} - \vec{b}|^2 = 2 \left( 2\sin^2\frac{\theta}{2} \right) = 4\sin^2\frac{\theta}{2}\)
Taking the square root on both sides:
\(|\vec{a} - \vec{b}| = 2\sin\frac{\theta}{2}\)
Dividing by 2:
\(\sin\frac{\theta}{2} = \frac{1}{2}|\vec{a} - \vec{b}|\)
Hence proved.
In simple words: Expand the squared difference formula using unit magnitudes. Convert the cosine term using a half-angle identity, then take the square root.

Exam Tip: Memorize the standard half-angle identities \(1 - \cos\theta = 2\sin^2(\theta/2)\) and \(1 + \cos\theta = 2\cos^2(\theta/2)\) as they are frequently used in vector derivations.

Question 2. If \(|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|\) , then find the angle between \(\vec{a}\) and \(\vec{b}\).
Answer: Square both sides of the given equation:
\(|\vec{a} + \vec{b}|^2 = |\vec{a} - \vec{b}|^2\)
\(|\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b})\)
Subtracting \(|\vec{a}|^2 + |\vec{b}|^2\) from both sides:
\(2(\vec{a} \cdot \vec{b}) = -2(\vec{a} \cdot \vec{b})\)
\(4(\vec{a} \cdot \vec{b}) = 0 \implies \vec{a} \cdot \vec{b} = 0\)
Since the dot product of the two vectors is zero, they are perpendicular to each other.
Therefore, the angle between \(\vec{a}\) and \(\vec{b}\) is \(90^\circ\) (or \(\frac{\pi}{2}\)).
In simple words: Squaring the equation shows that adding or subtracting the vectors results in the same length only if they are at a right angle (\(90\) degrees) to each other.

Exam Tip: This geometric result means that the diagonals of a parallelogram are equal in length if and only if the parallelogram is a rectangle.

Question 3. For what values of \(\lambda\), vectors \(\vec{a} = 3\hat{i} - 2\hat{j} + 4\hat{k}\) and \(\vec{a} = \lambda\hat{i} - 4\hat{j} + 8\hat{k}\) are (i) Orthogonal (ii) Parallel.
Answer: Let the second vector be denoted as \(\vec{b} = \lambda\hat{i} - 4\hat{j} + 8\hat{k}\).
(i) For the vectors to be orthogonal (perpendicular), their dot product must be zero:
\(\vec{a} \cdot \vec{b} = 0 \implies (3)(\lambda) + (-2)(-4) + (4)(8) = 0\)
\(3\lambda + 8 + 32 = 0 \implies 3\lambda + 40 = 0 \implies \lambda = -\frac{40}{3}\)
(ii) For the vectors to be parallel, their component ratios must be equal:
\(\frac{3}{\lambda} = \frac{-2}{-4} = \frac{4}{8} \implies \frac{3}{\lambda} = \frac{1}{2} \implies \lambda = 6\)
In simple words: For perpendicular vectors, the dot product equation gives \(\lambda = -40/3\). For parallel vectors, comparing the ratios of components gives \(\lambda = 6\).

Exam Tip: This question tests two separate vector relationships. Keep the methods distinct - use dot product for orthogonality and component ratios for parallel vectors.

Question 4. Find \(|\vec{x}|\), if for a unit vector \(\vec{a}\) , \((\vec{x} - \vec{a})\cdot(\vec{x} + \vec{a}) = 15\).
Answer: Expand the given dot product relation:
\((\vec{x} - \vec{a})\cdot(\vec{x} + \vec{a}) = 15\)
\(\vec{x}\cdot\vec{x} + \vec{x}\cdot\vec{a} - \vec{a}\cdot\vec{x} - \vec{a}\cdot\vec{a} = 15\)
Since the dot product is commutative (\(\vec{x}\cdot\vec{a} = \vec{a}\cdot\vec{x}\)):
\(|\vec{x}|^2 - |\vec{a}|^2 = 15\)
Since \(\vec{a}\) is a unit vector, \(|\vec{a}| = 1\):
\(|\vec{x}|^2 - 1 = 15 \implies |\vec{x}|^2 = 16\)
Since magnitude is always non-negative:
\(|\vec{x}| = 4\)
In simple words: The product acts like the algebraic difference of squares: \(|\vec{x}|^2 - |\vec{a}|^2 = 15\). Since \(\vec{a}\) has a length of 1, adding it to 15 gives 16, whose square root is 4.

Exam Tip: Remember to state that \(|\vec{a}| = 1\) explicitly to justify the substitution during the solution process.

Question 5. If \(\vec{a} = 5\hat{i} - \hat{j} + 7\hat{k}\) and \(\vec{b} = \hat{i} - \hat{j} + \mu\hat{k}\), find \(\mu\), such that \(\vec{a} + \vec{b}\) and \(\vec{a} - \vec{b}\) are orthogonal.
Answer: First, find the vector sum and difference:
\(\vec{a} + \vec{b} = (5 + 1)\hat{i} + (-1 - 1)\hat{j} + (7 + \mu)\hat{k} = 6\hat{i} - 2\hat{j} + (7 + \mu)\hat{k}\)
\(\vec{a} - \vec{b} = (5 - 1)\hat{i} + (-1 - (-1))\hat{j} + (7 - \mu)\hat{k} = 4\hat{i} + (7 - \mu)\hat{k}\)
Since the two resulting vectors are orthogonal, their dot product is zero:
\((\vec{a} + \vec{b})\cdot(\vec{a} - \vec{b}) = 0\)
\((6)(4) + (-2)(0) + (7 + \mu)(7 - \mu) = 0\)
\(24 + (49 - \mu^2) = 0\)
\(73 - \mu^2 = 0 \implies \mu = \pm\sqrt{73}\)
In simple words: Add and subtract the vectors to form two new ones. For them to be perpendicular, set their dot product to 0 and solve to find \(\mu = \pm\sqrt{73}\).

Exam Tip: Write both positive and negative values for \(\mu\) since squaring removes the sign information.

Question 6. Show that the vector \(2\hat{i} - \hat{j} + \hat{k}\) , \(-3\hat{j} - 5\hat{k}\) and \(3\hat{i} - 4\hat{j} - 4\hat{k}\) form sides of a right angled triangle.
Answer: Let the three vectors representing the side paths of the triangle be:
\(\vec{A} = 2\hat{i} - \hat{j} + \hat{k}\)
\(\vec{B} = \hat{i} - 3\hat{j} - 5\hat{k}\) (taking the standard textbook vector correction)
\(\vec{C} = 3\hat{i} - 4\hat{j} - 4\hat{k}\)
Notice that \(\vec{A} + \vec{B} = (2\hat{i} - \hat{j} + \hat{k}) + (\hat{i} - 3\hat{j} - 5\hat{k}) = 3\hat{i} - 4\hat{j} - 4\hat{k} = \vec{C}\). This confirms they form a closed triangle.
Now, let us calculate the square of the lengths of these sides:
\(|\vec{A}|^2 = 2^2 + (-1)^2 + 1^2 = 4 + 1 + 1 = 6\)
\(|\vec{B}|^2 = 1^2 + (-3)^2 + (-5)^2 = 1 + 9 + 25 = 35\)
\(|\vec{C}|^2 = 3^2 + (-4)^2 + (-4)^2 = 9 + 16 + 16 = 41\)
Since \(|\vec{A}|^2 + |\vec{B}|^2 = 6 + 35 = 41 = |\vec{C}|^2\), Pythagoras' theorem is satisfied.
Hence, these vectors form the sides of a right-angled triangle.
In simple words: Find the squared lengths of all three sides. Since \(6 + 35 = 41\), the side lengths satisfy Pythagoras' theorem, proving it is a right-angled triangle.

Exam Tip: Always verify that the sum of two side vectors equals the third to confirm they form a triangle before checking Pythagoras' theorem.

Question 7. Let \(\vec{a} = \hat{i} + 4\hat{j} + 2\hat{k}\) , \(\vec{b} = 3\hat{i} - 2\hat{j} + 7\hat{k}\) and \(\vec{c} = 2\hat{i} - \hat{j} + 4\hat{k}\). Find a vector \(\vec{d}\) which is perpendicular to both \(\vec{a}\) and \(\vec{b}\) and \(\vec{c} \cdot \vec{d} = 18\).
Answer: Since \(\vec{d}\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\), it must be parallel to their cross product \(\vec{a} \times \vec{b}\).
Let us calculate \(\vec{a} \times \vec{b}\):
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & 2 \\ 3 & -2 & 7 \end{vmatrix} = \hat{i}(28 - (-4)) - \hat{j}(7 - 6) + \hat{k}(-2 - 12) = 32\hat{i} - \hat{j} - 14\hat{k} \] Thus, we can represent \(\vec{d}\) as \(\lambda (32\hat{i} - \hat{j} - 14\hat{k})\) for some scalar \(\lambda\).
We are given that \(\vec{c} \cdot \vec{d} = 18\):
\((2\hat{i} - \hat{j} + 4\hat{k}) \cdot \lambda (32\hat{i} - \hat{j} - 14\hat{k}) = 18\)
\(\lambda [ 2(32) + (-1)(-1) + 4(-14) ] = 18\)
\(\lambda [ 64 + 1 - 56 ] = 18\)
\(\lambda [ 9 ] = 18 \implies \lambda = 2\)
Substitute \(\lambda = 2\) back to find \(\vec{d}\):
\vec{d} = 2(32\hat{i} - \hat{j} - 14\hat{k}) = 64\hat{i} - 2\hat{j} - 28\hat{k}
In simple words: Find a common perpendicular vector using the cross product. Scale this vector by a factor of \(\lambda\) and use the dot product condition to solve for \(\lambda = 2\).

Exam Tip: Using cross product to establish a perpendicular direction is a highly reliable technique in 3D vector algebra.

Question 8. If \(\vec{a}\) , \(\vec{b}\) , \(\vec{c}\) are three mutually perpendicular vectors of equal magnitudes, prove that \(\vec{a} + \vec{b} + \vec{c}\) is equally inclined with the vectors \(\vec{a}\) , \(\vec{b}\) , \(\vec{c}\).
Answer: Let \(|\vec{a}| = |\vec{b}| = |\vec{c}| = k\) (where \(k \neq 0\)).
Since the vectors are mutually perpendicular, their dot products are all zero:
\(\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{c} = \vec{c}\cdot\vec{a} = 0\)
Let \(\vec{s} = \vec{a} + \vec{b} + \vec{c}\). Find the square of its magnitude:
\(|\vec{s}|^2 = (\vec{a} + \vec{b} + \vec{c})\cdot(\vec{a} + \vec{b} + \vec{c}) = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a})\)
\(|\vec{s}|^2 = k^2 + k^2 + k^2 + 0 = 3k^2 \implies |\vec{s}| = \sqrt{3}k\)
Let \(\alpha, \beta, \gamma\) be the angles that \(\vec{s}\) makes with \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) respectively:
\(\cos\alpha = \frac{\vec{s}\cdot\vec{a}}{|\vec{s}||\vec{a}|} = \frac{(\vec{a} + \vec{b} + \vec{c})\cdot\vec{a}}{\sqrt{3}k \cdot k} = \frac{|\vec{a}|^2 + 0 + 0}{\sqrt{3}k^2} = \frac{k^2}{\sqrt{3}k^2} = \frac{1}{\sqrt{3}}\)
Similarly, calculate \(\cos\beta\) and \(\cos\gamma\):
\(\cos\beta = \frac{\vec{s}\cdot\vec{b}}{|\vec{s}||\vec{b}|} = \frac{|\vec{b}|^2}{\sqrt{3}k^2} = \frac{1}{\sqrt{3}}\)
\(\cos\gamma = \frac{\vec{s}\cdot\vec{c}}{|\vec{s}||\vec{c}|} = \frac{|\vec{c}|^2}{\sqrt{3}k^2} = \frac{1}{\sqrt{3}}\)
Since \(\cos\alpha = \cos\beta = \cos\gamma = \frac{1}{\sqrt{3}}\), the angles are equal: \(\alpha = \beta = \gamma\).
Thus, the vector \(\vec{a} + \vec{b} + \vec{c}\) is equally inclined to \(\vec{a}, \vec{b}, \vec{c}\).
In simple words: Find the length of the combined vector first. Then, calculate the angle it makes with each individual vector. Since all three cosines are equal to \(1/\sqrt{3}\), the angles are also equal.

Exam Tip: Showing that the cosines of the angles are identical is sufficient to prove that the angles themselves are equal.

Question 9. Let \(\vec{a}\) , \(\vec{b}\) , \(\vec{c}\) be three vectors such that \(|\vec{a}| = 3, |\vec{b}| = 4, |\vec{c}| = 5\) and each of them being perpendicular to the sum of the other two, find \(|\vec{a} + \vec{b} + \vec{c}|\).
Answer: Based on the perpendicular conditions:
1) \(\vec{a} \cdot (\vec{b} + \vec{c}) = 0 \implies \vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} = 0\)
2) \(\vec{b} \cdot (\vec{a} + \vec{c}) = 0 \implies \vec{b}\cdot\vec{a} + \vec{b}\cdot\vec{c} = 0\)
3) \(\vec{c} \cdot (\vec{a} + \vec{b}) = 0 \implies \vec{c}\cdot\vec{a} + \vec{c}\cdot\vec{b} = 0\)
Adding these three equations together:
\(2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 0 \implies \vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a} = 0\)
Now, consider the square of the magnitude of the sum:
\(|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a})\)
Substitute the given magnitudes and the zero sum of the dot products:
\(|\vec{a} + \vec{b} + \vec{c}|^2 = 3^2 + 4^2 + 5^2 + 0 = 9 + 16 + 25 = 50\)
Taking the square root of both sides:
\(|\vec{a} + \vec{b} + \vec{c}| = \sqrt{50} = 5\sqrt{2}\)
In simple words: Adding the equations together shows that the mixed dot products sum to 0. Expanding the squared length of the sum leaves only the individual squared lengths, giving a total of 50. The square root of 50 is \(5\sqrt{2}\).

Exam Tip: The algebraic simplification \(2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 0\) is the key step that simplifies this problem.

(iv) Projection of a vector

LEVEL I

Question 1. Find the projection of \(\vec{a}\) on \(\vec{b}\) if \(\vec{a} \cdot \vec{b} = 8\) and \(\vec{b} = 2\hat{i} + 6\hat{j} + 3\hat{k}\) .
Answer: The projection of vector \(\vec{a}\) on \(\vec{b}\) is given by the formula:
\(\text{Projection} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\)
First, find the magnitude of \(\vec{b}\):
\(|\vec{b}| = \sqrt{2^2 + 6^2 + 3^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7\)
Now, substitute the values:
\(\text{Projection} = \frac{8}{7}\)
In simple words: Divide the given dot product value, 8, by the total length of vector \(\vec{b}\), which is 7.

Exam Tip: Be sure to divide by the magnitude of the vector on which the projection is being taken (in this case, \(\vec{b}\)).

Question 2. Write 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 projection of \(\vec{a}\) on \(\vec{b}\) is:
\(\text{Projection} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{(1)(1) + (-1)(1)}{\sqrt{1^2 + 1^2}} = \frac{1 - 1}{\sqrt{2}} = 0\)
In simple words: Since the two vectors are perpendicular (their dot product is 0), the projection of one onto the other is exactly 0.

Exam Tip: If the dot product of two vectors is 0, their projection on each other is always 0. This is a common shortcut to remember.

Question 3. Find the angle between the vectors \(\hat{i} - 2\hat{j} + 3\hat{k}\) and \(3\hat{i} - 2\hat{j} + \hat{k}\).
Answer: Let \(\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}\) and \(\vec{b} = 3\hat{i} - 2\hat{j} + \hat{k}\).
Find the magnitudes and dot product:
\(|\vec{a}| = \sqrt{1^2 + (-2)^2 + 3^2} = \sqrt{14}\)
\(|\vec{b}| = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{14}\)
\(\vec{a} \cdot \vec{b} = (1)(3) + (-2)(-2) + (3)(1) = 3 + 4 + 3 = 10\)
Now, find the angle \(\theta\):
\(\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{10}{\sqrt{14}\sqrt{14}} = \frac{10}{14} = \frac{5}{7}\)
\(\theta = \cos^{-1}\left(\frac{5}{7}\right)\)
In simple words: Calculate the dot product (10) and divide it by the product of their lengths (14) to find the cosine. This gives an angle of \(\cos^{-1}(5/7)\).

Exam Tip: If the final cosine value is not a standard angle, write your answer clearly in inverse cosine notation.

Question 4. Find the projection of the vector \(\hat{i} + 3\hat{j} + 7\hat{k}\) on the vector \(7\hat{i} - \hat{j} + 8\hat{k}\).
Answer: Let \(\vec{a} = \hat{i} + 3\hat{j} + 7\hat{k}\) and \(\vec{b} = 7\hat{i} - \hat{j} + 8\hat{k}\).
Calculate the dot product of the two vectors:
\(\vec{a} \cdot \vec{b} = (1)(7) + (3)(-1) + (7)(8) = 7 - 3 + 56 = 60\)
Calculate the magnitude of \(\vec{b}\):
\(|\vec{b}| = \sqrt{7^2 + (-1)^2 + 8^2} = \sqrt{49 + 1 + 64} = \sqrt{114}\)
The projection of \(\vec{a}\) on \(\vec{b}\) is:
\(\text{Projection} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{60}{\sqrt{114}}\)
In simple words: Multiply the corresponding components to get a dot product of 60. Divide this by the total length of the target vector, which is \(\sqrt{114}\).

Exam Tip: Ensure you do not confuse which vector's magnitude goes in the denominator; it is always the vector being projected *onto*.

LEVEL II

Question 1. Three vertices of a triangle are A(0, -1, -2), B(3,1,4) and C(5,7,1). Show that it is a right angled triangle. Also find the other two angles.
Answer: The position vectors are \(\vec{OA} = -\hat{j} - 2\hat{k}\), \(\vec{OB} = 3\hat{i} + \hat{j} + 4\hat{k}\), and \(\vec{OC} = 5\hat{i} + 7\hat{j} + \hat{k}\).
The vectors representing the sides of the triangle are:
\(\vec{AB} = \vec{OB} - \vec{OA} = 3\hat{i} + 2\hat{j} + 6\hat{k}\)
\(\vec{BC} = \vec{OC} - \vec{OB} = 2\hat{i} + 6\hat{j} - 3\hat{k}\)
\(\vec{CA} = \vec{OA} - \vec{OC} = -5\hat{i} - 8\hat{j} - 3\hat{k}\)
Now, find the squared lengths of these side vectors:
\(|\vec{AB}|^2 = 3^2 + 2^2 + 6^2 = 9 + 4 + 36 = 49\)
\(|\vec{BC}|^2 = 2^2 + 6^2 + (-3)^2 = 4 + 36 + 9 = 49\)
\(|\vec{CA}|^2 = (-5)^2 + (-8)^2 + (-3)^2 = 25 + 64 + 9 = 98\)
Since \(|\vec{AB}|^2 + |\vec{BC}|^2 = 49 + 49 = 98 = |\vec{CA}|^2\), this satisfies Pythagoras' Theorem. Thus, ABC is a right-angled triangle, right-angled at B (\(\angle B = 90^\circ\)).
Also, since \(|\vec{AB}| = |\vec{BC}| = 7\), the triangle is an isosceles right-angled triangle.
Therefore, the other two angles are equal:
\(\angle A = \angle C = 45^\circ\) (or \(\frac{\pi}{4}\))
In simple words: Find the squared lengths of the three sides. Since \(49 + 49 = 98\), it is a right-angled triangle. Because two sides are equal, the other two angles must both be \(45\) degrees.

Exam Tip: Showing that the sum of the squares of two side lengths equals the square of the third is a standard and elegant way to prove a right-angled triangle.

Question 2. Show that the angle between any two diagonals of a cube is \(\cos^{-1}\left(\frac{1}{3}\right)\).
Answer: Let us consider a cube of side length \(a\) with one vertex at the origin \(O(0, 0, 0)\) and three edges along the coordinate axes.
The diagonals of this cube can be represented by vectors. Let us choose two of the diagonals:
\(\vec{d_1} = a\hat{i} + a\hat{j} + a\hat{k}\)
\(\vec{d_2} = -a\hat{i} + a\hat{j} + a\hat{k}\)
The angle \(\theta\) between these two diagonals is given by:
\(\cos\theta = \frac{\vec{d_1} \cdot \vec{d_2}}{|\vec{d_1}||\vec{d_2}|}\)
Calculate the dot product and magnitudes:
\(\vec{d_1} \cdot \vec{d_2} = a(-a) + a(a) + a(a) = a^2\)
\(|\vec{d_1}| = \sqrt{a^2 + a^2 + a^2} = \sqrt{3}a\)
\(|\vec{d_2}| = \sqrt{(-a)^2 + a^2 + a^2} = \sqrt{3}a\)
Substitute these into the cosine formula:
\(\cos\theta = \frac{a^2}{(\sqrt{3}a)(\sqrt{3}a)} = \frac{a^2}{3a^2} = \frac{1}{3}\)
\(\theta = \cos^{-1}\left(\frac{1}{3}\right)\)
Hence proved.
In simple words: Represent the diagonals of a cube as vectors with components \(a\). Calculate their dot product and lengths to find that the cosine of the angle between them is always \(1/3\).

Exam Tip: Choosing a side length of \(a\) (or simply \(1\)) simplifies the vector math significantly and is a completely rigorous way to handle proofs involving cubes.

Question 3. If \(\vec{a}\) , \(\vec{b}\) , \(\vec{c}\) are non - zero and non – coplanar vectors, prove that \(\vec{a} – 2\vec{b} + 3\vec{c}\) , \(– 3\vec{b} + 5\vec{c}\) and \(– 2\vec{a} + 3\vec{b} – 4\vec{c}\) are also coplanar.
Answer: Let us write the three vectors as: \(\vec{u} = \vec{a} - 2\vec{b} + 3\vec{c}\) \(\vec{v} = 0\vec{a} - 3\vec{b} + 5\vec{c}\) \(\vec{w} = -2\vec{a} + 3\vec{b} - 4\vec{c}\) The coplanarity of these vectors depends on their scalar triple product, which can be evaluated using the determinant of their coefficients: \[ D = \begin{vmatrix} 1 & -2 & 3 \\ 0 & -3 & 5 \\ -2 & 3 & -4 \end{vmatrix} \] Expanding the determinant along the first column: \[ D = 1 \left[ (-3)(-4) - (5)(3) \right] - 0 + (-2) \left[ (-2)(5) - (3)(-3) \right] \] \(D = 1(12 - 15) - 2(-10 + 9) = -3 - 2(-1) = -3 + 2 = -1\) Since the determinant is non-zero (\(D = -1 \neq 0\)), these vectors are linearly independent. Therefore, the three vectors are non-coplanar.
In simple words: Put the coefficients of the three vector combinations into a matrix and compute the determinant. Since the determinant is \(-1\) (not zero), the vectors do not lie in the same flat plane.

Exam Tip: If a test question asks you to prove coplanarity but the determinant is non-zero, write out your calculations clearly to show the scalar triple product is non-zero.

LEVEL III

Question 1. If a unit vector \(\vec{a}\) makes angles \(\pi /4\), with \(\hat{i}\), \(\pi/3\) with \(\hat{j}\) and an acute angle \(\theta\) with \(\hat{k}\) , then find the component of \(\vec{a}\) and angle \(\theta\) .
Answer: The direction angles are \(\alpha = \frac{\pi}{4}\), \(\beta = \frac{\pi}{3}\), and \(\gamma = \theta\). The direction cosines are: \(l = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\) \(m = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\) \(n = \cos\theta\) Using the direction cosine identity \(l^2 + m^2 + n^2 = 1\): \(\left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{2}\right)^2 + \cos^2\theta = 1\) \(\frac{1}{2} + \frac{1}{4} + \cos^2\theta = 1 \implies \frac{3}{4} + \cos^2\theta = 1\) \(\cos^2\theta = \frac{1}{4} \implies \cos\theta = \pm \frac{1}{2}\) Since \(\theta\) is given as an acute angle, \(\cos\theta\) must be positive: \(\cos\theta = \frac{1}{2} \implies \theta = \frac{\pi}{3}\) (or \(60^\circ\)) The components of the unit vector \(\vec{a}\) are its direction cosines: \(\vec{a} = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{2}\hat{j} + \frac{1}{2}\hat{k}\) Thus, the components are \(\left( \frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2} \right)\) and the angle \(\theta = \frac{\pi}{3}\).
In simple words: Use the direction cosine formula where the sum of squared values is 1. Solve to find the third angle cosine is \(1/2\), which means the angle is \(60\) degrees and the components are the cosines of these three angles.

Exam Tip: Since \(\vec{a}\) is a unit vector, its components along the axes are exactly equal to its direction cosines \(l\), \(m\), and \(n\).

Question 2. If \(\vec{a}\) , \(\vec{b}\) , \(\vec{c}\) are three mutually perpendicular vectors of equal magnitudes, prove that \(\vec{a} + \vec{b} + \vec{c}\) is equally inclined with the vectors \(\vec{a}\) , \(\vec{b}\) , \(\vec{c}\).
Answer: Let \(|a| = |b| = |c| = k\) with \(k \neq 0\). Since the vectors are mutually perpendicular, their dot products are all zero: \(\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{c} = \vec{c}\cdot\vec{a} = 0\). Let \(\vec{r} = \vec{a} + \vec{b} + \vec{c}\). The square of its magnitude is: \(|\vec{r}|^2 = (\vec{a} + \vec{b} + \vec{c})\cdot(\vec{a} + \vec{b} + \vec{c}) = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 3k^2 \implies |\vec{r}| = \sqrt{3}k\). Now, find the angles \(\alpha, \beta, \gamma\) that \(\vec{r}\) makes with \(\vec{a}, \vec{b}, \vec{c}\): \(\cos\alpha = \frac{\vec{r}\cdot\vec{a}}{|\vec{r}||\vec{a}|} = \frac{(\vec{a}+\vec{b}+\vec{c})\cdot\vec{a}}{\sqrt{3}k \cdot k} = \frac{|\vec{a}|^2 + 0 + 0}{\sqrt{3}k^2} = \frac{k^2}{\sqrt{3}k^2} = \frac{1}{\sqrt{3}}\) \(\cos\beta = \frac{\vec{r}\cdot\vec{b}}{|\vec{r}||\vec{b}|} = \frac{|\vec{b}|^2}{\sqrt{3}k^2} = \frac{1}{\sqrt{3}}\) \(\cos\gamma = \frac{\vec{r}\cdot\vec{c}}{|\vec{r}||\vec{c}|} = \frac{|\vec{c}|^2}{\sqrt{3}k^2} = \frac{1}{\sqrt{3}}\) Since \(\cos\alpha = \cos\beta = \cos\gamma = \frac{1}{\sqrt{3}}\), the angles are equal. Therefore, the vector \(\vec{a} + \vec{b} + \vec{c}\) is equally inclined to the three vectors. In simple words: Find the length of the total vector. Calculate the cosine of the angle with each individual vector. Since they all come out as \(1/\sqrt{3}\), the angles are all equal.

Exam Tip: This is a classic vector proof. Start by writing down the given perpendicular conditions as dot products equal to zero to lay a solid foundation for your proof.

Question 3. If with reference to the right handed system of mutually perpendicular unit vectors \(\hat{i}, \hat{j}\),and \(\hat{k}\), \(\vec{\alpha} = 3\hat{i} - \hat{j}\), \(\vec{\beta} = 2\hat{i} + \hat{j} - 3\hat{k}\) then express \(\vec{\beta}\) in the form of \(\vec{\beta}_1 + \vec{\beta}_2\), 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: \(\vec{\beta}_1 = \lambda \vec{\alpha} = \lambda(3\hat{i} - \hat{j}) = 3\lambda\hat{i} - \lambda\hat{j}\) We are given \(\vec{\beta} = \vec{\beta}_1 + \vec{\beta}_2\), so: \(\vec{\beta}_2 = \vec{\beta} - \vec{\beta}_1 = (2\hat{i} + \hat{j} - 3\hat{k}) - (3\lambda\hat{i} - \lambda\hat{j}) = (2 - 3\lambda)\hat{i} + (1 + \lambda)\hat{j} - 3\hat{k}\) Since \(\vec{\beta}_2\) is perpendicular to \(\vec{\alpha}\), their dot product must be zero: \(\vec{\beta}_2 \cdot \vec{\alpha} = 0 \implies ((2 - 3\lambda)\hat{i} + (1 + \lambda)\hat{j} - 3\hat{k}) \cdot (3\hat{i} - \hat{j}) = 0\) \(3(2 - 3\lambda) - 1(1 + \lambda) = 0 \implies 6 - 9\lambda - 1 - \lambda = 0\) \(5 - 10\lambda = 0 \implies \lambda = \frac{1}{2}\) Now substitute \(\lambda = \frac{1}{2}\) to find \(\vec{\beta}_1\) and \(\vec{\beta}_2\): \(\vec{\beta}_1 = \frac{3}{2}\hat{i} - \frac{1}{2}\hat{j}\) \(\vec{\beta}_2 = \left(2 - \frac{3}{2}\right)\hat{i} + \left(1 + \frac{1}{2}\right)\hat{j} - 3\hat{k} = \frac{1}{2}\hat{i} + \frac{3}{2}\hat{j} - 3\hat{k}\) Thus, \(\vec{\beta} = \left(\frac{3}{2}\hat{i} - \frac{1}{2}\hat{j}\right) + \left(\frac{1}{2}\hat{i} + \frac{3}{2}\hat{j} - 3\hat{k}\right)\). In simple words: Write \(\vec{\beta}_1\) as a multiple of \(\vec{\alpha}\). Use subtraction to find \(\vec{\beta}_2\) in terms of \(\lambda\). Set their dot product to 0 to find \(\lambda = 1/2\), and plug it back in.

Exam Tip: This problem can also be solved by projecting \(\vec{\beta}\) along \(\vec{\alpha}\) to get \(\vec{\beta}_1\) directly. Choose whichever method you find more intuitive.

Question 4. Show that the points A, B, C with position vectors \(\vec{a} = 3\hat{i} - 4\hat{j} - 4\hat{k}\) , \(\vec{b} = 2\hat{i} - \hat{j} + \hat{k}\) and \(\vec{c} = \hat{i} - 3\hat{j} - 5\hat{k}\) respectively form the vertices of a right angled triangle.
Answer: Let us find the side vectors of the triangle: \(\vec{AB} = \vec{b} - \vec{a} = (2\hat{i} - \hat{j} + \hat{k}) - (3\hat{i} - 4\hat{j} - 4\hat{k}) = -\hat{i} + 3\hat{j} + 5\hat{k}\) \(\vec{BC} = \vec{c} - \vec{b} = (\hat{i} - 3\hat{j} - 5\hat{k}) - (2\hat{i} - \hat{j} + \hat{k}) = -\hat{i} - 2\hat{j} - 6\hat{k}\) \(\vec{CA} = \vec{a} - \vec{c} = (3\hat{i} - 4\hat{j} - 4\hat{k}) - (\hat{i} - 3\hat{j} - 5\hat{k}) = 2\hat{i} - \hat{j} + \hat{k}\) Now, find the squared magnitudes of these side vectors: \(|\vec{AB}|^2 = (-1)^2 + 3^2 + 5^2 = 1 + 9 + 25 = 35\) \(|\vec{BC}|^2 = (-1)^2 + (-2)^2 + (-6)^2 = 1 + 4 + 36 = 41\) \(|\vec{CA}|^2 = 2^2 + (-1)^2 + 1^2 = 4 + 1 + 1 = 6\) Since \(|\vec{AB}|^2 + |\vec{CA}|^2 = 35 + 6 = 41 = |\vec{BC}|^2\), this satisfies Pythagoras' Theorem. Therefore, the points A, B, and C form the vertices of a right-angled triangle. In simple words: Find the side vectors by subtracting the position vectors. Work out their squared lengths. Since \(35 + 6 = 41\), the side lengths satisfy Pythagoras' theorem.

Exam Tip: Be methodical when subtracting coordinates to find the side vectors. One small arithmetic error can prevent Pythagoras' theorem from working out.

(vii) Cross product of two vectors

LEVEL I

Question 1. If \(|\vec{a}| = 3\) , \(|\vec{b}| = 5\) and \(\vec{a} \cdot \vec{b} = 9\). Find \(|\vec{a} \times \vec{b}|\).
Answer: We use Lagrange's Identity, which relates the dot product and cross product of two vectors:
\(|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2\)
Substitute the given values into the identity:
\(|\vec{a} \times \vec{b}|^2 + 9^2 = 3^2 \times 5^2\)
\(|\vec{a} \times \vec{b}|^2 + 81 = 9 \times 25 = 225\)
\(|\vec{a} \times \vec{b}|^2 = 225 - 81 = 144\)
Taking the square root:
\(|\vec{a} \times \vec{b}| = \sqrt{144} = 12\)
In simple words: Use the formula that links the dot product and cross product. Subtract the squared dot product (81) from the product of squared lengths (225) to get 144. The square root is 12.

Exam Tip: Lagrange's Identity is an extremely helpful formula that bypasses the need to find the angle \(\theta\) between the vectors first.

Question 2. Find \(|\vec{a} \times \vec{b}|\), if \(\vec{a} = \hat{i} - 7\hat{j} + 7\hat{k}\) and \(\vec{b} = 3\hat{i} + 2\hat{j} + 2\hat{k}\).
Answer: First, find the cross product of the two vectors using a determinant:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -7 & 7 \\ 3 & 2 & 2 \end{vmatrix} \] \[ \vec{a} \times \vec{b} = \hat{i}((-7)(2) - (7)(2)) - \hat{j}((1)(2) - (7)(3)) + \hat{k}((1)(2) - (-7)(3)) \] \(\vec{a} \times \vec{b} = -28\hat{i} + 19\hat{j} + 23\hat{k}\)
Now, find the magnitude of this cross product vector:
\(|\vec{a} \times \vec{b}| = \sqrt{(-28)^2 + 19^2 + 23^2} = \sqrt{784 + 361 + 529} = \sqrt{1674}\)
In simple words: Set up a 3 by 3 matrix to calculate the cross product vector. Then, calculate its magnitude using the square root of the sum of its squared parts.

Exam Tip: Take extra care when expanding the middle term (\(-\hat{j}\)) of the determinant, as sign errors are very common here.

Question 3. Find \(|\vec{x}|\), if \(\vec{p}\) is a unit vector and \( (\vec{x} - \vec{p})\cdot(\vec{x} + \vec{p}) = 80 \).
Answer: Expand the given dot product relation:
\((\vec{x} - \vec{p})\cdot(\vec{x} + \vec{p}) = 80 \implies |\vec{x}|^2 - |\vec{p}|^2 = 80\)
Since \(\vec{p}\) is a unit vector, \(|\vec{p}| = 1\):
\(|\vec{x}|^2 - 1 = 80 \implies |\vec{x}|^2 = 81\)
Taking the square root of both sides:
\(|\vec{x}| = 9\)
In simple words: Expand the equation to get \(|\vec{x}|^2 - 1 = 80\). Solving this gives \(|\vec{x}|^2 = 81\), so the magnitude is 9.

Exam Tip: This is a very clean question that tests vector expansion and the definition of a unit vector simultaneously.

Question 4. Find p, if \((2\hat{i} + 6\hat{j} + 27\hat{k}) \times (\hat{i} + 3\hat{j} + p\hat{k}) = \vec{0}\).
Answer: Since the cross product of two non-zero vectors is the zero vector, they must be parallel.
Therefore, their corresponding components are proportional:
\(\frac{2}{1} = \frac{6}{3} = \frac{27}{p}\)
\(2 = \frac{27}{p} \implies p = \frac{27}{2} = 13.5\)
In simple words: When the cross product is zero, the vectors are parallel. This means their parts are in the same ratio. The first vector is twice as big as the second, so \(2p = 27\), which means \(p = 13.5\).

Exam Tip: Using component ratios for parallel vectors is much faster than expanding the cross product determinant and setting it to zero.

LEVEL II

Question 1. Find \(\lambda\), if \((2\hat{i} + 6\hat{j} + 14\hat{k}) \times (\hat{i} - \lambda\hat{j} + 7\hat{k}) = \vec{0}\).
Answer: Since the cross product of these two vectors is the zero vector, they must be parallel.
For parallel vectors, the components are proportional:
\(\frac{2}{1} = \frac{6}{-\lambda} = \frac{14}{7}\)
From this proportional relation, we have:
\(2 = -\frac{6}{\lambda} \implies 2\lambda = -6 \implies \lambda = -3\)
In simple words: The cross product is zero because the vectors are parallel. Setting up the ratio of components gives \(2 = -6/\lambda\), which means \(\lambda = -3\).

Exam Tip: Remember to include the negative sign of the component (\(-\lambda\)) when setting up the proportion ratios.

Question 2. Show that \((\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = 2(\vec{a} \times \vec{b})\).
Answer: Expand the left-hand side of the equation using the distributive property of the cross product:
\((\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = \vec{a} \times \vec{a} + \vec{a} \times \vec{b} - \vec{b} \times \vec{a} - \vec{b} \times \vec{b}\)
We know that the cross product of a vector with itself is the zero vector:
\(\vec{a} \times \vec{a} = \vec{0}\) and \(\vec{b} \times \vec{b} = \vec{0}\)
Also, the cross product is anti-commutative (\(-\vec{b} \times \vec{a} = \vec{a} \times \vec{b}\)):
\(\text{L.H.S.} = \vec{0} + \vec{a} \times \vec{b} + \vec{a} \times \vec{b} - \vec{0} = 2(\vec{a} \times \vec{b}) = \text{R.H.S.}\)
Hence proved.
In simple words: Multiply out the terms. A vector cross product with itself is 0, and switching the order of a cross product flips its sign, leaving us with exactly two of \(\vec{a} \times \vec{b}\).

Exam Tip: Always state the vector identities \(\vec{a}\times\vec{a}=\vec{0}\) and \(\vec{b}\times\vec{a} = -\vec{a}\times\vec{b}\) during the steps to get full marks.

Question 3. Find the angle between two vectors \(\vec{a}\) and \(\vec{b}\) if \(|\vec{a}| = 3, |\vec{b}| = 4\) and \(|\vec{a} \times \vec{b}| = 6\).
Answer: The magnitude of the cross product of two vectors is given by:
\(|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta\)
Substitute the given values into the formula:
\(6 = (3)(4)\sin\theta \implies 6 = 12\sin\theta\)
\(\sin\theta = \frac{6}{12} = \frac{1}{2}\)
Thus, the angle is:
\(\theta = \frac{\pi}{6}\) (or \(30^\circ\)) or \(\theta = \frac{5\pi}{6}\) (or \(150^\circ\))
In simple words: Use the cross product formula to find that the sine of the angle is \(1/2\). The angles that satisfy this are \(30\) degrees and \(150\) degrees.

Exam Tip: Since sine is positive in both the first and second quadrants, always write both possible angle values (\(30^\circ\) and \(150^\circ\)) unless restricted by the question.

Question 4. Let \(\vec{a}\) , \(\vec{b}\) , \(\vec{c}\) be unit vectors such that \(\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} = 0\) and the angle between \(\vec{b}\) and \(\vec{c}\) is \(\pi /6\), prove that \(\vec{a} = \pm 2( \vec{b} \times \vec{c})\).
Answer: Since \(\vec{a} \cdot \vec{b} = 0\) and \(\vec{a} \cdot \vec{c} = 0\), \(\vec{a}\) is perpendicular to both \(\vec{b}\) and \(\vec{c}\). Therefore, \(\vec{a}\) is parallel to \(\vec{b} \times \vec{c}\). We can express this as: \(\vec{a} = \lambda (\vec{b} \times \vec{c})\) for some scalar \(\lambda\). Taking the magnitude on both sides: \(|\vec{a}| = |\lambda| |\vec{b} \times \vec{c}|\) Since \(\vec{a}\) is a unit vector, \(|\vec{a}| = 1\): \(1 = |\lambda| |\vec{b}| |\vec{c}| \sin\theta\) Since \(\vec{b}\) and \(\vec{c}\) are unit vectors, and the angle between them is \(\theta = \frac{\pi}{6}\): \(1 = |\lambda| (1)(1)\sin\left(\frac{\pi}{6}\right) \implies 1 = |\lambda| \cdot \frac{1}{2} \implies |\lambda| = 2\) Therefore, \(\lambda = \pm 2\). Substituting this value of \(\lambda\) back: \(\vec{a} = \pm 2(\vec{b} \times \vec{c})\) Hence proved.
In simple words: Since \(\vec{a}\) is perpendicular to both \(\vec{b}\) and \(\vec{c}\), it must point along their cross product. Calculating the length of their cross product gives \(1/2\), so we must multiply it by \(\pm 2\) to match the unit length of \(\vec{a}\).

Exam Tip: This question beautifully combines the concept of vector direction (orthogonality to two vectors) and magnitude scaling.

LEVEL III

Question 1. Find the value of the following: \(\hat{i}\cdot(\hat{j} \times \hat{k}) + \hat{j}\cdot(\hat{i} \times \hat{k}) + \hat{k}\cdot(\hat{i} \times \hat{j})\).
Answer: Recall the standard cross products of the unit coordinate vectors:
\(\hat{j} \times \hat{k} = \hat{i}\)
\(\hat{i} \times \hat{k} = -\hat{j}\)
\(\hat{i} \times \hat{j} = \hat{k}\)
Substitute these back into the expression:
\(\hat{i}\cdot(\hat{i}) + \hat{j}\cdot(-\hat{j}) + \hat{k}\cdot(\hat{k})\)
Since the dot product of any unit coordinate vector with itself is 1:
\(1 - 1 + 1 = 1\)
Therefore, the value is 1.
In simple words: Solve the cross products inside the parentheses first, which turn into coordinate unit vectors. Then perform the dot products to get \(1 - 1 + 1 = 1\).

Exam Tip: Pay close attention to the cyclic order of cross products; reversing the standard order introduces a negative sign (e.g., \(\hat{i} \times \hat{k} = -\hat{j}\)).

Question 2. Vectors \(\vec{a}\) and \(\vec{b}\) are such that \(|\vec{a}| = \sqrt{3}\) , \(|\vec{b}| = \frac{2}{3}\) , and \(\vec{a} \times \vec{b}\) is a unit vector. Write the angle between \(\vec{a}\) and \(\vec{b}\).
Answer: Since \(\vec{a} \times \vec{b}\) is a unit vector, its magnitude is 1:
\(|\vec{a} \times \vec{b}| = 1\)
Using the cross product magnitude formula:
\(|\vec{a}||\vec{b}|\sin\theta = 1\)
Substitute the given magnitudes:
\(\sqrt{3} \cdot \frac{2}{3} \cdot \sin\theta = 1 \implies \frac{2}{\sqrt{3}}\sin\theta = 1\)
\(\sin\theta = \frac{\sqrt{3}}{2}\)
The angle is:
\(\theta = \frac{\pi}{3}\) (or \(60^\circ\)) or \(\theta = \frac{2\pi}{3}\) (or \(120^\circ\))
In simple words: The cross product length formula gives \(\sin\theta = \sqrt{3}/2\). The angles with this sine value are \(60\) degrees and \(120\) degrees.

Exam Tip: Always check if there are multiple angles that satisfy the trigonometric equation, as sine is positive in both the first and second quadrants.

Question 3. If \(\vec{a} = \hat{i} + \hat{j} + \hat{k}\) and \(\vec{b} = \hat{j} - \hat{k}\) , find a vector \(\vec{c}\) such that \(\vec{a} \times \vec{c} = \vec{b}\) and \(\vec{a} \cdot \vec{c} = 3\).
Answer: Let the required vector be \(\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}\). From the dot product condition: \(\vec{a} \cdot \vec{c} = 3 \implies x + y + z = 3\) - (1) From the cross product condition \(\vec{a} \times \vec{c} = \vec{b}\): \[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix} = \hat{j} - \hat{k} \] \((z - y)\hat{i} + (x - z)\hat{j} + (y - x)\hat{k} = \hat{j} - \hat{k}\) Comparing the components on both sides: 1) \(z - y = 0 \implies y = z\) 2) \(x - z = 1 \implies x = z + 1\) 3) \(y - x = -1 \implies x = y + 1\) (which is consistent with \(y = z\)) Substitute \(x = z + 1\) and \(y = z\) into equation (1): \((z + 1) + z + z = 3 \implies 3z + 1 = 3 \implies 3z = 2 \implies z = \frac{2}{3}\) Thus, \(y = \frac{2}{3}\) and \(x = \frac{2}{3} + 1 = \frac{5}{3}\). The required vector is: \(\vec{c} = \frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k} = \frac{1}{3}(5\hat{i} + 2\hat{j} + 2\hat{k})\)
In simple words: Write \(\vec{c}\) as \(x, y, z\). Set up the cross product and dot product equations, then solve for the variables to find \(\vec{c} = \frac{1}{3}(5\hat{i} + 2\hat{j} + 2\hat{k})\).

Exam Tip: Be consistent when equating components from the cross product. Solving the resulting linear equations carefully will ensure a correct result.

Question 4. If \(\vec{a} \times \vec{b} = \vec{c} \times \vec{d}\) and \(\vec{a} \times \vec{c} = \vec{b} \times \vec{d}\) show that \((\vec{a} - \vec{d})\) is parallel to \(\vec{b} - \vec{c}\) , where \(\vec{a} \neq \vec{d}\) and \(\vec{b} \neq \vec{c}\).
Answer: To show that \((\vec{a} - \vec{d})\) is parallel to \((\vec{b} - \vec{c})\), we must prove that their cross product is the zero vector: \((\vec{a} - \vec{d}) \times (\vec{b} - \vec{c}) = \vec{0}\) Expanding the left-hand side using the distributive law of cross products: \((\vec{a} - \vec{d}) \times (\vec{b} - \vec{c}) = \vec{a} \times \vec{b} - \vec{a} \times \vec{c} - \vec{d} \times \vec{b} + \vec{d} \times \vec{c}\) Using anti-commutativity (\(-\vec{d} \times \vec{b} = \vec{b} \times \vec{d}\) and \(\vec{d} \times \vec{c} = -\vec{c} \times \vec{d}\)): \(= \vec{a} \times \vec{b} - \vec{a} \times \vec{c} + \vec{b} \times \vec{d} - \vec{c} \times \vec{d}\) Substitute the given relations \(\vec{a} \times \vec{b} = \vec{c} \times \vec{d}\) and \(\vec{a} \times \vec{c} = \vec{b} \times \vec{d}\) into the expression: \(= (\vec{c} \times \vec{d}) - (\vec{b} \times \vec{d}) + (\vec{b} \times \vec{d}) - (\vec{c} \times \vec{d}) = \vec{0}\) Since the cross product is the zero vector and both vectors are non-zero, the vector \((\vec{a} - \vec{d})\) is parallel to \((\vec{b} - \vec{c})\). Hence proved.
In simple words: Multiply out the cross product of the two differences. Substitute the given equal cross product terms, which makes all the terms cancel out to 0, proving they are parallel.

Exam Tip: Proving that the cross product of two non-zero vectors is zero is the standard way to show they are parallel in vector algebra.

Question 5. Express \(2\hat{i} - \hat{j} + 3\hat{k}\) as the sum of a vector parallel and perpendicular to \(2\hat{i} + 4\hat{j} - 2\hat{k}\).
Answer: Let \(\vec{\beta} = 2\hat{i} - \hat{j} + 3\hat{k}\) and \(\vec{\alpha} = 2\hat{i} + 4\hat{j} - 2\hat{k}\). We can express \(\vec{\beta}\) as \(\vec{\beta}_1 + \vec{\beta}_2\), where \(\vec{\beta}_1\) is parallel to \(\vec{\alpha}\) and \(\vec{\beta}_2\) is perpendicular to \(\vec{\alpha}\). The component parallel to \(\vec{\alpha}\) is the projection vector of \(\vec{\beta}\) along \(\vec{\alpha}\): \(\vec{\beta}_1 = \left( \frac{\vec{\beta} \cdot \vec{\alpha}}{|\vec{\alpha}|^2} \right) \vec{\alpha}\) Calculate \(\vec{\beta} \cdot \vec{\alpha}\) and \(|\vec{\alpha}|^2\): \(\vec{\beta} \cdot \vec{\alpha} = (2)(2) + (-1)(4) + (3)(-2) = 4 - 4 - 6 = -6\) \(|\vec{alpha}|^2 = 2^2 + 4^2 + (-2)^2 = 4 + 16 + 4 = 24\) So, \(\vec{\beta}_1 = \frac{-6}{24} \vec{\alpha} = -\frac{1}{4}(2\hat{i} + 4\hat{j} - 2\hat{k}) = -\frac{1}{2}\hat{i} - \hat{j} + \frac{1}{2}\hat{k}\) Now, find the perpendicular component \(\vec{\beta}_2\): \(\vec{\beta}_2 = \vec{\beta} - \vec{\beta}_1 = (2\hat{i} - \hat{j} + 3\hat{k}) - \left(-\frac{1}{2}\hat{i} - \hat{j} + \frac{1}{2}\hat{k}\right) = \frac{5}{2}\hat{i} + \frac{5}{2}\hat{k}\) Thus, the vector is decomposed as: \(\vec{\beta} = \left(-\frac{1}{2}\hat{i} - \hat{j} + \frac{1}{2}\hat{k}\right) + \left(\frac{5}{2}\hat{i} + \frac{5}{2}\hat{k}\right)\). In simple words: Project the vector onto the target to find the parallel component. Subtract this from the original vector to find the perpendicular component.

Exam Tip: This decomposition is extremely useful in physics and geometry for splitting forces or motions into parallel and perpendicular directions.

(viii) Area of a triangle & Area of a parallelogram

LEVEL I

Question 1. Find the area of Parallelogram whose adjacent sides are represented by the vectors \(\vec{a} = 3\hat{i} + \hat{j} - 2\hat{k}\) and \(\vec{b} = \hat{i} - 3\hat{j} + 4\hat{k}\).
Answer: The area of a parallelogram with adjacent sides \(\vec{a}\) and \(\vec{b}\) is equal to \(|\vec{a} \times \vec{b}|\).
First, find the cross product \(\vec{a} \times \vec{b}\):
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 1 & -3 & 4 \end{vmatrix} = \hat{i}(4 - 6) - \hat{j}(12 - (-2)) + \hat{k}(-9 - 1) = -2\hat{i} - 14\hat{j} - 10\hat{k} \] Now, compute its magnitude:
\(|\vec{a} \times \vec{b}| = \sqrt{(-2)^2 + (-14)^2 + (-10)^2} = \sqrt{4 + 196 + 100} = \sqrt{300} = 10\sqrt{3}\)
Therefore, the area of the parallelogram is \(10\sqrt{3}\) square units.
In simple words: Find the cross product vector of the two adjacent sides. The total length of this cross product vector is the area of the parallelogram, which is \(10\sqrt{3}\).

Exam Tip: Always state "square units" in the final answer for any area calculation to ensure you do not lose presentation marks.

Question 2. If \(\vec{a}\) and \(\vec{b}\) represent the two adjacent sides of a Parallelogram, then write the area of parallelogram in terms of \(\vec{a}\) and \(\vec{b}\).
Answer: The area of a parallelogram whose adjacent sides are represented by vectors \(\vec{a}\) and \(\vec{b}\) is given by:
\(\text{Area} = |\vec{a} \times \vec{b}|\)
In simple words: The area is simply the magnitude of the cross product of the two side vectors.

Exam Tip: This is a fundamental formula. Remember that using diagonals instead of sides would introduce a factor of \(1/2\).

Question 3. Find the area of triangle having the points A(1,1,1) , B(1,2,3) and C(2,3,1) as its vertices.
Answer: First, find two vectors representing adjacent sides of the triangle, such as \(\vec{AB}\) and \(\vec{AC}\):
\(\vec{AB} = (1-1)\hat{i} + (2-1)\hat{j} + (3-1)\hat{k} = \hat{j} + 2\hat{k}\)
\(\vec{AC} = (2-1)\hat{i} + (3-1)\hat{j} + (1-1)\hat{k} = \hat{i} + 2\hat{j}\)
Now, calculate their cross product \(\vec{AB} \times \vec{AC}\):
\[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 2 \\ 1 & 2 & 0 \end{vmatrix} = \hat{i}(0 - 4) - \hat{j}(0 - 2) + \hat{k}(0 - 1) = -4\hat{i} + 2\hat{j} - \hat{k} \] Find the magnitude of this cross product vector:
\(|\vec{AB} \times \vec{AC}| = \sqrt{(-4)^2 + 2^2 + (-1)^2} = \sqrt{16 + 4 + 1} = \sqrt{21}\)
The area of the triangle is half of this magnitude:
\(\text{Area} = \frac{1}{2}|\vec{AB} \times \vec{AC}| = \frac{\sqrt{21}}{2}\) square units.
In simple words: Find two side vectors from point A. Work out their cross product, find its length, and divide by 2 to get the area of the triangle.

Exam Tip: Be sure to divide by 2, as the cross product magnitude alone represents the area of a parallelogram, not a triangle.

LEVEL II

Question 1. Show that the area of the Parallelogram having diagonals ( \(3\hat{i} + \hat{j} - 2\hat{k}\) ) and ( \(\hat{i} - 3\hat{j} + 4\hat{k}\) ) is \(5\sqrt{3}\) Sq units.
Answer: Let the diagonals be represented by \(\vec{d_1} = 3\hat{i} + \hat{j} - 2\hat{k}\) and \(\vec{d_2} = \hat{i} - 3\hat{j} + 4\hat{k}\).
The area of a parallelogram in terms of its diagonals is:
\(\text{Area} = \frac{1}{2}|\vec{d_1} \times \vec{d_2}|\)
First, find the cross product \(\vec{d_1} \times \vec{d_2}\):
\[ \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 1 & -3 & 4 \end{vmatrix} = -2\hat{i} - 14\hat{j} - 10\hat{k} \] Now, calculate the magnitude of this cross product:
\(|\vec{d_1} \times \vec{d_2}| = \sqrt{(-2)^2 + (-14)^2 + (-10)^2} = \sqrt{4 + 196 + 100} = \sqrt{300} = 10\sqrt{3}\)
Calculate the area:
\(\text{Area} = \frac{1}{2} (10\sqrt{3}) = 5\sqrt{3}\) square units.
Hence proved.
In simple words: Find the cross product of the two diagonals. The area of the parallelogram is half of the length of this cross product vector, which evaluates to \(5\sqrt{3}\).

Exam Tip: Remember that when adjacent sides are given, the area is \(|\vec{a

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