ML Aggarwal Class 12 Maths Solutions Section B Chapter 02 Straight Lines In Space

Introduction

In Class XI, we studied Analytical Geometry in two dimensions and touched on 3-dimensional geometry, using only the cartesian method. Now we will apply vector algebra to 3-dimensional geometry, which simplifies and refines its study. In this chapter, we will cover:

• the vector and cartesian equations of a line in various forms
• angle between two lines when their equations are known
• coplanar and skew lines
• shortest distance between two lines
• conditions for intersection of two lines
• condition for coplanarity of two lines

2.1 Straight Line (In Space)

In space, a straight line is uniquely fixed if:

• it goes through a given point and moves in a given direction, or
• it goes through two given points

We will work out the vector equation and cartesian equation of a line (in space).

2.1.1 Equation of a Straight Line Passing Through a Given Point and Parallel to a Given Vector

Suppose the line goes through the given point A with position vector \( \vec{a} \) and is parallel to the given vector \( \vec{b} \).

Let P with position vector \( \vec{r} \) be any point. Then P lies on the line if and only if \( \vec{AP} \) is parallel to the vector \( \vec{b} \); that is, if and only if

\( \vec{AP} = \lambda \vec{b} \), where \( \lambda \) is a real number (called parameter)

i.e. if and only if \( \vec{OP} - \vec{OA} = \lambda \vec{b} \)

i.e. if and only if \( \vec{r} - \vec{a} = \lambda \vec{b} \)

i.e. if and only if \( \vec{r} = \vec{a} + \lambda \vec{b} \) ... (1)

which is the required vector equation of the line.

Cartesian Form

Let the given point be A \( (x_1, y_1, z_1) \) and \( \langle a, b, c \rangle \) be the direction numbers of the line and P be \( (x, y, z) \). Then

\( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \), \( \vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k} \) and \( \vec{b} = a\hat{i} + b\hat{j} + c\hat{k} \).

Substituting these values of \( \vec{r}, \vec{a} \), and \( \vec{b} \) in (1), we get

\( x\hat{i} + y\hat{j} + z\hat{k} = (x_1 + a\lambda)\hat{i} + (y_1 + b\lambda)\hat{j} + (z_1 + c\lambda)\hat{k} \)

Equating the coefficients of the base vectors \( \hat{i}, \hat{j}, \hat{k} \), we get

\( x = x_1 + a\lambda \)
\( y = y_1 + b\lambda \)
\( z = z_1 + c\lambda \) ...(2)

These are called the parametric (or one point) form of the line. Any point on the line is P\( (x_1 + a\lambda, y_1 + b\lambda, z_1 + c\lambda) \). The point A is called the base point.

Removing \( \lambda \) from equations (2), we get

\( \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \) ...(3)

These are called the symmetrical form of the line.

2.1.2 Equation of a Straight Line Passing Through Two Given Points

Suppose the line goes through two given points A and B with position vectors \( \vec{a} \) and \( \vec{b} \) respectively.

Let P with position vector \( \vec{r} \) be any point. Then P lies on the line AB if and only if

\( \vec{AP} = \lambda \vec{AB} \), where \( \lambda \) is real number (called parameter)

i.e. if and only if \( \vec{OP} - \vec{OA} = \lambda(\vec{OB} - \vec{OA}) \)

i.e. if and only if \( \vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a}) \) ...(1)

which is the required vector equation of the line.

Cartesian Form

Let the given points be A \( (x_1, y_1, z_1) \), B \( (x_2, y_2, z_2) \) and the point P be \( (x, y, z) \). From (1), we get

\( x\hat{i} + y\hat{j} + z\hat{k} = (x_1\hat{i} + y_1\hat{j} + z_1\hat{k}) + t[(x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}] \).

Equating the coefficients of base vectors, we get

\( x = x_1 + \lambda(x_2 - x_1) \)
\( y = y_1 + \lambda(y_2 - y_1) \)
\( z = z_1 + \lambda(z_2 - z_1) \) ...(2)

These are called the parametric (or two points) form of the line. Removing \( \lambda \) from the equations (2), we get

\( \frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1} \) ...(3)

These are called the symmetrical form.

Illustrative Examples

Example 1. Find the vector equation of the line which is parallel to the vector \( 2\hat{i} - \hat{j} + 3\hat{k} \) and which passes through the point (5, -2, 4). Also find its cartesian equations.
Answer: Let \( \vec{r} \) be the position vector of any point P(x, y, z) on the given line. The vector equation of the line is \( \vec{r} = 5\hat{i} - 2\hat{j} + 4\hat{k} + \lambda(2\hat{i} - \hat{j} + 3\hat{k}) \).

Since \( \vec{r} \) is the position vector of P(x, y, z), we get \( x\hat{i} + y\hat{j} + z\hat{k} = (5 + 2\lambda)\hat{i} + (-2 - \lambda)\hat{j} + (4 + 3\lambda)\hat{k} \).

Equating coefficients of the base vectors, we obtain x = 5 + 2\lambda, y = -2 - \lambda, z = 4 + 3\lambda.

Eliminating the parameter \( \lambda \) from these equations, we get \( \frac{x - 5}{2} = \frac{y + 2}{-1} = \frac{z - 4}{3} \), which are the required cartesian equations.
In simple words: The vector equation shows where a point sits on the line using a parameter. To get the cartesian form, remove that parameter by making all three fractions equal to each other.

Exam Tip: Always write the vector equation as position vector of point plus parameter times direction vector. When converting to cartesian form, make sure the denominators match the direction numbers exactly.

Example 2. If the cartesian equations of a line are \( \frac{x - 1}{2} = \frac{y + 2}{3} = \frac{z - 5}{-1} \), find its vector equation.
Answer: The given line passes through the point (1, -2, 5) and is parallel to the vector \( 2\hat{i} + 3\hat{j} - \hat{k} \).

Let \( \vec{r} \) be the position vector of any point on the given line. The vector equation of the line is \( \vec{r} = \hat{i} - 2\hat{j} + 5\hat{k} + \lambda(2\hat{i} + 3\hat{j} - \hat{k}) \) where \( \lambda \) is a parameter.
In simple words: From the cartesian form, pick out the base point and the direction numbers, then build the vector equation by combining them with a parameter.

Exam Tip: When reading off direction numbers from the cartesian form, they are always the denominators. The numerators show you which point the line passes through.

Example 3. Find the vector and the cartesian equations of the line which passes through the point (-2, 4, -5) and is parallel to the line given by \( \frac{x + 3}{3} = \frac{y - 4}{5} = \frac{z + 8}{6} \).
Answer: Direction ratios of the given line are \( \langle 3, 5, 6 \rangle \). The required line goes through the point A(-2, 4, -5) and is parallel to the vector \( 3\hat{i} + 5\hat{j} + 6\hat{k} \). Its vector equation is \( \vec{r} = -2\hat{i} + 4\hat{j} - 5\hat{k} + \lambda(3\hat{i} + 5\hat{j} + 6\hat{k}) \), where \( \lambda \) is a parameter.

Cartesian equations of the required line are \( \frac{x - (-2)}{3} = \frac{y - 4}{5} = \frac{z - (-5)}{6} \), i.e. \( \frac{x + 2}{3} = \frac{y - 4}{5} = \frac{z + 5}{6} \).
In simple words: Read the direction numbers from the given line's equation. Use those same direction numbers for your new line, but change the base point to the one given.

Exam Tip: Direction numbers stay the same for parallel lines. Only the base point changes based on which specific line you are asked about.

Example 4. Find the vector equation of a line passing through the point with position vector \( 2\hat{i} - \hat{j} + \hat{k} \) and parallel to the line joining the points \( -\hat{i} + 4\hat{j} + \hat{k} \) and \( \hat{i} + 2\hat{j} + 2\hat{k} \). Also find cartesian equations of the line.
Answer: Let A, B and C be the points with position vectors \( 2\hat{i} - \hat{j} + \hat{k} \), \( -\hat{i} + 4\hat{j} + \hat{k} \) and \( \hat{i} + 2\hat{j} + 2\hat{k} \) respectively.

\( \vec{BC} = \) P.V. of C - P.V. of B = \( (\hat{i} + 2\hat{j} + 2\hat{k}) - (-\hat{i} + 4\hat{j} + \hat{k}) = 2\hat{i} - 2\hat{j} + \hat{k} \).

We need to find the vector equation of a line through A and parallel to \( \vec{BC} \). The vector equation of the required line is \( \vec{r} = 2\hat{i} - \hat{j} + \hat{k} + \lambda(2\hat{i} - 2\hat{j} + \hat{k}) \) where \( \lambda \) is a parameter.

Since \( \vec{r} \) is the position vector of any point P(x, y, z) on the line, we get \( x\hat{i} + y\hat{j} + z\hat{k} = 2\hat{i} - \hat{j} + \hat{k} + \lambda(2\hat{i} - 2\hat{j} + \hat{k}) \).

Equating the coefficients of the base vectors, we obtain x = 2 + 2\lambda, y = -1 - 2\lambda, z = 1 + \lambda.

Removing the parameter \( \lambda \) from these equations, we get \( \frac{x - 2}{2} = \frac{y + 1}{-2} = \frac{z - 1}{1} \), which are the required cartesian equations.
In simple words: Find the direction vector by subtracting position vectors of the two given points. Use this direction with the given base point to build your line equations.

Exam Tip: Always compute the direction vector first by subtracting the earlier point from the later point. Verify that your final cartesian equations use these same direction numbers in the denominators.

Example 5. Find the vector and the cartesian equations of the line that passes through the points A(3, -2, -5) and B(3, -2, 6).
Answer: Let \( \vec{a} \) and \( \vec{b} \) be the position vectors of the points A(3, -2, -5) and B(3, -2, 6) respectively. Then \( \vec{a} = 3\hat{i} - 2\hat{j} - 5\hat{k} \) and \( \vec{b} = 3\hat{i} - 2\hat{j} + 6\hat{k} \). Therefore \( \vec{b} - \vec{a} = 11\hat{k} \).

Let \( \vec{r} \) be the position vector of any point P(x, y, z) on the line passing through the points A and B. The vector equation of the line is \( \vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a}) \), i.e. \( \vec{r} = 3\hat{i} - 2\hat{j} - 5\hat{k} + \lambda(11\hat{k}) \), i.e. \( \vec{r} = 3\hat{i} - 2\hat{j} - 5\hat{k} + 11\lambda\hat{k} \), where \( \lambda \) is a parameter.

Since \( \vec{r} \) is the position vector of P(x, y, z), we get \( x\hat{i} + y\hat{j} + z\hat{k} = 3\hat{i} - 2\hat{j} - 5\hat{k} + 11\lambda\hat{k} \), i.e. \( x\hat{i} + y\hat{j} + z\hat{k} = 3\hat{i} - 2\hat{j} + (11\lambda - 5)\hat{k} \).

Equating the coefficients of base vectors, we obtain x = 3, y = -2, z = 11\lambda - 5, which are the required cartesian equations (in parametric form).
In simple words: When two points have the same x and y coordinates, the line is vertical (along the z-axis). This means x and y stay fixed while z changes.

Exam Tip: If two direction numbers turn out to be zero, the line is parallel to the remaining axis. The cartesian equations will show two coordinates as fixed constants.

Example 6. The points A(1, 2, 3), B(-1, -2, -1) and C(2, 3, 2) are three vertices of a parallelogram ABCD. Find vector and cartesian equations of the sides AB and BC. Also find the coordinates of D.
Answer: Given A(1, 2, 3), B(-1, -2, -1) and C(2, 3, 2).

Direction numbers of the line AB are \( \langle -1 - 1, -2 - 2, -1 - 3 \rangle \), i.e. \( \langle -2, -4, -4 \rangle \) or equivalently \( \langle 1, 2, 2 \rangle \). The vector equation of the side AB is \( \vec{r} = \hat{i} + 2\hat{j} + 3\hat{k} + \lambda(\hat{i} + 2\hat{j} + 2\hat{k}) \), where \( \lambda \) is a parameter. The cartesian equations of this line are \( \frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{2} \).

Direction numbers of the side BC are \( \langle 2 + 1, 3 + 2, 2 + 1 \rangle \), i.e. \( \langle 3, 5, 3 \rangle \). The vector equation of the side BC is \( \vec{r} = -\hat{i} - 2\hat{j} - \hat{k} + \mu(3\hat{i} + 5\hat{j} + 3\hat{k}) \), where \( \mu \) is a parameter. The cartesian equations of this line are \( \frac{x + 1}{3} = \frac{y + 2}{5} = \frac{z + 1}{3} \).

Let D be \( (\alpha, \beta, \gamma) \). Since ABCD is a parallelogram, diagonals AC and BD bisect each other; that is, the mid-point of segment AC is the same as the mid-point of segment BD. This gives us \( \left(\frac{1 + 2}{2}, \frac{2 + 3}{2}, \frac{3 + 2}{2}\right) = \left(\frac{-1 + \alpha}{2}, \frac{-2 + \beta}{2}, \frac{-1 + \gamma}{2}\right) \).

Therefore \( \alpha - 1 = 3, \beta - 2 = 5, \gamma - 1 = 5 \), which gives \( \alpha = 4, \beta = 7, \gamma = 6 \). Hence, the point D is (4, 7, 6).
In simple words: For a parallelogram, the two diagonals cut each other exactly at their midpoints. Use this property to find the fourth vertex by setting the midpoints equal.

Exam Tip: When asked to find a fourth vertex of a parallelogram given three, always use the property that diagonals bisect each other. Set up the mid-point formulas and solve for the unknown coordinates.

Example 7. Find the coordinates of the point where the line through the points (5, 1, 6) and (3, 4, 1) crosses the yz-plane.
Answer: Direction numbers of the line passing through the points (5, 1, 6) and (3, 4, 1) are \( \langle 3 - 5, 4 - 1, 1 - 6 \rangle \), i.e. \( \langle -2, 3, -5 \rangle \) or equivalently \( \langle 2, -3, 5 \rangle \).

The equations of the line through the given points are \( \frac{x - 5}{-2} = \frac{y - 1}{3} = \frac{z - 6}{-5} \) ...(i)

Any point on the line (i) is (5 + 2\lambda, 1 - 3\lambda, 6 + 5\lambda). When this point lies on the yz-plane, its x-coordinate is zero. Therefore 5 + 2\lambda = 0, which gives \( \lambda = -\frac{5}{2} \).

The required point is \( \left(0, 1 + 3 \cdot \frac{5}{2}, 6 - 5 \cdot \frac{5}{2}\right) \), i.e. \( \left(0, \frac{17}{2}, -\frac{13}{2}\right) \).
In simple words: Set x equal to zero (since points on the yz-plane have x-coordinate zero), solve for the parameter, then substitute back to find y and z.

Exam Tip: A point lies on the xy-plane when z = 0, on the yz-plane when x = 0, and on the xz-plane when y = 0. Always set the appropriate coordinate to zero first.

Example 8. Show that the points whose position vectors are given by \( -2\hat{i} + 3\hat{j} + 5\hat{k}, \hat{i} + 2\hat{j} + 3\hat{k} \) and \( 7\hat{i} - \hat{k} \) are collinear.
Answer: Let A, B and C be the points whose position vectors are \( -2\hat{i} + 3\hat{j} + 5\hat{k}, \hat{i} + 2\hat{j} + 3\hat{k} \) and \( 7\hat{i} - \hat{k} \) respectively.

Equation of the line passing through points A and B is \( \vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a}) \), i.e. \( \vec{r} = -2\hat{i} + 3\hat{j} + 5\hat{k} + \lambda[(\hat{i} + 2\hat{j} + 3\hat{k}) - (-2\hat{i} + 3\hat{j} + 5\hat{k})] \), i.e. \( \vec{r} = -2\hat{i} + 3\hat{j} + 5\hat{k} + \lambda(3\hat{i} - \hat{j} - 2\hat{k}) \) ...(i)

The points A, B and C are collinear if the point C lies on (i), i.e. if \( 7\hat{i} - \hat{k} = -2\hat{i} + 3\hat{j} + 5\hat{k} + \lambda(3\hat{i} - \hat{j} - 2\hat{k}) \) is consistent.

This requires 7 = -2 + 3\lambda, 0 = 3 - \lambda and -1 = 5 - 2\lambda to all be true. Solving: \( \lambda = 3, \lambda = 3 \) and \( \lambda = 3 \), which are consistent. Hence, the given points are collinear.

Alternative method (cartesian approach): The coordinates of the points A, B and C are (-2, 3, 5), (1, 2, 3) and (7, 0, -1) respectively. Direction numbers of the line AB are \( \langle 1 + 2, 2 - 3, 3 - 5 \rangle \), i.e. \( \langle 3, -1, -2 \rangle \). The equations of the line AB are \( \frac{x + 2}{3} = \frac{y - 3}{-1} = \frac{z - 5}{-2} \) ...(ii)

The points A, B and C are collinear if the point C(7, 0, -1) lies on (ii), i.e. if \( \frac{7 + 2}{3} = \frac{0 - 3}{-1} = \frac{-1 - 5}{-2} \) is consistent. Checking: \( \frac{9}{3} = 3, \frac{-3}{-1} = 3 \) and \( \frac{-6}{-2} = 3 \), so all three equal 3, which is true. Hence, the given points are collinear.
In simple words: Three points are collinear if you can find a single parameter value that makes the third point sit on the line through the first two. Test whether all three fractions give the same number.

Exam Tip: Always try both vector and cartesian methods to show your work completely. If all fractions (or all parameter equations) give the same consistent value, the points are collinear.

Example 9. Find the values of p and q by using vector method such that the points A(5, 0, 5), B(2, 1, 3) and C(-4, p, q) are collinear.
Answer: Direction numbers of the line AB are \( \langle 2 - 5, 1 - 0, 3 - 5 \rangle \), i.e. \( \langle -3, 1, -2 \rangle \) or equivalently \( \langle 3, -1, 2 \rangle \). The vector equation of the line AB is \( \vec{r} = 5\hat{i} + 0\hat{j} + 5\hat{k} + \lambda(3\hat{i} - \hat{j} + 2\hat{k}) \).

The position vector of the point C is \( -4\hat{i} + p\hat{j} + q\hat{k} \). Since the points A, B and C are collinear, C lies on the line AB. Therefore \( -4\hat{i} + p\hat{j} + q\hat{k} = 5\hat{i} + 5\hat{k} + \lambda(3\hat{i} - \hat{j} + 2\hat{k}) \).

Equating the coefficients of the base vectors, we obtain -4 = 5 + 3\lambda, p = -\lambda and q = 5 + 2\lambda. From the first equation, \( \lambda = -3 \). Therefore p = -(-3) = 3 and q = 5 + 2(-3) = -1. Hence, p = 3 and q = -1.
In simple words: Set up the line through the first two points. Then make the third point's position vector equal to that line equation. Solve for the unknowns by equating coefficients.

Exam Tip: When finding unknown coordinates, always equate coefficients of \( \hat{i}, \hat{j}, \hat{k} \) separately. You will get enough equations to solve for all the unknowns.

Example 10. The cartesian equations of a line are 2x - 3 = 3y + 1 = 5 - 6z. Find the direction ratios of the line and write down the vector and cartesian equations of the line through (7, -5, 0) which is parallel to the given line.
Answer: The equations of the line are 2x - 3 = 3y + 1 = 5 - 6z, i.e. \( 2\left(x - \frac{3}{2}\right) = 3\left(y + \frac{1}{3}\right) = -6\left(z - \frac{5}{6}\right) \).

This gives \( \frac{x - \frac{3}{2}}{1/2} = \frac{y + \frac{1}{3}}{1/3} = \frac{z - \frac{5}{6}}{-1/6} \), i.e. \( \frac{x - \frac{3}{2}}{3} = \frac{y + \frac{1}{3}}{2} = \frac{z - \frac{5}{6}}{-1} \).

Therefore, direction ratios of the given line are \( \langle 3, 2, -1 \rangle \). The vector equation of the line through (7, -5, 0) and parallel to the given line is \( \vec{r} = 7\hat{i} - 5\hat{j} + \lambda(3\hat{i} + 2\hat{j} - \hat{k}) \), where \( \lambda \) is a parameter. The cartesian equations of this line are \( \frac{x - 7}{3} = \frac{y + 5}{2} = \frac{z}{-1} \).
In simple words: To convert the "equal fractions" form to standard form, divide through by the coefficient in front of each variable term. The resulting denominators are your direction numbers.

Exam Tip: Be careful when converting compound equations like 2x - 3 = 3y + 1. Factor out coefficients to get the standard \( \frac{x - a}{l} = \frac{y - b}{m} \) form before reading off direction numbers.

Example 11. Find the points on the line through the points A(1, 2, 3) and B(3, 5, 9) at a distance of 14 units from the mid-point of segment AB.
Answer: Direction numbers of the line AB are \( \langle 3 - 1, 5 - 2, 9 - 3 \rangle \), i.e. \( \langle 2, 3, 6 \rangle \). The equations of the line AB are \( \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{6} \).

Any point on the line AB is P(1 + 2\lambda, 2 + 3\lambda, 3 + 6\lambda). The mid-point of AB is M\( \left(\frac{1 + 3}{2}, \frac{2 + 5}{2}, \frac{3 + 9}{2}\right) \), i.e. M\( \left(2, \frac{7}{2}, 6\right) \).

According to the given condition, \( |MP| = 14 \). Therefore \( \sqrt{(2\lambda + 1 - 2)^2 + \left(3\lambda + 2 - \frac{7}{2}\right)^2 + (6\lambda + 3 - 6)^2} = 14 \).

Simplifying: \( (2\lambda - 1)^2 + \left(3\lambda - \frac{3}{2}\right)^2 + (6\lambda - 3)^2 = 196 \).

Expanding: 4\(\lambda^2\) - 4\(\lambda\) + 1 + 9\(\lambda^2\) - 9\(\lambda\) + \(\frac{9}{4}\) + 36\(\lambda^2\) - 36\(\lambda\) + 9 = 196.

Combining: 49\(\lambda^2\) - 49\(\lambda\) - \(\frac{735}{4}\) = 0.

Multiplying by 4: 196\(\lambda^2\) - 196\(\lambda\) - 735 = 0, which simplifies to 4\(\lambda^2\) - 4\(\lambda\) - 15 = 0. Factoring: (2\(\lambda\) - 5)(2\(\lambda\) + 3) = 0.

Therefore \( \lambda = \frac{5}{2} \) or \( \lambda = -\frac{3}{2} \).

The required points are \( \left(1 + 2 \cdot \frac{5}{2}, 2 + 3 \cdot \frac{5}{2}, 3 + 6 \cdot \frac{5}{2}\right) \) and \( \left(1 + 2 \cdot \left(-\frac{3}{2}\right), 2 + 3 \cdot \left(-\frac{3}{2}\right), 3 + 6 \cdot \left(-\frac{3}{2}\right)\right) \), i.e. \( \left(6, \frac{19}{2}, 18\right) \) and \( (-2, -\frac{5}{2}, -6\right) \).
In simple words: Use the distance formula between two points. Set it equal to 14 and solve the resulting equation for the parameter value. Substitute back to find the actual coordinates.

Exam Tip: When the distance is measured from a special point like the mid-point, always calculate that point first. Then use the distance formula in parametric form and expand carefully.

Exercise 2.1

1. Find the vector equation of the line passing through the point (2, -3, 5) and parallel to the vector \( 3\hat{i} + 2\hat{j} - 7\hat{k} \).

2. Find the cartesian equation of the line which passes through the point with position vector \( 2\hat{i} - \hat{j} + 4\hat{k} \) and is in the direction of the vector \( \hat{i} + 2\hat{j} - \hat{k} \).

3. Find the vector equation of the line \( \frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2} \).

4. Find the cartesian equations of the line which passes through the point (-2, 4, -5) and parallel to the line \( \frac{x + 3}{3} = \frac{y - 4}{5} = \frac{z + 8}{6} \).

5. Find the vector equation of a line passing through (1, -2, 3) and parallel to the line \( \frac{x - 1}{3} = \frac{y + 2}{-5} = \frac{z + 5}{2} \).

6. Find the vector equation of the line passing through the points (-1, 0, 2) and (3, 4, 6).

7. Find the vector and the cartesian equations of the line that passes through origin and the point (5, -2, 3).

8. The cartesian equations of a line AB are \( \frac{2x - 1}{2} = \frac{4 - y}{7} = \frac{z + 1}{2} \). Write the direction ratios of a line parallel to AB.

9. The cartesian equations of a line are 2x - 3 = 3y + 1 = 5 - 6z. Write the direction ratios of the line.

10. Write the vector equation of the x-axis.

11. Find the vector and the cartesian equations of the line through the point (5, 2, -4) and which is parallel to the vector \( 3\hat{i} + 2\hat{j} - 8\hat{k} \).

12. Find the vector equation of the line passing through the point A(2, -1, 1) and parallel to the line joining the points B(-1, 4, 1) and C(1, 2, 2). Also find the cartesian equation of the line.

13. Find the vector equation of a line passing through the point with position vector \( \hat{i} - 2\hat{j} - 3\hat{k} \) and parallel to the line joining the points with position vectors \( \hat{i} - \hat{j} + 4\hat{k} \) and \( 2\hat{i} + \hat{j} + 2\hat{k} \). Also find the cartesian equations of the line.

14. Find the vector equation of the line joining the points whose position vectors are \( 2\hat{i} - \hat{j} + \hat{k} \) and \( \hat{i} + 2\hat{j} - 3\hat{k} \). Also find its cartesian equations.

15. The cartesian equations of a line are 6x - 2 = 3y + 1 = 2z - 2. Find direction ratios of the line and write down the vector equation of the line through (2, -1, -1) which is parallel to the given line.

2.2 Angle Between Two Lines

The angle between two lines is the same as the angle between the two vectors along their directions. Notice that the angle between a pair of straight lines depends only on their directions and not on their positions. In this section, we will find the angle between two lines when their equations are known.

2.2.1 To Find Angle Between Two Lines

Let \( \vec{r} = \vec{a} + \lambda \vec{b} \) ...(1) and \( \vec{r} = \vec{a'} + \mu \vec{c} \) ...(2) be the vector equations of the two lines.

These lines are in the directions of \( \vec{b} \) and \( \vec{c} \). If \( \theta \) is the angle between these lines, then \( \theta \) is the angle between the vectors \( \vec{b} \) and \( \vec{c} \).

\( \therefore \cos \theta = \frac{\vec{b} \cdot \vec{c}}{|\vec{b}| |\vec{c}|} \) ...(3)

Cartesian Form

Let the equations of the given lines in cartesian form be \( \frac{x - x_1}{a_1} = \frac{y - y_1}{b_1} = \frac{z - z_1}{c_1} \) ...(4)