**Chapter 6 Line and Plane Miscellaneous Exercise 6A**

## Chapter 6 Line and Plane Miscellaneous Exercise 6A

**Question 1.**

**Question 2.**

**Question 3.**

**Question 4.**

**Question 5.Find the vector equation of the line which passes through the origin and the point (5, -2, 3).Solution:**

**Question 6.Find the Cartesian equations of the line which passes through points (3, -2, -5) and (3, -2, 6).Solution:**

Let A = (3, -2, -5), B = (3, -2, 6)

The direction ratios of the line AB are

3 – 3, -2 – (-2), 6 – (-5) i.e. 0, 0, 11.

The parametric equations of the line passing through (x

_{1}, y

_{1}, z

_{1}) and having direction ratios a, b, c are

x = x

_{1}+ aλ, y = y

_{1}+ bλ, z = z

_{1}+ cλ

∴ the parametric equattions of the line passing through (3, -2, -5) and having direction ratios are 0, 0, 11 are

x = 3 + (0)λ, y = -2 + 0(λ), z = -5 + 11λ

i.e. x = 3, y = -2, z = 11λ – 5

∴ the cartesian equations of the line are

x = 3, y = -2, z = 11λ – 5, λ is a scalar.

**Question 7.Find the Cartesian equations of the line passing through A(3, 2, 1) and B(1, 3, 1).Solution:**

The direction ratios of the line AB are 3 – 1, 2 – 3, 1 – 1 i.e. 2, -1, 0.

The parametric equations of the line passing through (x

_{1}, y

_{1}, z

_{1}) and having direction ratios a, b, c are

x = x

_{1}+ aλ, y = y

_{1}+ bλ, z = z

_{1}+ cλ

∴ the parametric equattions of the line passing through (3, 2, 1) and having direction ratios 2, -1, 0 are

x = 3 + 2λ, y = 2 – λ, z = 1 + 0(λ)

x – 3 = 2λ, y – 2 = -λ, z = 1

**Question 8.**

∴ the required line has direction ratios -1, 1, -1.

The cartesian equations of the line passing through (x

_{1}, y

_{1}, z

_{1}) and having direction ratios a, b, c are

∴ the cartesian equations of the line passing through the point (1, 1, 2) and having direction ratios -1, 1, -1 are

**Question 9.**

∴ p + 2q + r = 0, 3 + 2q – r = 0

∴ the required line has direction ratios 2, -7, 4.

The cartesian equations of the line passing through (x_{1}, y_{1}, z_{1}) and having direction ratios a, b, c are

∴ the cartesian equation of the line passing through the point (2, -7, 4) and having directions ratios 2, -7, 4 are

**Question 10.Find the vector equation of the line which passes through the origin and intersect the line x – 1 = y – 2 = z – 3 at right angle.Solution:**

**Question 11.Solution:**

The equations of the given lines are

**Question 12.**

**Solution:**

**Question 13.Find the acute angle between lines x = y, z = 0 and x = 0, z = 0.Solution:**

The equations x = y, z = 0 can be written as x/1=y/1, z = 0

∴ the direction ratios of the line are 1, 1, 0.

The direction ratios of the line x = 0, z = 0, i.e., Y-axis J are 0, 1, 0.

∴ its directiton ratios are 0, 1, 0.

Let a¯ and b¯ be the vectors in the direction of the lines x = y, z = 0 and x = 0, z = 0.

If θ is the acute angle between the lines, then

**Question 14.Find the acute angle between lines x = -y, z = 0 and x = 0, z = 0.Solution:**
The equations x = -y, z = 0 can be written as x/1=y/1, z = 0.

∴ the direction ratios of the line are 1, 1, 0.

The direction ratios of the line x = 0, z = 0, i.e., Y-axis are 0, 1, 0.

∴ its direction ratios are 0, 1, 0.

Let a¯ and b¯ be the vectors in the direction of the lines x = y, z = 0 and x = 0, z = 0

**Question 15.**

Solution:

Solution:

Let P = (0, 2, 3)

Let M be the foot of the perpendicular drawn from P to the line

The coordinates of any point on the line are given by

x = 5λ – 3, y = 2λ + 1, z = 3λ – 4

Let M = (5λ – 3, 2λ + 1, 3λ – 4) …(1)

The direction ratios of PM are

5λ – 3 – 0, 2λ + 1 – 2, 3λ – 4 – 3 i.e. 5λ – 3, 2λ – 1, 3λ – 7

Since, PM is perpendicular to the line whose direcction ratios are 5, 2, 3,

5(5λ – 3) + 2(2λ – 1) + 3(3λ – 7) = 0

25λ – 15 + 4λ – 2 + 9λ – 21 =0

38λ – 38 = 0 ∴ λ = 1

Substituting λ = 1 in (1), we get.

M = (5 – 3, 2 + 1, 3 – 4) = (2, 3, -1).

Hence, the coordinates of the foot of perpendicular are (2, 3, – 1).

**Question 16.By computing the shortest distance determine whether following lines intersect each other.**

**Solution:**

The shortest distance between the lines

Shortest distance between the lines is 0.

∴ the lines intersect each other.

Solution:

The shortest distance between the lines

∴ x_{1} = 5, y_{1} = 7, z_{1} = 3, x_{2} = 6, y_{2} = 8, z_{2} = 2,

l_{1} = 4, m_{1} = 5, n_{1} = 1, l_{2} = 1, m_{2} = -2, n_{2} = 1

= 4(-6 + 2) – 6(7 – 1) + 8(-14 + 6)

= -16 – 36 – 64

= -116

and

(m_{1}n_{2} – m_{2}n_{1})^{2} + (l_{2}n_{1} – l_{1}n_{2})^{2} + (l_{1}m_{2} – l_{2}m_{1})^{2}

= (-6 + 2)^{2} + (1 – 7)^{2} + (1 – 7)^{2} + (-14 + 6)^{2}

= 16 + 36 + 64

= 116

Hence, the required shortest distance between the given lines

or

Shortest distance between the lines is 0.

∴ the lines intersect each other.

**Question 17.Solution:**

Here, (x

_{1}, y

_{1}, z

_{1}) ≡ (1, -1, 1),

(x

_{2}, y

_{2}, z

_{2}) ≡ (2, -m, 2),

a

_{1}= 2, b

_{1}= 3, c

_{1}= 4,

a

_{2}= 1, b

_{2}= 2, c

_{2}= 1

Substituting these values in (1), we get

∴ 1(3 – 8) – (1 – m)(2 – 4) + 1 (4 – 3) = 0

∴ -5 + 2 – 2m + 1 = 0

∴ -2m = 2

∴ m = -1.

**Question 18.Find the vector and Cartesian equations of the line passing through the point (-1, -1, 2) and parallel to the line 2x – 2 = 3y + 1 = 6z – 2.Solution:**

Let a¯ be the position vector of the point A (-1, -1, 2) w.r.t. the origin.

The equation of given line is

x – 2 = 3y + 1 = 6z – 2.

The direction ratios of this line are

**Question 19.Solution:**

**Question 20.Solution:**

Let the required line have direction ratios a, b, c

Since the line passes through the origin, its cartesian equations are

This line is perpendicular to the line

x – 1 = y – 2 = z – 1 whose direction ratios are 1, 1, 1.

∴ 1(4b – 3c) + 1(4a – 2c) + 1(3a – 2b) = 0

∴ 4b – 3c + 4a – 2c + 3a – 2b = 0

∴ 7a + 2b – 5c = 0

From (2) and (3), we get

**Question 21.Write the vector equation of the line whose Cartesian equations are y = 2 and 4x – 3z + 5 = 0.Solution:**

4x – 3z + 5 = 0 can be written as

**Question 22.**

When λ = 1, M = (1 + 1, -2 + 2, 2 + 3) … [By (1)]

i. e. M = (2, 0, 5)

When λ = -1, M = (1 – 1, 2 + 2, -2 + 3) … [By (1)]

i. e. M = (0, 4, 1)

Hence, the coordinates of the required points are (2, 0, 5) and (0, 4, 1).