**Chapter 6 Line and Plane Ex 6.2**

## Chapter 6 Line and Plane Ex 6.2

**Question 1.**

Solution:

Solution:

Let PM be the perpendicular drawn from the point P (2, -3, 1) to the line

The coordinates of any point on the line are given by x = -1 + 2λ, y = 3 + 3λ, z = -1 – λ

Let the coordinates of M be

(-1 + 2λ, 3 + 3λ, -1 – λ) … (1)

The direction ratios of PM are

-1 + 2λ – 2, 3 + 3λ + 3, -1 – λ – 1

i.e. 2λ – 3, 3λ + 6, -λ – 2

The direction ratios of the given line are 2, 3, -1.

Since PM is perpendicular to the given line, we get

2(2λ – 3) + 3(3λ + 6) – 1(-λ – 2) = 0

∴ 4λ – 6 + 9λ + 18 + λ + 2 = 0

∴ 14λ + 14 = 0 ∴ λ = -1.

Put λ = -1 in (1), the coordinats of M are

(-1 – 2, 3 – 3, -1 + 1) i.e. (-3, 0,0).

∴ length of perpendicular from P to the given line

Alternative Method:

Substituting these values in (1), we get

length of perpendicular from P to given line

**Question 2.Solution:**

Hence, the coordinates of the foot of perpendicular are (1,2, 3) and length of perpendicular = √14 units.

**Question 3.Find the shortest distance between the **

**Question 4.Solution:**

The shortest distance between the lines

= 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}+ (-14 + 6)

^{2}

= 16 + 36 + 64 = 116

**Question 5.**

The coordinates of any point on the line are given by x = -1 + 2λ, y = 3 + 2λ, z = 8 – λ

Let the coordinates of M be

(-1 + 2λ, 3 + 3λ, -1 – λ) …..(1)

The direction ratios of PM are

-1 + 2λ – 2, 3 + 3λ + 3, -1 – λ – 1

i.e. 2λ – 3, 3λ = 6, -λ – 2

The direction ratios of the given line are 2, 3, 8.

Since PM is perpendicular to the given line, we get

2(2λ – 3) + 3(3λ + 6) – 1(-λ – 2) = O

∴ 4λ – 6 + 9λ + 18 + λ + 2 = 0

∴ 14λ + 14 = 0

∴ λ = -1

Put λ in (1), the coordinates of M are

(-1 – 2, 3 – 3, -1 + 1) i.e. (-3, 0, 0).

∴ length of perpendicular from P to the given line

= PM

Alternative Method :

We know that the perpendicular distance from the point

Substitutng tese values in (1), w get

length of perpendicular from P to given line

= PM

**Question 6.A(1, 0, 4), B(0, -11, 13), C(2, -3, 1) are three points and D is the foot of the perpendicular from A to BC. Find the co-ordinates of D.Solution:**
Equation of the line passing through the points (x

_{1}, y

_{1}, z

_{1}) and (x

_{2}, y

_{2}, z

_{2}) is

AD is the perpendicular from the point A (1, 0, 4) to the line BC.

The coordinates of any point on the line BC are given by x = 2λ, y = -11 + 8λ, z = 13 – 12λ

Let the coordinates of D be (2λ, -11 + 8λ, 13 – 12λ) … (1)

∴ the direction ratios of AD are

2λ – 1, -1λ + 8λ – 0, 13 – 12λ – 4 i.e.

2λ – 1, -11 + 8λ, 9 – 12λ

The direction ratios of the line BC are 2, 8, -12.

Since AD is perpendicular to BC, we get

2(2λ – 1) + 8(-11 + 8λ) – 12(9 – 12λ) = 0

∴ 42λ – 2 – 88 + 64λ – 108 + 144λ = 0

∴ 212λ – 198 = 0

**Question 7.By computing the shortest distance, determine whether following lines intersect each other.Solution:**

The shortest distance between the lines

Hence, the given lines do not intersect.

Solution:

The shortest distance between the lines

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

l_{1} = 7, m_{1} = -6, 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} + (-14 + 6)^{2}

= 16 + 36 + 64

= 116

Hence, the required shortest distance between the given lines

**Question 8.Solution:**

The lines

∴ x

_{1}= 1, y

_{1}= -1, z

_{1}= 1, x

_{2}= 3, y

_{2}= k, z

_{2}= 0,

l

_{1}= 2, m

_{1}= 3, n

_{1}= 4, l

_{2}= 1, m

_{2}= 2, n

_{2}= 1.

Since these lines intersect, we get