Class 12 RD Sharma Solutions Chapter 29 The Plane Exercise 29.14 (original) (raw)

Last Updated : 27 Aug, 2024

Question 1. Find the shortest distance between the lines \frac{x-2}{-1}=\frac{y-5}{2}=\frac{z-0}{3} and \frac{x-0}{2}=\frac{y+1}{-1}=\frac{z-1}{2} .

**Solution:

Let us consider

P_1=\frac{x-2}{-1}=\frac{y-5}{2}=\frac{z-0}{3}\\ P_2=\frac{x-0}{2}=\frac{y+1}{-1}=\frac{z-1}{2}

According to the equations line P1 passes through the point P(2, 5, 0)

And the equation of a plane containing line P2 is

a(x - 0) + b(y + 1) + c(z - 1) = 0 -(1)

Where 2a - b + 2c = 0

If it is parallel to line P1 then

-a + 2b + 3c = 0

So,

\frac{a}{-7}=\frac{b}{-8}=\frac{c}{3}

Now, substitute the value of a, b, c in the eq(1) we get

a(x - 0) + b(y + 1) + c(z - 1) = 0

-7(x - 0) - 8(y + 1) + 3(z - 1) = 0

-7x - 8y - 8 + 3z - 3 = 0

7x + 8y - 3z + 11 = 0 -(2)

So, this is the equation of the plane that contain line P2 and parallel to line P1.

Hence, the shortest distance between P1 and P2 = Distance between point P(2, 5, 0) and plane (2)

=\left|\frac{14+40+11}{\sqrt{7^2+8^2+(-3)^2}}\right|=\frac{65}{\sqrt{122}}

Question 2. Find the shortest distance between the lines \frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1} and \frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1} .

**Solution:

Let us consider

P_1:\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}

P_2:\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}

Let us assume the equation of the plane containing P1 is a(x + 1) + b(y + 1) + c(z+1) = 0

Plane is parallel to P1 = 7a - 6b + c = 0 -(1)

Plane is parallel to P2 = a - 2b + c = 0 -(2)

On solving eq(1) and eq(2), we get,

\frac{a}{-6+2}=\frac{b}{1-7}=\frac{c}{-14+6}\\ \frac{a}{-4}=\frac{b}{-6}=\frac{c}{-8}

The equation of the plane is -4(x + 1) - 6(y + 1) - 8(z + 1) = 0

Final equation of plane is 4(x + 1) + 6(y + 1) + 8(z + 1) = 0

Question 3. Find the shortest distance between the lines \frac{x-1}{2}=\frac{y-3}{4}=\frac{z+2}{1} and 3x - y - 2z + 4 = 0, 2x + y + z + 1 = 0.

**Solution:

The equation of a plane containing the line 3x - y - 2z + 4 = 0, 2x + y + z + 1 = 0 is

x(2λ + 3) + y(λ - 1) + z(λ - 2) + λ + 4 = 0 -(1)

If it is parallel to the line \frac{x-1}{2}=\frac{y-3}{4}=\frac{z+2}{1} then,

2(2λ + 3) + 4(λ - 1) + (λ - 2) = 0

λ = 0

On putting λ = 0 in eq(1) we get,

3x - y - 2z + 4 = 0 -(2)

As this equation of the plane consist the second line and parallel to the first line.

It is clear that the line \frac{x-1}{2}=\frac{y-3}{4}=\frac{z+2}{1} passes through the point (1, 3, -2)

So, the shortest distance 'D' between the given lines is equal to the

length of perpendicular from point (1, 3, -2) on the plane (2)

D = \left|\frac{3-3+4+4}{\sqrt{1+9+4}}\right|=\frac{8}{\sqrt{14}}

Summary

Exercise 29.14 in RD Sharma Class 12 typically deals with finding the equation of a plane under various conditions. Key concepts include: