Skip to content
Related Articles

Related Articles

Class 12 RD Sharma Solutions – Chapter 29 The Plane – Exercise 29.14
  • Last Updated : 11 Feb, 2021

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}}

My Personal Notes arrow_drop_up
Recommended Articles
Page :