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# 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 and .

Solution:

Let us consider 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, 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) ### Question 2. Find the shortest distance between the lines and .

Solution:

Let us consider  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, 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 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 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 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 = My Personal Notes arrow_drop_up
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