RD Sharma Class 12 Ex 29.14 Solutions Chapter 29 The Plane

Here we provide RD Sharma Class 12 Ex 29.14 Solutions Chapter 29 The Plane for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 29.14 Solutions Chapter 29 The Plane book pdf download. Now you will get step-by-step solutions to each question.

RD Sharma Class 12 Ex 29.14 Solutions Chapter 29 The Plane

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 Pand 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 P= 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 =

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