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.

Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 29 |

Exercise | 29.14 |

Category | RD Sharma Solutions |

Table of Contents

**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 P_{1} passes through the point P(2, 5, 0)

And the equation of a plane containing line P_{2} is

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

Where 2a – b + 2c = 0

If it is parallel to line P_{1} 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 P_{2 }and parallel to line P_{1}.

Hence, the shortest distance between P_{1} and P_{2} = 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 P_{1} is a(x + 1) + b(y + 1) + c(z+1) = 0

Plane is parallel to P_{1 }= 7a – 6b + c = 0 -(1)

Plane is parallel to P_{2} = 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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