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Find the equation of the plane which is perpendicular to the plane 5x + 3y + 6z + 8 = 0 and which contains the line of intersection of the planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0.

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Question

Find the equation of the plane which is perpendicular to the plane 5x + 3y + 6z + 8 = 0 and which contains the line of intersection of the planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0.

Sum
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Solution

The given planes are

P1: 5x + 3y + 6z + 8 = 0

P2: x + 2y + 3z – 4 = 0

P3: 2x + y – z + 5 = 0

Now, the equation of the plane passing through the line of intersection of P1 and P3 is

(x + 2y + 3z – 4) + λ(2x + y – z + 5) = 0

(1 + 2λ)x + (2 + λ)y + (3 – λ)z – 4 + 5λ = 0   ......(i)

From the question its understood that plane (i) is perpendicular to P1, then

5(1 + 2λ) + 3(2 + λ) + 6(3 – λ) = 0

5 + 10λ + 6 + 3λ + 18 – 6λ = 0

7λ + 29 = 0

λ = `(-29)/7`

Putting the value of the equation (i), we get

`[1 + 2((-29)/7)]x + [2 - 29/7]y + [3 + 29/7]z - 4 + 5((-29)/7)` = 0

⇒ `(-15)/7x - 15/7y + 50/7z - 4 - 145/7` = 0

–15x – 15y + 50z – 28 – 145 = 0

–15x – 15y + 50z – 173 = 0

⇒ 51x + 15y – 50z + 173 = 0

Thus, the required equation of plane is 51x + 15y – 50z + 173 = 0.

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Chapter 12: Introduction to Three Dimensional Geometry - Exercise [Page 237]

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NCERT Exemplar Mathematics [English] Class 11
Chapter 12 Introduction to Three Dimensional Geometry
Exercise | Q 22 | Page 237

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