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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.
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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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