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Question
Find the equation of the plane through the points (2, 1, –1) and (–1, 3, 4), and perpendicular to the plane x – 2y + 4z = 10.
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Solution
We know that, equation of the plane passing through two points (x1, y1, z1) and (x2, y2, z2) with its normal’s direction ratios is
a(x – x1) + b(y – y1) + c(z – z1) = 0 ......(i)
Now, if the plane is passing through two points (2, 1, –1) and (–1, 3, 4) then
a(x2 – x1) + b(y2 – y1) + c(z2 – z1) = 0
a(–1 – 2) + b(3 – 1) + c(4 + 1) = 0
–3a + 2b + 5c = 0 .......(ii)
As the required plane is perpendicular to the given plane x – 2y + 4z = 10, then
1.a – 2.b + 4.c = 10 ......(iii)
On solving (ii) and (iii) we get,
`a/(8 + 10) = (-b)/(-2 - 5) = c/(6 - 2) = lambda`
So, a = 18λ, b = 17λ and c = 4λ
Thus, the required plane is
18λ(x – 2) + 17λ(y – 1) + 4λ(z + 1) = 0
18x – 36 + 17y – 17 + 4z + 4 = 0
⇒ 18x + 17y + 4z – 49 = 0
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