Question

Find the equation of the plane passing through the points (2, 1, 0), (3, 2, 2) and parallel to the line $fraction numerator straight x minus 1 over denominator 2 end fraction space equals space fraction numerator straight y minus 2 over denominator 3 end fraction space equals space fraction numerator straight z minus 3 over denominator 1 end fraction.$

Solution

The equation of any plane through (2, 1, 0) is
A(x – 2) + B(– 1) + C(z – 0) = 0 ...(1)
∴ it passes through (3, 2, 2)
∴ A(3 – 2) + B(2 – 1) + C(2 – 0) = 0
∴ A + B + 2C = 0 ...(2)
Since plane (1) is parallel to the line $fraction numerator straight x minus 1 over denominator 2 end fraction space equals space fraction numerator straight y minus 2 over denominator 3 end fraction space equals space fraction numerator straight z minus 3 over denominator 1 end fraction$
∴ normal to the plane with direction ratios A, B, C is perpendicular to the line with direction ratios 2, 3, 1.
∴ (2) (A) + (3) (B) + (1) (C) = 0
∴ 2A + 3B + C = 0 ...(3)
Solving (2) and (3), we get,
 $fraction numerator straight A over denominator 1 minus 6 end fraction space equals space fraction numerator straight B over denominator 4 minus 1 end fraction space equals space fraction numerator straight C over denominator 3 minus 2 end fraction$
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∴ A = 5 k, B = – 3 k, C = – k
Putting these values of A, B, C in (1), we get,
5 k (x – 2) – 3 k (y – 1) – k (z – 0) = 0
or 5 (x – 2) – 3(y – 1) – z = 0
or 5 x – 10 – 3 y + 3 – z = 0
or 5 x – 3 y – z – 7 = 0
which is required equation of plane.

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