Question

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

Solution

The equation of any plane through (1, 2, 3) is
A(x – 1)+B(y – 2) + C(z – 3) = 0    ...(1)
∵ it passes through (0, – 1, 0)
∴ A (0 – 1) + B (– 1 – 2) + C (0 – 3) = 0
∴ – A – 3B – 3C = 0    ⇒ A + 3B + 3C = 0    ...(2)
Since plane (1) is parallel to the line $fraction numerator straight x minus 1 over denominator 2 end fraction space equals space fraction numerator straight y plus 2 over denominator 3 end fraction space equals space fraction numerator straight z over denominator negative 3 end fraction$
∴ normal to the plane with direction ratios A, B, C is perpendicular to the line with direction ratios 2, 3, – 3.
∴ A(2) + B(3) + C(– 3) = 0    [∵ a₁a₂+ b₁b₂ + c₁c₂ = 0]
∴ 2A + 3B-3C = 0
Solving (2) and (3), we get,
$fraction numerator straight A over denominator negative 9 minus 9 end fraction space equals space fraction numerator straight B over denominator 6 plus 3 end fraction space equals space fraction numerator straight C over denominator 3 minus 6 end fraction$
$therefore space space space space space space space fraction numerator straight A over denominator negative 18 end fraction space equals space straight B over 9 space equals space fraction numerator straight C over denominator negative 3 end fraction therefore space space space space space space straight A over 6 space equals space fraction numerator straight B over denominator negative 3 end fraction space equals space straight C over 1 space equals space straight k space left parenthesis say right parenthesis therefore space space space space straight A space equals space 6 space straight k comma space space space straight B space equals space minus 3 space straight k comma space space space straight C space equals space straight k$
Putting values of A, B, C in (1), we get,
6k(x – 1) + (–3 k)(y – 2) + k(z – 3) = 0 or    6 (x – 1) –3 (y – 2) + (z – 3) = 0
or    6x – 6 – 3y + 6 + z – 3 = 0
or    6x-3y + z=3
which is required equation of plane.

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