Find the points where the line $fraction numerator straight x minus 1 over denominator 2 end fraction space equals fraction numerator straight y minus 2 over denominator negative 3 end fraction space equals fraction numerator straight z plus 3 over denominator 4 end fraction space meets space the space plane space 2 straight x plus 4 straight y minus straight z space equals 1$ .

Solution

The equation of plane is
2x + 4 y – z = 1 ...(1)
The equation of line is
$fraction numerator straight x minus 1 over denominator 2 end fraction space equals space fraction numerator straight y minus 2 over denominator negative 3 end fraction space equals space fraction numerator straight z plus 3 over denominator 4 end fraction equals space straight r space left parenthesis say right parenthesis$
Any point on it is (2 r + 1, – 3 r + 2, 4 r – 3)
Let it lie on (1).
∴ 2 (2 r + 1) + 4 (–3 r + 2) – (4 r – 3) = 1
∴ 4 r + 2 – 12 r + 8 – 4r + 3 = 1
∴ – 12 r = – 12 ⇒ r = 1
∴ point is (2 + 1,–3 + 2,4 – 4) i.e. (3,–1,1).

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Three Dimensional Geometry

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