Find the co-ordinates of the point where the line $fraction numerator straight x plus 1 over denominator 2 end fraction space equals space fraction numerator straight y plus 2 over denominator 3 end fraction space equals space fraction numerator straight z plus 3 over denominator 4 end fraction$ meets the plane x + y + 4 z =6.

Solution

The equation of plane is
x + y+ 4 z = 6 ...(1)
The equation of line is
$fraction numerator straight x plus 1 over denominator 2 end fraction space equals space fraction numerator straight y plus 2 over denominator 3 end fraction space equals space fraction numerator straight z plus 3 over denominator 4 end fraction space equals space straight r space left parenthesis say right parenthesis$
Any point on it is (2 r – 1, 3 r – 2, 4 r – 3)
Let it lie on (1)
∴ (2 r – 1) + (3 r – 2) + 4 (4 r – 3) = 6
∴ 2r – 1 + 3r – 2 +16r – 12 = 6
∴ 21 r = 21 ⇒ r = 1
∴ point is (2 – 1, 3 – 2, 4 – 3) i.e. (1, 1, 1).

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Three Dimensional Geometry

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