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प्रश्न
Find the distance of the point (1, -5, 9) from the plane
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उत्तर
The equation of line parallel to the line x = y = z and passing through the point (1, −5, 9) is
Any point on this line is of the form (k + 1, k − 5, k + 9).
If (k + 1, k − 5, k + 9) be the point of intersection of line (1) and the given plane, then
(k + 1) − (k − 5) + (k + 9) = 5
⇒ k = −10
So, the point of intersection of line (1) and the given plane is (−10 + 1, −10 − 5, −10 + 9) i.e. (−9, −15, −1).
∴ Required distance = Distance between (1, −5, 9) and (−9, −15, −1) =
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