The distance of the point (1, 0, 2) from the point of intersection of the line \frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12} and the plane x

Engineering

Mathematics

Equation of Straight Line in 3D

Question

The distance of the point (1, 0, 2) from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane x – y + z = 16, is

$3 \sqrt{21}$

$2 \sqrt{14}$

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General point on the line is (3r + 2, 4r – 1, 12r + 2) which satisfies the plane x – y + z = 16.

3r + 2 – 4r + 1 +12r +2 = 16 ⇒ r = 1

Required distance between points (5, 3, 14) and (1, 0, 2) is 13.