Three Dimensional Geometry Question - CBSE

Question

Foot of perpendicular from point $(0, 2, 3)$ on the line $\dfrac{x+3}{5} = \dfrac{y-1}{2} = \dfrac{z+4}{3}$ is:
(a) $(2, 3, -1)$
(b) $(-3, 1, -4)$
(c) $(5, 2, 3)$
(d) $(1, 2, 3)$

Accepted Answer

$38\lambda - 38 = 0 \Rightarrow \lambda = 1 \Rightarrow M = (2, 3, -1)$. [1.0 Mark]

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🎯 Official CBSE Marking Scheme:
Evaluating foot of perpendicular: 1.0 Mark

Question:

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