JEE Advanced · Mathematics · 31. 3D Geometry

In $R^{3}$ , let L be a straight line passing through the origin. Suppose that all the points on L are at a constant distance from the two planes $P_{1} : x + 2 y - z + 1 = 0$ and $P_{2} : 2 x - y + z - 1 = 0 .$ Let M be the locus of the feet of the perpendiculars drawn from the points on L to the plane $P_{1}$ . Which of the following points lie (s) on M ?

A $(0, - \frac{5}{6}, - \frac{2}{3})$
B $(- \frac{1}{6}, - \frac{1}{3}, \frac{1}{6})$
C $(- \frac{5}{6}, 0, \frac{1}{6})$
D $(- \frac{1}{3}, 0, \frac{2}{3})$

Verified Solution

Answer & Solution

Correct Answer

(B) $(- \frac{1}{6}, - \frac{1}{3}, \frac{1}{6})$

Step-by-step Solution

Detailed explanation

Line L will lie on one of the angle bisector planes and will be parallel to line of intersection of the given planes. Also locus of foot of perpendicular of line L in plane

P_{1}

will be line M, which will be parallel L.
Foot of perpendicular from (0, 0, 0) on plane

P_{1}

(- \frac{1}{6}, - \frac{1}{3}, \frac{1}{6})

Hence equation of line M

\frac{x + \frac{1}{6}}{1} = \frac{y + \frac{1}{3}}{- 3} = \frac{z - \frac{1}{6}}{- 5}

Where

\hat{i} - 3 \hat{j} - 5 \hat{k}

is vector parallel to line of intersection of

P_{1}

and

P_{2}

.
On checking

(0, - \frac{5}{6} - \frac{2}{3}), (- \frac{1}{6}, - \frac{1}{3}, \frac{1}{6})

satisfy the above equation.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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