TS EAMCET · Maths · Three Dimensional Geometry

$π_{1}$ is a plane passing through the point $(1, 2, 3)$ and perpendicular to the planes $x + 2 y + 3 z - 6 = 0, x + 2 y + 2 z - 5 = 0$ . If $(- 1, 2, - 3)$ is the foot of the perpendicular drawn from the point $(1, 3, 2)$ on to a plane $π_{2},$ then the angle between the planes $π_{1}$ and $π_{2}$ is

A $\cos^{- 1} (\frac{9}{\sqrt{255}})$
B $\frac{π}{4}$
C $\cos^{- 1} (\frac{\sqrt{6}}{10})$
D $\frac{π}{2}$

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Answer & Solution

Correct Answer

(C) $\cos^{- 1} (\frac{\sqrt{6}}{10})$

Step-by-step Solution

Detailed explanation

We know that angle between two planes is the same as the angle between normals of these planes So normal vector of plane π2 is n2→=-i^+2j^-3k^-i^+3j^+2k^=-2i^-j^-5k^ Now let direction ratios of normal of plane π1 are a, b, c Given that π1 is…

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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π1 is a plane passing through the point (1,2,3) and perpendicular to the planes x+2y+3z-6=0, x+2y+2z-5=0. If (-1,2,-3) is the foot of the perpendicular drawn from the point (1,3,2) on to a plane π2, then the angle between the planes π1 and π2 is

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