MHT CET · Maths · Three Dimensional Geometry

The direction ratios of the normal to the plane passing through origin and the line of intersection of the planes $x + 2 y + 3 z = 4$ and $4 x + 3 y + 2 z = 1$ are …

A $2,3, 1$
B $1,2, 3$
C $3,1, 2$
D $3,2, 1$

Verified Solution

Answer & Solution

Correct Answer

(D) $3,2, 1$

Step-by-step Solution

Detailed explanation

We have, line of intersection of the planes

x + 2 y + 3 z = 4

and

4 x + 3 y + 2 z = 1

∴

Equation of plane passing through the given planes is

(x + 2 y + 3 z - 4) + λ (4 x + 3 y + 2 z - 1) = 0

\Rightarrow (1 + 4 λ) x + (2 + 3 λ) y + (3 + 2 λ) + (- 4 - λ) = 0

Since, plane passing through origin

∴ - 4 - λ = 0 \Rightarrow λ = - 4

Now, equation of plane is

(1 - 16) x + (2 - 12) y + (3 - 8) z + 0 = 0

\Rightarrow - 15 x - 10 y - 5 z = 0

\Rightarrow 3 x + 2 y + z = 0

∴

Direction rations of the normal to the plane are 3, 2, 1.

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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The direction ratios of the normal to the plane passing through origin and the line of intersection of the planes x+2y+3z=4 and 4x+3y+2z=1 are …

Step-by-step Solution

The direction ratios of the normal to the plane passing through origin and the line of intersection of the planes $x + 2 y + 3 z = 4$ and $4 x + 3 y + 2 z = 1$ are …