Question

The d.r.s. of the normal to the plane passing through the origin and the line of intersection of the planes $x+2y+3z=4$ and $4x+3y+2z=1$ are

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

Hint:

Since the target plane passes through the origin, its constant term $D$ must be 0. Look at the constant terms of the two planes ($-4$ and $-1$). To make them cancel out to 0, you must multiply the second plane by $-4$ and add it to the first: $P_1 - 4P_2 = 0$. This lets you find $\lambda = -4$ mentally!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need the direction ratios of the normal vector to a plane that passes through the origin and contains the intersection line of two given planes.

Step 2: Key Formula or Approach:
The family of planes through the intersection line is P₁ + λP₂ = 0. Substitute the origin (0,0,0) to find λ, then extract the coefficients of x, y, z.

Step 3: Detailed Explanation:
Family: (x+2y+3z–4) + λ(4x+3y+2z–1)=0. At (0,0,0): –4 – λ = 0 → λ = –4. Substituting back: –15x – 10y – 5z = 0 → 3x + 2y + z = 0. Direction ratios are (3, 2, 1).

Step 4: Final Answer:
The direction ratios are 3, 2, 1, matching option (A).