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 to find the direction ratios (d.r.s.) of the normal vector to a specific plane. This plane satisfies two conditions: it passes through the origin $(0,0,0)$ and it contains the line of intersection of the two given planes $x+2y+3z-4=0$ and $4x+3y+2z-1=0$.

Step 2: Key Formula or Approach:
The equation of any plane passing through the line of intersection of two planes $P_1 = 0$ and $P_2 = 0$ can be represented using a scalar parameters family equation: $$P_1 + \lambda P_2 = 0$$ Once we substitute our equations, we use the constraint that the plane crosses the origin $(0,0,0)$ to solve for $\lambda$. The coefficients of $x$, $y$, and $z$ in the resulting equation represent the direction ratios of its normal vector.

Step 3: Detailed Explanation:
Write the equation of the family of planes: $$(x + 2y + 3z - 4) + \lambda(4x + 3y + 2z - 1) = 0$$ Group the terms by variables: $$(1 + 4\lambda)x + (2 + 3\lambda)y + (3 + 2\lambda)z - (4 + \lambda) = 0 \quad \text{--- (Equation 1)}$$ Since this target plane passes through the origin $(0,0,0)$, substitute $x = 0$, $y = 0$, and $z = 0$ into Equation 1: $$(1 + 4\lambda)(0) + (2 + 3\lambda)(0) + (3 + 2\lambda)(0) - (4 + \lambda) = 0$$ $$-(4 + \lambda) = 0 \implies \lambda = -4$$ Now, substitute $\lambda = -4$ back into Equation 1 to find the specific plane: $$(1 + 4(-4))x + (2 + 3(-4))y + (3 + 2(-4))z - (4 + (-4)) = 0$$ $$(1 - 16)x + (2 - 12)y + (3 - 8)z - 0 = 0$$ $$-15x - 10y - 5z = 0$$ Divide the entire equation by $-5$ to simplify to its irreducible form: $$3x + 2y + z = 0$$ The direction ratios of the normal vector are given by the coefficients of $x$, $y$, and $z$, which are $(3, 2, 1)$.

Step 4: Final Answer:
The direction ratios of the normal to the plane are $3, 2, 1$, which corresponds to option (A).