Question

If the plane \( 3x + y + 2z + 6 = 0 \) { is parallel to the line} \[ \frac{3x - 1}{2b} = \frac{3 - y}{1} = \frac{z - 1}{a}, \] {then the value of \( 3a + 3b \) is:}

• A. \( \frac{1}{2} \)
• B. \( \frac{3}{2} \)
• C. 3
• D. 4

Hint:

When finding the condition for parallelism between a plane and a line, use the dot product of the normal vector of the plane and the direction ratios of the line. The condition for parallelism is that the dot product must be zero.

Answer 1

The Correct Option is B

We are given the equation of the plane: \[ 3x + y + 2z + 6 = 0 \] The equation of the line is given in symmetric form: \[ \frac{3x - 1}{2b} = \frac{3 - y}{1} = \frac{z - 1}{a} \] Step 1: Direction ratios of the line We can write the direction ratios of the line from the symmetric form: \[ {Direction ratios} = \left( 2b, -1, a \right) \] Step 2: Normal to the plane The normal to the plane is given by the coefficients of \( x, y, \) and \( z \) in the plane equation: \[ {Normal vector} = (3, 1, 2) \] Step 3: Condition for parallelism For the plane and the line to be parallel, the direction ratios of the line must be perpendicular to the normal vector of the plane. The condition for perpendicularity is that the dot product of the direction ratios of the line and the normal vector of the plane must be zero: \[ (3, 1, 2) \cdot (2b, -1, a) = 0 \] Taking the dot product: \[ 3(2b) + 1(-1) + 2(a) = 0 \] Simplifying: \[ 6b - 1 + 2a = 0 \] \[ 6b + 2a = 1 \] Step 4: Find \( 3a + 3b \) We are asked to find \( 3a + 3b \). From the equation \( 6b + 2a = 1 \), divide through by 2: \[ 3b + a = \frac{1}{2} \] Now, multiply both sides by 3: \[ 3a + 3b = \frac{3}{2} \] Thus, the correct answer is Option B.