Question

If \[ \frac{x+3}{a}=\frac{y-1}{b}=\frac{z}{c} \] is the common line of the two perpendicular planes \[ 2x+3y+4z+d=0 \] and \[ x+py+z+5=0, \] then \(p+d=\)

• A. \(-5\)
• B. \(1\)
• C. \(-4\)
• D. \(2\)

Hint:

If a line is common to two planes, every point on the line satisfies both plane equations. Substitute the given point first, then use the perpendicular condition \[ \boxed{\vec n_1\cdot\vec n_2=0} \] to verify the result.

Answer 1

The Correct Option is B

Step 1: Use the point on the common line. The common line passes through \[ (-3,1,0). \] Since this point lies on both planes, For \[ 2x+3y+4z+d=0, \] \[ 2(-3)+3(1)+d=0, \] \[ -6+3+d=0, \] \[ d=3. \] For \[ x+py+z+5=0, \] \[ -3+p+5=0, \] \[ p=-2. \]

Step 2: Verify that the planes are perpendicular. The normal vectors are \[ (2,3,4) \] and \[ (1,-2,1). \] Their dot product is \[ 2(1)+3(-2)+4(1) =2-6+4 =0. \] Hence, the planes are perpendicular.

Step 3: Find the required value. Therefore, \[ p+d=-2+3=1. \] Hence, \[ \boxed{1}. \] Thus, \[ \boxed{(B)} \] is the correct answer.