Question

Let the plane \(\pi\) pass through the point \((1,0,1)\) and perpendicular to the planes \[ 2x+3y-z=2 \] and \[ x-y+2z=1. \] Let the equation of the plane passing through the point \((11,7,5)\) and parallel to the plane \(\pi\) be \[ ax+by-z+d=0. \] Then \[ \frac{a}{b}+\frac{b}{d}= \]

• A. \(3\)
• B. \(0\)
• C. \(2\)
• D. \(-2\)

Hint:

If a plane is perpendicular to two planes, its normal vector is perpendicular to the normal vectors of both planes. Hence, use the cross product of the two given normal vectors.

Answer 1

The Correct Option is D

Step 1: Find the normal of plane $\pi$.
$\pi$ is perpendicular to planes with normals $\mathbf{n}_1=(2,3,-1)$ and $\mathbf{n}_2=(1,-1,2)$. So $\mathbf{n}=\mathbf{n}_1\times\mathbf{n}_2$.
Step 2: Compute the cross product.
\[\mathbf{n}=\mathbf{i}(6-1)-\mathbf{j}(4+1)+\mathbf{k}(-2-3)=(5,-5,-5).\] Simplified direction: $(1,-1,-1)$.
Step 3: Equation of $\pi$ through $(1,0,1)$.
$(x-1)-(y)-(z-1)=0 \implies x-y-z=0$.
Step 4: Parallel plane through $(11,7,5)$.
$x-y-z+d'=0$. At $(11,7,5)$: $11-7-5+d'=0 \implies d'=1$. So $x-y-z+1=0$.
Step 5: Identify $a,b,d$.
Writing as $1\cdot x+(-1)\cdot y-z+1=0$ in form $ax+by-z+d=0$: $a=1$, $b=-1$, $d=1$.
Step 6: Compute the expression.
\[\frac{a}{b}+\frac{b}{d}=\frac{1}{-1}+\frac{-1}{1}=-1-1=-2.\] \[ \boxed{-2} \]