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 normal vectors of the given planes.
The given planes are \[ 2x+3y-z=2 \] and \[ x-y+2z=1 \] Their normal vectors are \[ \vec{n}_1=(2,3,-1) \] and \[ \vec{n}_2=(1,-1,2) \]

Step 2: Find the normal vector of plane \(\pi\).
Since plane \(\pi\) is perpendicular to both given planes, its normal vector must be perpendicular to both \(\vec{n}_1\) and \(\vec{n}_2\).
Therefore, the normal vector of \(\pi\) is proportional to \[ \vec{n}_1\times \vec{n}_2 \] Now, \[ \vec{n}_1\times \vec{n}_2= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 1 & -1 & 2 \end{vmatrix} \] \[ = \hat{i}(3\cdot 2-(-1)(-1))-\hat{j}(2\cdot 2-(-1)(1))+\hat{k}(2(-1)-3(1)) \] \[ = \hat{i}(6-1)-\hat{j}(4+1)+\hat{k}(-2-3) \] \[ =5\hat{i}-5\hat{j}-5\hat{k} \] So, \[ \vec{n}_\pi=(1,-1,-1) \]

Step 3: Write the equation of the plane parallel to \(\pi\).
The required plane is parallel to \(\pi\), so it has the same normal vector.
Thus, its equation is of the form \[ x-y-z+d=0 \] Comparing with \[ ax+by-z+d=0, \] we get \[ a=1,\qquad b=-1 \]

Step 4: Use the point \((11,7,5)\).
Since the plane passes through \((11,7,5)\), substitute this point in \[ x-y-z+d=0 \] \[ 11-7-5+d=0 \] \[ -1+d=0 \] \[ d=1 \]

Step 5: Evaluate the required expression.
Now, \[ \frac{a}{b}+\frac{b}{d} = \frac{1}{-1}+\frac{-1}{1} \] \[ =-1-1 \] \[ =-2 \]

Step 6: Final conclusion.
Therefore, \[ \boxed{-2} \]