Question

The equation of a plane passing through the intersection of two planes \(\vec{r}\cdot (\hat{i}-2\hat{j}+3\hat{k})=4\) and \(\vec{r}\cdot (2\hat{i}+\hat{j}-3\hat{k})=7\) and parallel to Y-axis is

• A. \(\vec{r}\cdot (2\hat{i}-2\hat{k})=11\)
• B. \(\vec{r}\cdot (\hat{i}+\hat{j}-2\hat{k})=3\)
• C. \(\vec{r}\cdot (5\hat{i}-3\hat{k})=18\)
• D. \(\vec{r}\cdot (4\hat{i}-3\hat{k})=28\)

Hint:

Form the family n1 + lambda n2 and choose lambda so that the j component of the normal is zero.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
A plane through the intersection is \(\vec{r}\cdot(\vec{n}_{1}+\lambda\vec{n}_{2})=4+7\lambda\).

Step 2: Combine the normals
\(\vec{n}_{1}+\lambda\vec{n}_{2}=(1+2\lambda)\hat{i}+(-2+\lambda)\hat{j}+(3-3\lambda)\hat{k}\).

Step 3: Parallel to Y-axis
The normal must be perpendicular to \(\hat{j}\), so the j component is zero: \(-2+\lambda=0\Rightarrow\lambda=2\).

Step 4: Substitute
Normal: \(5\hat{i}-3\hat{k}\). Right side: \(4+14=18\).
The plane is \(\vec{r}\cdot(5\hat{i}-3\hat{k})=18\).

Final Answer:
The plane is r · (5i − 3k) = 18. \[ \boxed{\text{(C) }\vec{r}\cdot (5\hat{i}-3\hat{k})=18} \]