Question

The acute angle between the line \(\bar{r}=2\hat{i}+2\hat{j}+\hat{k}+\lambda (2\hat{i}-3\hat{j}+2\hat{k})\) and the plane \(\bar{r}\cdot (3\hat{i}-2\hat{j}+5\hat{k})=4\) is

• A. \(\sin ^{-1}\frac{22}{\sqrt{466}}\)
• B. \(\cos ^{-1}\frac{22}{\sqrt{646}}\)
• C. \(\tan ^{-1}\frac{22}{9\sqrt{2}}\)
• D. \(\sin ^{-1}\frac{22}{\sqrt{664}}\)

Hint:

Find the sine of the angle from the dot product of the direction vector and the plane normal, then convert to tangent.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
With direction \(\vec{d}=2\hat{i}-3\hat{j}+2\hat{k}\) and normal \(\vec{n}=3\hat{i}-2\hat{j}+5\hat{k}\), \(\sin\theta=\frac{|\vec{d}\cdot\vec{n}|}{|\vec{d}||\vec{n}|}\).

Step 2: Compute
\(\vec{d}\cdot\vec{n}=6+6+10=22\), \(|\vec{d}|=\sqrt{17}\), \(|\vec{n}|=\sqrt{38}\).
So \(\sin\theta=\frac{22}{\sqrt{646}}\).

Step 3: Convert to tangent
\(\cos\theta=\sqrt{1-\frac{484}{646}}=\frac{\sqrt{162}}{\sqrt{646}}=\frac{9\sqrt{2}}{\sqrt{646}}\), so \(\tan\theta=\frac{22}{9\sqrt{2}}\).

Final Answer:
The angle is \(\tan^{-1}\frac{22}{9\sqrt{2}}\). \[ \boxed{\text{(C) }\tan ^{-1}\dfrac{22}{9\sqrt{2}}} \]