Q. The angle between $\widehat{i}$ line of the intersection of the plane $\overrightarrow{r} \bullet \left( \widehat{i} + 2\widehat{j} + 3\widehat{k} \right) = 0$ and$\ \overrightarrow{r} \bullet \left( 3\widehat{i} + 3\widehat{j} + \widehat{k} \right) = 0$, is
A) $\cos^{- 1}\left( \frac{1}{3} \right)$
B) $\cos^{- 1}\left( \frac{1}{\sqrt{3}} \right)$
C) $\cos^{- 1}\left( \frac{2}{\sqrt{3}} \right)$
D) None of these
Explanation
(d)
Line of intersection of $\overrightarrow{r} \bullet \left( \widehat{i} + 2\widehat{j} + 3\widehat{k} \right) = 0$ and $\overrightarrow{r} \bullet \left( 3\widehat{i} + 3\widehat{j} + \widehat{k} \right) = 0$ will be parallel to $\left( 3\widehat{i} + 3\widehat{j} + \widehat{k} \right) \times \left( \widehat{i} + 2\widehat{j} + 3\widehat{k} \right)$, i.e., $7\widehat{i} - 8\widehat{j} + 3\widehat{k}$
If the required angle is $\theta$, then
$\cos\theta = \frac{7}{\sqrt{49 + 64 + 9}} = \frac{7}{\sqrt{122}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.