Q. The distance between the line $\overrightarrow{\mathbf{r}} = 2\widehat{\dot{\mathbf{i}}} - 2\widehat{\dot{\mathbf{j}}} + 3\widehat{\mathbf{k}} + \lambda\ (\widehat{\dot{\mathbf{i}}} - \widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}})$ and the plane $\overrightarrow{\mathbf{r}} \bullet \left( \widehat{\dot{\mathbf{i}}} + 5\widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}} \right) = 5$ is
A) $\frac{10}{9}$
B) $\frac{3}{10}$
C) $\frac{10}{3\sqrt{3}}$
D) $\frac{10}{9}$
Explanation
(c)
Here, $(\widehat{\dot{\mathbf{i}}} - \widehat{\dot{\mathbf{j}}} + 4\widehat{\mathbf{k}}$)∙($\ \widehat{\dot{\mathbf{i}}} + 5\widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}}) = 0$
It means line is parallel to the plane
General point on the line is $(\lambda + 2, - \lambda - 2,4\lambda + 3)$
For $\lambda = 0,$ point on this line is $(2, - 2,3)$ and distance from
$\overrightarrow{\mathbf{r}} \bullet \left( \widehat{\dot{\mathbf{i}}} + 5\widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}} \right) = 5$ is
$d = \left| \frac{2 + 5( - 2) + 3 - 5}{\sqrt{(1)^{2} + (5)^{2} + (1)^{2}}} \right| = \frac{10}{3\sqrt{3}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.