Q. The position vector of the point where the line $\overrightarrow{\mathbf{r}} = \widehat{\dot{\mathbf{i}}} - \widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}} + t(\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} - \widehat{\mathbf{k}})$ meets the plane $\overrightarrow{\mathbf{r}}$∙($\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}}) = 5$ is
A) $5\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} - \widehat{\mathbf{k}}$
B) $5\widehat{\dot{\mathbf{i}}} + 3\widehat{\dot{\mathbf{j}}} - 3\widehat{\mathbf{k}}$
C) $2\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}}$
D) $5\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}}$
Explanation
(b)
Given, $\overrightarrow{\mathbf{r}} = (1 + t)\widehat{\dot{\mathbf{i}}} - (1 - t)\widehat{\dot{\mathbf{j}}} + (1 - t)\widehat{\mathbf{k}}$
$and\ \overrightarrow{\mathbf{r}} \bullet \left( \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}} \right) = 5$
Since, they intersect, therefore
$(1 + t) - (1 - t) + (1 - t) = 5$
$\Longrightarrow t = 4$
$\therefore\overrightarrow{\mathbf{r}} = (1 + 4)\widehat{\dot{\mathbf{i}}} - (1 - 4)\widehat{\dot{\mathbf{j}}} + (1 - 4)\widehat{\mathbf{k}}$
$= 5\widehat{\dot{\mathbf{i}}} + 3\widehat{\dot{\mathbf{j}}} - 3\widehat{\mathbf{k}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.