Q. If $\overrightarrow{\mathbf{A}} = \widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} + 3\widehat{\mathbf{k}}\ ,\ \overrightarrow{\mathbf{B}} = \widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}}$ and $\overrightarrow{\mathbf{C}} = 3\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}},$ evaluate $t$, if the vector $(\overrightarrow{\mathbf{A}} + t\overrightarrow{\mathbf{B}})$ and $\overrightarrow{\mathbf{C}}\$are mutually perpendicular.
A) 5
B) 4
C) 1
D) 2
Explanation
(a)
Since, $\left( \overrightarrow{\mathbf{A}} + t\overrightarrow{\mathbf{B}} \right) \bullet \overrightarrow{\mathbf{C}} = 0$ [given]
$\Longrightarrow \left\lbrack (1 - t)\widehat{\dot{\mathbf{i}}} + (2 + 2t)\widehat{\dot{\mathbf{j}}} + (3 + t)\widehat{\mathbf{k}} \right\rbrack \bullet \left( 3\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} \right) = 0$
$\Longrightarrow 3(1 - t) + (2 + 2t) = 0 \Longrightarrow t = 5$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.