Q. The two vectors $\{\ \overrightarrow{\mathbf{a}}\mathbf{=}2\widehat{\mathbf{i}}\mathbf{+}\widehat{\mathbf{j}}\mathbf{+}3\widehat{\mathbf{k}}\mathbf{,\ }\overrightarrow{\mathbf{b}}\mathbf{=}4\widehat{\mathbf{i}}\mathbf{-}\lambda\widehat{\mathbf{j}}\mathbf{+}6\widehat{\mathbf{k}}\ \}$ are parallel, if $\lambda$ is equal to
A) 2
B) $- 3$
C) 3
D) $- 2$
Explanation
(d)
For a parallel $\overrightarrow{\mathbf{a}}\mathbf{\times}\overrightarrow{\mathbf{b}} = 0$
$\left| \begin{matrix}
\widehat{\mathbf{i}} & \widehat{\mathbf{j}} & \widehat{\mathbf{k}} \\
2 & 1 & 3 \\
4 & - \lambda & 6
\end{matrix} \right| = 0$
$\Rightarrow \widehat{\mathbf{i}}(6 + 3\lambda) - \widehat{\mathbf{j}}(0) + \widehat{\mathbf{k}}( - 2\lambda - 4) = 0$
$= 0 \bullet \widehat{\mathbf{i}} + 0 \bullet \widehat{\mathbf{j}} + 0 \bullet \widehat{\mathbf{k}}$
$\therefore 6 + 3\lambda = 0\ \Rightarrow \ \lambda = - 2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.