Question

Let $\overrightarrow{a}=a_1\hat{i}+{\alpha }_2\hat{j}+a_3\hat{k},\overrightarrow{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}and\overrightarrow{c}=c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$ be three non-zero vectors such that $\overrightarrow{c}$ is a unit vector perpendicular to both the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$. If the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is $\frac{\pi }{6},$
then ${\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|}^2$ is equal to

• A. 0
• B. 1
• C. $\frac{1}{4}(a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)$
• D. $\frac{3}{4}(a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)(c_1^2+c_2^2+c_3^2)$

Answer 1

The correct option is C

$\frac{1}{4}(a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)$

Since, $(\overrightarrow{a}\times \overrightarrow{b})=|\overrightarrow{a}||\overrightarrow{b}|sin\frac{\pi }{6}.\hat{n}(\overrightarrow{a}\times \overrightarrow{b}).\overrightarrow{c}=\frac{1}{2}|\overrightarrow{a}||\overrightarrow{b}|.\hat{n}.\overrightarrow{c}\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=\frac{1}{2}|\overrightarrow{a}||\overrightarrow{b}|.cos{0}^{\circ }$
$\because \hat{n}$ is perpendicular to both $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ is also a unit vector perpendicular to both $\overrightarrow{a}$ and $\overrightarrow{b}.$
$\therefore {\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|}^2={\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}^2=\frac{1}{4}.|\overrightarrow{a}{|}^2|\overrightarrow{b}{|}^2=\frac{1}{4}(a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)$