Question

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

• (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

Explanation:
Given:$\overrightarrow{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k},\overrightarrow{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$ be three non-zero vectors such that $\overrightarrow{c}$ is a vector perpendicular to both $\overrightarrow{a}$ and $\overrightarrow{b}$.Angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ = $\frac{\pi }{6}$ We know that $\left|\begin{array}{lll}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|=[(\overrightarrow{a}\times \overrightarrow{b})\cdot \overrightarrow{c}{]}^2$So,$={[|\overrightarrow{a}\times \overrightarrow{b}||\overrightarrow{c}|\cos 0^{\circ }]}^2$$(\because \overrightarrow{a}\times \overrightarrow{b}\text{ is parellel to vector }\overrightarrow{c}\text{ as }\overrightarrow{c}\text{ is perpendicular to both }\overrightarrow{a}\text{ and }\overrightarrow{b})$$={(|\overrightarrow{a}||\overrightarrow{b}|\sin \frac{\pi }{6})}^2$$=|\overrightarrow{a}{|}^2|\overrightarrow{b}{|}^2{\left(\frac{1}{2}\right)}^2$$=\frac{1}{4}|\overrightarrow{a}{|}^2|\overrightarrow{b}{|}^2$We know that,$|\overrightarrow{a}{|}^2=(a_1^2+a_2^2+a_3^2)$$|\overrightarrow{b}{|}^2=(b_1^2+b_2^2+b_3^2)$So,$\frac{1}{4}|\overrightarrow{a}{|}^2|\overrightarrow{b}{|}^2=\frac{1}{4}\left[(a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)\right]$Hence, the correct option is (C).