Question

Let $a=a_{1}i+a_{2}j+a_{3}k,b=b_{1}i+b_{2}j+b_{3}k,$ and $c=c_{1}i+c_{2}j+c_{3}k$ be three non-zero vectors such that $c$ is unit vector perpendicular to both vectors $a$ and $b$. If the angle between vectors $a$ and $b$ 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 $\frac{1}{4}(a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)$

We have, ${\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|}^2={\left[abc\right]}^2={[(a\times b).c]}^2$

$={[\left|a\right|\left|b\right|\sin \frac{\pi }{6}c.c]}^2={\left|a\right|}^2{\left|b\right|}^2{\sin }^2\frac{\pi }{6}.1(\because \left|c\right|=1)$

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