Question

Let $\overrightarrow{ a }=a_{1} \hat{ i }+a_{2} \hat{ j }+a_{3} \hat{ k }, \overrightarrow{ a }=b_{1} \hat{ i }+b_{2} \hat{ j }+b_{3} \hat{ k }$ and $\overrightarrow{ a }=c_{1} \hat{ i }+c_{2} \hat{ j }+c_{3} \hat{ k }$ be three non-zero vectors such that $\overrightarrow{ c }$ is a unit vector perpendicular to both the vectors $\overrightarrow{ c }$ and $\overrightarrow{ b }$. If the angle between $\overrightarrow{ a }$ and $\overrightarrow{ b }$ is $\frac{\pi}{6}$, then $\begin{vmatrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3}\end{vmatrix}^2$ 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

Answer: (C)

Since, $(\overrightarrow{ a } \times \overrightarrow{ b })=|\overrightarrow{ a }||\overrightarrow{ b }| \sin \frac{\pi}{6} \cdot \hat{ n }$
$(\overrightarrow{ a } \times \overrightarrow{ b }) \cdot \overrightarrow{ c } =\frac{1}{2}|\overrightarrow{ a }||\overrightarrow{ b }| \cdot \hat{ n } \cdot \overrightarrow{ c } $
${[\overrightarrow{ a } \overrightarrow{ b } \overrightarrow{ c }] } =\frac{1}{2}|\overrightarrow{ a }||\overrightarrow{ b }| \cdot \cos 0^{\circ}$
$\because \hat{ n }$ 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 \begin{vmatrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3}\end{vmatrix}^{2} =[\overrightarrow{ a } \overrightarrow{ b } \overrightarrow{ c }]^{2}=\frac{1}{4} \cdot|\overrightarrow{ a }|^{2}|\overrightarrow{ b }|^{2} $
$=\frac{1}{4}\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)\left(b_{1}^{2}+b_{2}^{2}+b_{3}^{2}\right) $