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

$0$

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$1$

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$\frac{1}{4} ({a_{1}}^{2} + {a_{2}}^{2} + {a_{3}}^{2}) ({b_{1}}^{2} + {b_{2}}^{2} + {b_{3}}^{2})$

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$\frac{3}{4} ({a_{1}}^{2} + {a_{2}}^{2} + {a_{3}}^{2}) ({b_{1}}^{2} + {b_{2}}^{2} + {b_{3}}^{2})$

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Solution

The correct option is C
$\frac{1}{4} ({a_{1}}^{2} + {a_{2}}^{2} + {a_{3}}^{2}) ({b_{1}}^{2} + {b_{2}}^{2} + {b_{3}}^{2})$

Explanation for Correct Answer:
Step 1:Given Data
Let $\vec{a}$ , $\vec{b}$ , $\vec{c}$ be three non zero vectors.
$c$ is a unit vector perpendicular to both $a$ and $b$ .
$\Rightarrow a . c = 0, b . c = 0$ and $|\vec{c}| = {c_{1}}^{2} + {c_{2}}^{2} + {c_{3}}^{2} = 1$
$\vec{a} . \vec{b} = |\vec{a}| |\vec{b}| \cos θ \Rightarrow a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3} = (\sqrt{{a_{1}}^{2} + {a_{2}}^{2} + {a_{3}}^{2}}) (\sqrt{{b_{1}}^{2} + {b_{2}}^{2} + {b_{3}}^{2}}) \cos (\frac{π}{6}) \Rightarrow a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3} = \frac{\sqrt{3}}{2} (\sqrt{{a_{1}}^{2} + {a_{2}}^{2} + {a_{3}}^{2}}) (\sqrt{{b_{1}}^{2} + {b_{2}}^{2} + {b_{3}}^{2}}) [∵ \cos (\frac{π}{6}) = \frac{\sqrt{3}}{2}] \Rightarrow {(a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3})}^{2} = \frac{3}{4} ({a_{1}}^{2} + {a_{2}}^{2} + {a_{3}}^{2}) ({b_{1}}^{2} + {b_{2}}^{2} + {b_{3}}^{2}) \to (i) [squaringonbothsides]$
Step 2: Finding ${|\begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix}|}^{2}$
${|\begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix}|}^{2} = |\begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix}| |\begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix}| = |\begin{matrix} {a_{1}}^{2} + {a_{2}}^{2} + {a_{3}}^{2} & a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3} & a_{1} c_{1} + a_{2} c_{2} + a_{3} c_{3} \\ a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3} & {b_{1}}^{2} + {b_{2}}^{2} + {b_{3}}^{2} & b_{1} c_{1} + b_{2} c_{2} + b_{3} c_{3} \\ c_{1} a_{1} + c_{2} a_{2} + c_{3} a_{3} & c_{1} b_{1} + c_{2} b_{2} + c_{3} b_{3} & {c_{1}}^{2} + {c_{2}}^{2} + {c_{3}}^{2} \end{matrix}| = |\begin{matrix} {a_{1}}^{2} + {a_{2}}^{2} + {a_{3}}^{2} & a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3} & 0 \\ a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3} & {b_{1}}^{2} + {b_{2}}^{2} + {b_{3}}^{2} & 0 \\ 0 & 0 & {c_{1}}^{2} + {c_{2}}^{2} + {c_{3}}^{2} \end{matrix}| [∵ a . c = 0, b . c = 0] = ({c_{1}}^{2} + {c_{2}}^{2} + {c_{3}}^{2}) [({b_{1}}^{2} + {b_{2}}^{2} + {b_{3}}^{2}) ({a_{1}}^{2} + {a_{2}}^{2} + {a_{3}}^{2}) - {(a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3})}^{2}]$
substituting equation $(i)$
$= 1 [({b_{1}}^{2} + {b_{2}}^{2} + {b_{3}}^{2}) ({a_{1}}^{2} + {a_{2}}^{2} + {a_{3}}^{2}) - \frac{3}{4} ({a_{1}}^{2} + {a_{2}}^{2} + {a_{3}}^{2}) ({b_{1}}^{2} + {b_{2}}^{2} + {b_{3}}^{2})] = \frac{1}{4} ({a_{1}}^{2} + {a_{2}}^{2} + {a_{3}}^{2}) ({b_{1}}^{2} + {b_{2}}^{2} + {b_{3}}^{2})$
Hence, the correct answer is option (C).

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Similar questions

Q. Let

\to a = a_{1}^i + a_{2}^j + a_{3}^k, \to b = b_{1}^i + b_{2}^j + b_{3}^k

and

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be three non-zero vectors such that

\to c

is a unit vector perpendicular to both

\to a

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and angle between

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{⎡ ⎢ ⎣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ⎤ ⎥ ⎦}^{2} =

Let $\to a = a_{1}^i + α_{2}^j + a_{3}^k, \to b = b_{1}^i + b_{2}^j + b_{3}^k a n d \to c = c_{1}^i + c_{2}^j + c_{3}^k$ be three non-zero vectors such that $\to c$ is a unit vector perpendicular to both the vectors $\to a$ and $\to b$ . If the angle between $\to a$ and $\to b$ is $\frac{π}{6},$
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