Vectors 3 Question 15
15. Let $\overrightarrow{\mathbf{a}}=a _1 \hat{\mathbf{i}}+a _2 \hat{\mathbf{j}}+a _3 \hat{\mathbf{k}}, \overrightarrow{\mathbf{a}}=b _1 \hat{\mathbf{i}}+b _2 \hat{\mathbf{j}}+b _3 \hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{a}}=c _1 \hat{\mathbf{i}}+c _2 \hat{\mathbf{j}}+c _3 \hat{\mathbf{k}}$ be three non-zero vectors such that $\overrightarrow{\mathbf{c}}$ is a unit vector perpendicular to both the vectors $\overrightarrow{\mathbf{c}}$ and $\overrightarrow{\mathbf{b}}$. If the angle between $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ is $\frac{\pi}{6}$, then $\left|\begin{array}{lll}a _1 & a _2 & a _3 \ b _1 & b _2 & b _3 \ c _1 & c _2 & c _3\end{array}\right|^{2}$ is equal to
(1986, 2M)
(a) 0
(b) 1
(c) $\frac{1}{4}\left(a _1^{2}+a _2^{2}+a _3^{2}\right)\left(b _1^{2}+b _2^{2}+b _3^{2}\right)$
(d) $\frac{3}{4}\left(a _1^{2}+a _2^{2}+a _3^{2}\right)\left(b _1^{2}+b _2^{2}+b _3^{2}\right)\left(c _1^{2}+c _2^{2}+c _3^{2}\right)$
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Answer:
Correct Answer: 15. (c)
Solution:
- Since, $\quad(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}})=|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}| \sin \frac{\pi}{6} \cdot \hat{\mathbf{n}}$
$$ \begin{aligned} (\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \cdot \overrightarrow{\mathbf{c}} & =\frac{1}{2}|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}| \cdot \hat{\mathbf{n}} \cdot \overrightarrow{\mathbf{c}} \\ {[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}] } & =\frac{1}{2}|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}| \cdot \cos 0^{\circ} \end{aligned} $$
$\because \hat{\mathbf{n}}$ is perpendicular to both $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ is also a unit vector perpendicular to both $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$.
$\begin{aligned} \quad\left|\begin{array}{lll}a _1 & a _2 & a _3 \ b _1 & b _2 & b _3 \ c _1 & c _2 & c _3\end{array}\right|^{2} & =[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]^{2}=\frac{1}{4} \cdot|\overrightarrow{\mathbf{a}}|^{2}|\overrightarrow{\mathbf{b}}|^{2} \ & =\frac{1}{4}\left(a _1^{2}+a _2^{2}+a _3^{2}\right)\left(b _1^{2}+b _2^{2}+b _3^{2}\right)\end{aligned}$
