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If $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ are three vectors such that $|\mathbf{a}|=1,|\mathbf{b}|=2$, $|\mathbf{c}|=3$ and $\mathbf{a} \cdot \mathbf{b}=\mathbf{b} \cdot \mathbf{c}=\mathbf{c} \cdot \mathbf{a}=0$, then $\|[\mathbf{a} \mathbf{b} \mathbf{c}] \mid$ is equal to
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The correct answer is:
$6$
For three vectors a, b and c
$|\mathrm{a}|=1,|\mathrm{~b}|=2,|\mathrm{c}|=3$
and $\mathrm{a} \cdot \mathrm{b}=\mathrm{b} \cdot \mathrm{c}=\mathrm{c} \cdot \mathrm{a}=0$
$=\left|\begin{array}{ccc}|\mathrm{a}|^2 & 0 & 0 \\ 0 & |\mathrm{~b}|^2 & 0 \\ 0 & 0 & |\mathrm{c}|^2\end{array}\right| \quad[\because \mathrm{a} \cdot \mathrm{a}=|\mathrm{a}|]$
$\begin{aligned} & =\left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 9 & 9\end{array}\right| \\ & =1(36-0) \\ & \Rightarrow \quad \mid\left[\left.\begin{array}{lll}a & b & c\end{array}\right|^2=36\right. \\ & \therefore \quad\left|\left[\begin{array}{lll}a & b & c\end{array}\right]\right|=6\end{aligned}$
$|\mathrm{a}|=1,|\mathrm{~b}|=2,|\mathrm{c}|=3$
and $\mathrm{a} \cdot \mathrm{b}=\mathrm{b} \cdot \mathrm{c}=\mathrm{c} \cdot \mathrm{a}=0$
$=\left|\begin{array}{ccc}|\mathrm{a}|^2 & 0 & 0 \\ 0 & |\mathrm{~b}|^2 & 0 \\ 0 & 0 & |\mathrm{c}|^2\end{array}\right| \quad[\because \mathrm{a} \cdot \mathrm{a}=|\mathrm{a}|]$
$\begin{aligned} & =\left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 9 & 9\end{array}\right| \\ & =1(36-0) \\ & \Rightarrow \quad \mid\left[\left.\begin{array}{lll}a & b & c\end{array}\right|^2=36\right. \\ & \therefore \quad\left|\left[\begin{array}{lll}a & b & c\end{array}\right]\right|=6\end{aligned}$
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