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Let $\vec{c}$ be a vector coplanar with the unit vectors $\vec{a}, \vec{b}$ and let $\vec{d}$ be the unit vector perpendicular to $\vec{a}, \vec{b}$ and $\vec{c}$. If $[\vec{a} \vec{b} \vec{d}] \vec{c}-[\vec{a} \vec{b} \vec{c}] \vec{d}=\hat{i}+2 \hat{j}+2 \hat{k}$ and the angle between $\vec{a}$ and $\vec{b}$ is $30^{\circ}$, then $|\vec{c}|=$
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The correct answer is:
6
$\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are coplaner
$\vec{d} \perp \vec{a}, \vec{d} \perp \vec{b}, \vec{d} \perp \vec{c}$
$\begin{aligned} & [\vec{a} \vec{b} \vec{d}]] \vec{c}-[\vec{a} \vec{b} \vec{c}] \vec{d} \\ & =(\vec{a} \times \vec{b}) \times(\vec{c} \times \vec{d})\end{aligned}$
$\therefore[\vec{a} \vec{b} \vec{c}]=0$ (as vectors are coplaner)
also $[\vec{a} \vec{b} \vec{d}]=\vec{a} \cdot(\vec{b} \times \vec{d})=(\vec{a} \times \vec{b}) \cdot \vec{d}$
$\begin{aligned} & =|\vec{a} \times \vec{b}| \cdot|\vec{d}| \cdot \cos 0^{\circ} \quad(\because \vec{a} \times \vec{b} \perp \vec{a}, \vec{b} \Rightarrow \vec{a} \times \vec{b} \| \vec{d}) \\ & =|\vec{a}||\vec{b}| \sin 30^{\circ} \cdot|\vec{d}|=\frac{1}{2} \\ & \therefore \frac{1}{2} \vec{c}=\hat{i}+2 \hat{j}+2 \hat{k} \Rightarrow \vec{c}=2 \hat{i}+4 \hat{j}+4 \hat{k} \\ & \therefore|\vec{c}|=6\end{aligned}$
$\vec{d} \perp \vec{a}, \vec{d} \perp \vec{b}, \vec{d} \perp \vec{c}$
$\begin{aligned} & [\vec{a} \vec{b} \vec{d}]] \vec{c}-[\vec{a} \vec{b} \vec{c}] \vec{d} \\ & =(\vec{a} \times \vec{b}) \times(\vec{c} \times \vec{d})\end{aligned}$
$\therefore[\vec{a} \vec{b} \vec{c}]=0$ (as vectors are coplaner)
also $[\vec{a} \vec{b} \vec{d}]=\vec{a} \cdot(\vec{b} \times \vec{d})=(\vec{a} \times \vec{b}) \cdot \vec{d}$
$\begin{aligned} & =|\vec{a} \times \vec{b}| \cdot|\vec{d}| \cdot \cos 0^{\circ} \quad(\because \vec{a} \times \vec{b} \perp \vec{a}, \vec{b} \Rightarrow \vec{a} \times \vec{b} \| \vec{d}) \\ & =|\vec{a}||\vec{b}| \sin 30^{\circ} \cdot|\vec{d}|=\frac{1}{2} \\ & \therefore \frac{1}{2} \vec{c}=\hat{i}+2 \hat{j}+2 \hat{k} \Rightarrow \vec{c}=2 \hat{i}+4 \hat{j}+4 \hat{k} \\ & \therefore|\vec{c}|=6\end{aligned}$
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