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Let $\vec{a}=2 \hat{i}+\hat{j}-2 \hat{k}, \vec{b}=\hat{i}+\hat{j}$. If $\vec{c}$ is a vector such that $\vec{a} \bullet \vec{c}=|\vec{c}|,|\vec{c}-\vec{a}|=2 \sqrt{2}$ and the angle between $\vec{a} \times \vec{b}$ and $\vec{c}$ is $30^{\circ}$, then $|(\vec{a} \times \vec{b}) \times \vec{c}|$ equals:
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Verified Answer
The correct answer is:
$\frac{3}{2}$
$\frac{3}{2}$
$\vec{a}=2 \hat{i}+\hat{j}-2 \hat{k}, \vec{b}=\hat{i}+\hat{j}$
$\Rightarrow|\vec{a}|=3$
and $\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 1 & 0\end{array}\right|=2 \hat{i}-2 \hat{j}+\hat{k}$
$|\vec{a} \times \vec{b}|=\sqrt{4+4+1}=3$
Now, $|\vec{c}-\vec{a}|=2 \sqrt{2} \Rightarrow|\vec{c}-\vec{a}|^2=8$
$\Rightarrow|\vec{c}-\vec{a}| \cdot(\vec{c}-\vec{a})=8$
$\Rightarrow|\vec{c}|^2+|\vec{a}|^2-2 \vec{c} \cdot \vec{a}=8$
$\Rightarrow|\vec{c}|^2+9-2|\vec{c}|=8$
$\Rightarrow(|\vec{c}|-1)^2=0 \Rightarrow|\vec{c}|=1$
$\therefore|(\vec{a} \times \vec{b}) \times \vec{c}|=|\vec{a} \times \vec{b}||\vec{c}| \sin 30^{\circ}=3 \times 1 \times \frac{1}{2}=\frac{3}{2}$
$\Rightarrow|\vec{a}|=3$
and $\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 1 & 0\end{array}\right|=2 \hat{i}-2 \hat{j}+\hat{k}$
$|\vec{a} \times \vec{b}|=\sqrt{4+4+1}=3$
Now, $|\vec{c}-\vec{a}|=2 \sqrt{2} \Rightarrow|\vec{c}-\vec{a}|^2=8$
$\Rightarrow|\vec{c}-\vec{a}| \cdot(\vec{c}-\vec{a})=8$
$\Rightarrow|\vec{c}|^2+|\vec{a}|^2-2 \vec{c} \cdot \vec{a}=8$
$\Rightarrow|\vec{c}|^2+9-2|\vec{c}|=8$
$\Rightarrow(|\vec{c}|-1)^2=0 \Rightarrow|\vec{c}|=1$
$\therefore|(\vec{a} \times \vec{b}) \times \vec{c}|=|\vec{a} \times \vec{b}||\vec{c}| \sin 30^{\circ}=3 \times 1 \times \frac{1}{2}=\frac{3}{2}$
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