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Let $\mathbf{a}=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=-\hat{\mathbf{j}}+\hat{\mathbf{k}}$. If $\mathbf{c}$ is a vector such that $\mathbf{a} \cdot \mathbf{c}=|\mathbf{c}|,|\mathbf{c}-\mathbf{a}|=2 \sqrt{2}$ and the angle between $\mathbf{a} \times \mathbf{b}$ and $\mathbf{c}$ is $\frac{\pi}{3}$, then $|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|=$
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Verified Answer
The correct answer is:
$\frac{3 \sqrt{3}}{2}$
Given, $\mathbf{a}=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$, then $|\mathbf{a}|=\sqrt{4+4+1}=3$
Now given $|\mathbf{c}-\mathbf{a}|=2 \sqrt{2}$
Squaring on both sides,
$|\mathbf{c}-\mathbf{a}|^2=8 \Rightarrow|\mathbf{c}|^2+|\mathbf{a}|^2-2 \mathbf{a} \cdot \mathbf{c}=8$
$|\mathbf{c}|^2+9-2|\mathbf{c}|=8 \quad[\because|\mathbf{a}|=3$ and $\mathbf{a} \cdot \mathbf{c}=|\mathbf{c}|]$
$\therefore \quad|\mathbf{c}|^2-2|\mathbf{c}|+1=0$
$(|\mathbf{c}|-1)^2=0 \Rightarrow|\mathbf{c}|-1=0 \Rightarrow|\mathbf{c}|=1$
Now, we find $(\mathbf{a} \times \mathbf{b})=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & -2 & 1 \\ 0 & -1 & 1\end{array}\right|$
$(\mathbf{a} \times \mathbf{b})=\hat{\mathbf{i}}(-2+1)-\hat{\mathbf{j}}(2-0)+\hat{\mathbf{k}}(-2)$
$(\mathbf{a} \times \mathbf{b})=-\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$
$\therefore \quad|\mathbf{a} \times \mathbf{b}|=\sqrt{1+4+4}=3$
Now, given angle between $\mathbf{a} \times \mathbf{b}$ and $\mathbf{c}$ is $\frac{\pi}{3}$.
$\therefore|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|=|\mathbf{a} \times \mathbf{b}||\mathbf{c}| \cdot \sin \frac{\pi}{3}=3 \times 1 \times \frac{\sqrt{3}}{2}=\frac{3 \sqrt{3}}{2}$
Now given $|\mathbf{c}-\mathbf{a}|=2 \sqrt{2}$
Squaring on both sides,
$|\mathbf{c}-\mathbf{a}|^2=8 \Rightarrow|\mathbf{c}|^2+|\mathbf{a}|^2-2 \mathbf{a} \cdot \mathbf{c}=8$
$|\mathbf{c}|^2+9-2|\mathbf{c}|=8 \quad[\because|\mathbf{a}|=3$ and $\mathbf{a} \cdot \mathbf{c}=|\mathbf{c}|]$
$\therefore \quad|\mathbf{c}|^2-2|\mathbf{c}|+1=0$
$(|\mathbf{c}|-1)^2=0 \Rightarrow|\mathbf{c}|-1=0 \Rightarrow|\mathbf{c}|=1$
Now, we find $(\mathbf{a} \times \mathbf{b})=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & -2 & 1 \\ 0 & -1 & 1\end{array}\right|$
$(\mathbf{a} \times \mathbf{b})=\hat{\mathbf{i}}(-2+1)-\hat{\mathbf{j}}(2-0)+\hat{\mathbf{k}}(-2)$
$(\mathbf{a} \times \mathbf{b})=-\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$
$\therefore \quad|\mathbf{a} \times \mathbf{b}|=\sqrt{1+4+4}=3$
Now, given angle between $\mathbf{a} \times \mathbf{b}$ and $\mathbf{c}$ is $\frac{\pi}{3}$.
$\therefore|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|=|\mathbf{a} \times \mathbf{b}||\mathbf{c}| \cdot \sin \frac{\pi}{3}=3 \times 1 \times \frac{\sqrt{3}}{2}=\frac{3 \sqrt{3}}{2}$
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