Q74 / 160 · TS EAMCET 2020 (11 Sep Shift 2) · Mathematics

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 $|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|=$

  • A
    $3 \sqrt{3}$
  • B
    $\frac{3}{2}$
  • Correct answer
    C
    $\frac{3 \sqrt{3}}{2}$
  • D
    0

Solution

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 $\frac{\pi}{3}$. $\therefore|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|=|\mathbf{a} \times \mathbf{b}||\mathbf{c}| \cdot \sin \frac{\pi}{3}=3 \times 1 \times \frac{\sqrt{3}}{2}=\frac{3 \sqrt{3}}{2}$

Hence, the correct answer is option: 'C'.

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