Let $\mathbf{a}=2 \mathbf{i}+\mathbf{j}-2 \mathbf{k}$ and $\mathbf{b}=\mathbf{i}+\mathbf{j}$ be two vectors.…

Let $\mathbf{a}=2 \mathbf{i}+\mathbf{j}-2 \mathbf{k}$ and $\mathbf{b}=\mathbf{i}+\mathbf{j}$ be two vectors. $\mathbf{c}$ is a vector such that $\mathbf{a} \cdot \mathbf{c}=|\mathbf{c}|$ and $|\mathbf{c}-\mathbf{a}|=2 \sqrt{2}$. If the angle between $\mathbf{a} \times \mathbf{b}$ and $\mathbf{c}$ is $30^{\circ}$, then $|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|$ is equal to
  1. $\frac{3}{2}$
  2. $\frac{2}{3}$
  3. 2
  4. $\frac{\sqrt{3}}{2}$

Solution

$\begin{aligned} & \text { Let } \mathbf{a}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}} \\ & \qquad \begin{aligned} |\mathbf{a}| & =\sqrt{2^2+1^2+(-2)^2}=3 \text { and } \mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}} \\ |\mathbf{b}| & =\sqrt{1+1}=\sqrt{2} \\ \mathbf{a} \times \mathbf{b} & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{array}\right|=\hat{\mathbf{i}}(2)-\hat{\mathbf{j}}(2)+\hat{\mathbf{k}}(1)=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}} \\ |\mathbf{a} \times \mathbf{b}| & =\sqrt{2^2+(-2)^2+(1)^2}=3 \end{aligned} \end{aligned}$ Now, $\mathbf{a} \cdot \mathbf{c}=|\mathbf{c}|$ and $|\mathbf{c}-\mathbf{a}|=2 \sqrt{2}$ Squaring both side $\begin{gathered} |\mathbf{c}-\mathbf{a}|^2=(2 \sqrt{2})^2 \\ \Rightarrow \quad\left|\mathbf{c}^2\right|+\left|\mathbf{a}^2\right|-2 \mathbf{c} \cdot \mathbf{a}=8 \\ \Rightarrow|\mathbf{c}|^2+9-2|\mathbf{c}|=8 \Rightarrow|\mathbf{c}|^2-2|\mathbf{c}|+1=0 \end{gathered}$ Let $|\mathbf{c}|=x$ $\begin{aligned} & \Rightarrow \quad x^2-2 x+1=0 \Rightarrow(x-1)^2=0 \\ & \Rightarrow \quad x=1 \Rightarrow|\mathbf{c}|=1 \\ & \text { Now, }|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|=|\mathbf{a} \times \mathbf{b}||\mathbf{c}| \sin 30^{\circ} \\ & \\ & \end{aligned}$

Asked in: MHT CET Full Test 9

Practice more Vector Algebra questions on Aicharya