Vectors 2 Question 7
7. Let $\mathbf{a}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}}$ and $\mathbf{c}$ be a vector such that $|\mathbf{c}-\mathbf{a}|=3,|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|=3$ and the angle between $\mathbf{c}$ and $\mathbf{a} \times \mathbf{b}$ is $30^{\circ}$. Then, $\mathbf{a} \cdot \mathbf{c}$ is equal to
(2017 Main)
(a) $\frac{25}{8}$
(b) 2
(c) 5
(d) $\frac{1}{8}$
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Answer:
Correct Answer: 7. (b)
Solution:
- We have, $\mathbf{a}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$
$\Rightarrow \quad|\mathbf{a}|=\sqrt{4+1+4}=3$
$ \text { and } \quad \mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}} $
$ \Rightarrow \quad|\mathbf{b}|=\sqrt{1+1}=\sqrt{2} $
Now, $\quad|\mathbf{c}-\mathbf{a}|=3 \Rightarrow|\mathbf{c}-\mathbf{a}|^{2}=9$
$\Rightarrow \quad(\mathbf{c}-\mathbf{a}) \cdot(\mathbf{c}-\mathbf{a})=9$
$\Rightarrow|\mathbf{c}|^{2}+|\mathbf{a}|^{2}-2 \mathbf{c} \cdot \mathbf{a}=9$
Again,
$ |(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|=3 $
$\Rightarrow|\mathbf{a} \times \mathbf{b}||\mathbf{c}| \sin 30^{\circ}=3$
$ \Rightarrow \quad|\mathbf{c}|=\frac{\mathbf{6}}{|\mathbf{a} \times \mathbf{b}|} $
But $\mathbf{a} \times \mathbf{b}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \ 2 & 1 & -2 \ 1 & 1 & 0\end{array}\right|=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$
$\therefore \quad|\mathbf{c}|=\frac{6}{\sqrt{4+4+1}}=2$
From Eqs. (i) and (ii), we get
$ \begin{array}{rr} & (2)^{2}+(3)^{2}-2 \mathbf{c} \cdot \mathbf{a}=9 \\ \Rightarrow & 4+9-2 \mathbf{c} \cdot \mathbf{a}=9 \\ \Rightarrow & \mathbf{c} \cdot \mathbf{a}=2 \end{array} $
