Vectors Ques 102
- Let $\overrightarrow{\mathbf{a}}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}$. If $\overrightarrow{\mathbf{c}}$ is a vector such that $\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{c}}=|\overrightarrow{\mathbf{c}}|,|\overrightarrow{\mathbf{c}}-\overrightarrow{\mathbf{a}}|=2 \sqrt{2}$ and the angle between $(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}})$ and $\overrightarrow{\mathbf{c}}$ is $30^{\circ}$, then $\left.\mid \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}\right) \times \overrightarrow{\mathbf{c}} \mid$ is equal to
(1999, 2M)
(a) $\frac{2}{3}$
(b) $\frac{3}{2}$
(c) 2
(d) 3
Show Answer
Answer:
Correct Answer: 102.(b)
Solution:
Formula:
- NOTE In this question, vector $\overrightarrow{\mathbf{c}}$ is not given, therefore, we cannot apply the formulae of $\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}$ (vector triple product)
Now, $|(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \times \overrightarrow{\mathbf{c}}|=|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}||\overrightarrow{\mathbf{c}}| \sin 30^{\circ}……(i)$
Again, $\quad|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 1 & -2 \\ 1 & 1 & 0\end{array}\right|=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$
$\Rightarrow|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|=\sqrt{2^{2}+(-2)^{2}+1}=\sqrt{4+4+1}=\sqrt{9}=3$
Since, $ |\overrightarrow{\mathbf{c}}-\overrightarrow{\mathbf{a}}|=2 \sqrt{2} \quad [given] $
$ \begin{array}{lc} \Rightarrow & |\overrightarrow{\mathbf{c}}-\overrightarrow{\mathbf{a}}|^{2}=8 \\ \Rightarrow & (\overrightarrow{\mathbf{c}}-\overrightarrow{\mathbf{a}}) \cdot(\overrightarrow{\mathbf{c}}-\overrightarrow{\mathbf{a}})=8 \\ \Rightarrow & \overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{c}}-\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{a}}=8 \\ \Rightarrow & |\overrightarrow{\mathbf{c}}|^{2}+|\overrightarrow{\mathbf{a}}|^{2}-2 \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{c}}=8 \\ \Rightarrow & |\overrightarrow{\mathbf{c}}|^{2}+9-2|\overrightarrow{\mathbf{c}}|=8 \\ \Rightarrow & |\overrightarrow{\mathbf{c}}|^{2}-2|\overrightarrow{\mathbf{c}}|+1=0 \\ \Rightarrow & (|\overrightarrow{\mathbf{c}}|-1)^{2}=0 \Rightarrow|\overrightarrow{\mathbf{c}}|=1 \end{array} $
From Eq. (i), $|(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \times \overrightarrow{\mathbf{c}}|=$ (3) (1) $\cdot\left(\frac{1}{2}\right)=\frac{3}{2}$
BETA
