Vectors 3 Question 13
13. Let $\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}-\hat{\mathbf{j}}, \overrightarrow{\mathbf{b}}=\hat{\mathbf{j}}-\hat{\mathbf{k}}, \overrightarrow{\mathbf{c}}=\hat{\mathbf{k}}-\hat{\mathbf{i}}$. If $\overrightarrow{\mathbf{d}}$ is a unit vector such that $\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{d}}=0=[\overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}} \overrightarrow{\mathbf{d}}]$, then $\overrightarrow{\mathbf{d}}$ equals
(1995, 2M)
(a) $\pm \frac{\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}}{\sqrt{6}}$
(b) $\pm \frac{\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}}{\sqrt{3}}$
(c) $\pm \frac{\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}}{\sqrt{3}}$
(d) $\pm \hat{\mathbf{k}}$
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Answer:
Correct Answer: 13. (a)
Solution:
- Let
$$ \overrightarrow{\mathbf{d}}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}} $$
where, $x^{2}+y^{2}+z^{2}=1$
$[\because \overrightarrow{\mathbf{d}}$ being unit vector]
Since, $\quad \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{d}}=0$
$\Rightarrow \quad x-y=0 \quad \Rightarrow \quad x=y$
Also, $\quad[\overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}} \overrightarrow{\mathbf{d}}]=0$
$\Rightarrow \quad\left|\begin{array}{ccc}0 & 1 & -1 \ -1 & 0 & 1 \ x & y & z\end{array}\right|=0 \Rightarrow x+y+z=0$
$$ \Rightarrow \quad 2 x+z=0 \quad \text { [from Eq. (ii)] …(iii) } $$
From Eqs. (i), (ii) and (iii),
$$ \begin{aligned} x^{2}+x^{2}+4 x^{2} & =1 \Rightarrow x= \pm \frac{1}{\sqrt{6}} \\ \therefore \quad \overrightarrow{\mathbf{d}} & = \pm \frac{1}{\sqrt{6}}(\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \end{aligned} $$
