Vectors 5 Question 6
6. A unit vector coplanar with $\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and perpendicular to $\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ is
(1992, 2M)
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Answer:
Correct Answer: 6. $\left(\frac{5}{3} \hat{\mathbf{i}}, \frac{2}{3} \hat{\mathbf{j}}, \frac{2}{3} \hat{\mathbf{k}}\right)$
Solution:
- Any vector coplanar with $\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ is given by
$$ \begin{aligned} \overrightarrow{\mathbf{a}} & =x(\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}})+y(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}) \\ & =(x+y) \hat{\mathbf{i}}+(x+2 y) \hat{\mathbf{j}}+(2 x+y) \hat{\mathbf{k}} \end{aligned} $$
This vector is perpendicular to $\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$, if
$$ \begin{array}{rlrl} & & (x+y) 1+(x+2 y) 1+(2 x+y) 1 & =0 \\ \Rightarrow & 4 x+4 y=0 \quad \Rightarrow \quad-x & =y \end{array} $$
$\therefore \quad \overrightarrow{\mathbf{a}}=-x \hat{\mathbf{j}}+x \hat{\mathbf{k}}=x(-\hat{\mathbf{j}}+\hat{\mathbf{k}}) \Rightarrow|\overrightarrow{\mathbf{a}}|=\sqrt{2}|x|$
Hence, the required unit vector is $\hat{\mathbf{a}}= \pm \frac{1}{\sqrt{2}} \cdot(-\hat{\mathbf{j}}+\hat{\mathbf{k}})$
