Vector Algebra Question 165
Question: If $ \mathbf{u}=2,\mathbf{i}+2\mathbf{j}-\mathbf{k} $ and $ \mathbf{v}=6,\mathbf{i}-3,\mathbf{j}+2,\mathbf{k}, $ then a unit vector perpendicular to both u and v is
[MP PET 1987]
Options:
A) $ \mathbf{i}-10\mathbf{j}-18\mathbf{k} $
B) $ \frac{1}{\sqrt{17}},( \frac{1}{5}\mathbf{i}-2\mathbf{j}-\frac{18}{5}\mathbf{k} ) $
C) $ \frac{1}{\sqrt{473}},(7\mathbf{i}-10\mathbf{j}-18\mathbf{k}) $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- $ \mathbf{u}=2\mathbf{i}+2\mathbf{j}-\mathbf{k} $ and $ \mathbf{v}=6\mathbf{i}-3\mathbf{j}+2\mathbf{k} $
Let vector $ \mathbf{c}=c_1\mathbf{i}+c_2\mathbf{j}+c_3\mathbf{k} $ is perpendicular to both $ \mathbf{u} $ and $ \mathbf{v}, $ then $ \mathbf{c},.,\mathbf{u}=0 $
$ \Rightarrow 2c_1+2c_2-c_3=0 $ …(i)
and $ \mathbf{c},.,\mathbf{v}=0\Rightarrow 6c_1-3c_2+2c_3=0 $ …(ii)
Solving equation (i) and (ii) by cross multiplication
$ \frac{c_1}{4-3}=\frac{c_2}{-6-4}=\frac{c_3}{-6-12}=\lambda $ , (say)
$ \Rightarrow \frac{c_1}{1}=\frac{c_2}{-10}=\frac{c_3}{-18}=\lambda $
$ \Rightarrow c_1=\lambda , $ $ c_2=-10\lambda $ and $ c_3=-18\lambda $
Thus $ \mathbf{c}=\lambda (\mathbf{i}-10\mathbf{j}-18\mathbf{k}) $
$ |\mathbf{c}|=\lambda \sqrt{1+100+324}=\lambda \sqrt{425} $
Hence required unit vector is, $ \frac{\mathbf{c}}{|\mathbf{c}|} $
$ =\frac{\lambda (\mathbf{i}-10\mathbf{j}-18\mathbf{k})}{\lambda \sqrt{425}}=\frac{1}{\sqrt{425}}(\mathbf{i}-10\mathbf{j}-18\mathbf{k}) $ = $ \frac{1}{5\sqrt{17}}(\mathbf{i}-10\mathbf{j}-18\mathbf{k})=\frac{1}{\sqrt{17}}( \frac{1}{5}\mathbf{i}-2\mathbf{j}-\frac{18}{5}\mathbf{k} ) $ Aliter : Required vector is $ \frac{\mathbf{u}\times \mathbf{v}}{|\mathbf{u}\times \mathbf{v}|}=\frac{\mathbf{i}-10\mathbf{j}-18\mathbf{k}}{\sqrt{425}}. $
BETA
