Vector Algebra Question 197
Question: Let $ \mathbf{a}=2\mathbf{i}+\mathbf{j}+\mathbf{k},\mathbf{b}=\mathbf{i}+2\mathbf{j}-\mathbf{k} $ and a unit vector c be coplanar. If c is perpendicular to a, then c =
[IIT 1999; Pb. CET 2003; DCE 2005]
Options:
A) $ \frac{1}{\sqrt{2}}(-\mathbf{j}+\mathbf{k}) $
B) $ \frac{1}{\sqrt{3}}(-\mathbf{i}-\mathbf{j}-\mathbf{k}) $
C) $ \frac{1}{\sqrt{5}},(\mathbf{i}-2\mathbf{j}) $
D) $ \frac{1}{\sqrt{3}}(\mathbf{i}-\mathbf{j}-\mathbf{k}) $
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Answer:
Correct Answer: A
Solution:
- $ \mathbf{c} $ is coplanar with $ \mathbf{a},,\mathbf{b} $
$ \therefore ,\mathbf{c}=x\mathbf{a}+y\mathbf{b} $
$ \Rightarrow \mathbf{c}=x(2\mathbf{i}+\mathbf{j}+\mathbf{k})+y(\mathbf{i}+2\mathbf{j}-\mathbf{k}) $
$ \Rightarrow \mathbf{c}=(2x+y)\mathbf{i}+(x+2y)\mathbf{j}+(x-y)\mathbf{k} $
$ \because ,\mathbf{a},.\mathbf{c}=0 $
$ \therefore ,2(2x+y)+x+2y+x-y=0 $
$ \Rightarrow y=-2x $
$ \mathbf{c}=-3x\mathbf{j}+3x\mathbf{k}=3x(-\mathbf{j}+\mathbf{k}) $
$ \because ,|\mathbf{c}|,=1 $
$ \therefore ,9x^{2}+9x^{2}=1 $
$ \Rightarrow x=\pm \frac{1}{3\sqrt{2}}\Rightarrow \mathbf{c}=\frac{1}{\sqrt{2}}(-\mathbf{j}+\mathbf{k}) $ .
BETA
