Vector Algebra Question 43
Question: Let $ \vec{a}=\hat{i}-\hat{j},\vec{b}=\hat{j}-\hat{k} $ and $ \vec{c}=\hat{k}-\hat{i} $ . If $ \vec{d} $ is a unit vector such that $ \vec{a}\cdot \vec{d}=0=[\vec{b}\vec{c}\vec{d}] $ , then $ \vec{d} $ equals
Options:
A) $ \pm \frac{\hat{i}+\hat{j}-2\hat{k}}{\sqrt{6}} $
B) $ \pm \frac{\hat{i}+\hat{j}-\hat{k}}{\sqrt{3}} $
C) $ \pm \frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}} $
D) $ \pm ,\hat{k} $
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Answer:
Correct Answer: A
Solution:
- [a] Let $ \vec{d}=x\hat{i}+y\hat{j}+z\hat{k} $ Where $ x^{2}+y^{2}+z^{2}=1 $ ?(i)
$ \therefore \vec{a}\cdot \vec{d}=0\Rightarrow x-y=0 $ or $ x=y $ ?(ii) $ [\vec{b},\vec{c},\vec{d}]=0\Rightarrow \begin{vmatrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ x & y & z \\ \end{vmatrix} =0 $ Or $ x+y+z=0 $ Or $ 2x+z=0 $ [Using (ii)] Or z = - 2x ?(iii) From (i), (ii) and (iii), we have $ x^{2}+x^{2}+4x^{2}=1 $
$ \Rightarrow x=\pm \frac{1}{\sqrt{6}} $
$ \therefore \vec{d}=\pm ( \frac{1}{\sqrt{6}}\hat{i}+\frac{1}{\sqrt{6}}\hat{j}-\frac{2}{\sqrt{6}}\vec{k} )=\pm ( \frac{\hat{i}+\hat{j}-2\hat{k}}{\sqrt{6}} ) $
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