Vector Algebra Question 180
Question: Let $ \mathbf{a}=\mathbf{i}-\mathbf{j},\mathbf{b}=\mathbf{j}-\mathbf{k},\mathbf{c}=\mathbf{k}-\mathbf{i}. $ If $ \mathbf{\hat{d}} $ is a unit vector such that $ \mathbf{a},.,\mathbf{\hat{d}}=0=
[\mathbf{b}\mathbf{c}\mathbf{\hat{d}}], $ then $ \mathbf{\hat{d}} $ is equal to [IIT 1995]
Options:
A) $ \pm \frac{\mathbf{i}+\mathbf{j}-\mathbf{k}}{\sqrt{3}} $
B) $ \pm \frac{\mathbf{i}+\mathbf{j}+\mathbf{k}}{\sqrt{3}} $
C) $ \pm \frac{\mathbf{i}+\mathbf{j}-2\mathbf{k}}{\sqrt{6}} $
D) $ \pm \mathbf{k} $
Show Answer
Answer:
Correct Answer: C
Solution:
- Let $ \mathbf{a}=\mathbf{i}-\mathbf{j}, $ $ \mathbf{b}=\mathbf{j}-\mathbf{k} $ and $ \mathbf{c}=\mathbf{k}-\mathbf{i} $
Let $ \mathbf{\hat{d}}=a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k}, $ $ |\mathbf{\hat{d}}|,=\sqrt{a_1^{2}+a_2^{2}+a_3^{2}}=1 $
$ \Rightarrow a_1^{2}+a_2^{2}+a_3^{2}=1 $ ……(i)
$ \mathbf{a},.\mathbf{\hat{d}}=0\Rightarrow a_1-a_2=0 $ ..(ii)
$ [\mathbf{b},\mathbf{c},\mathbf{\hat{d}}]=0\Rightarrow \mathbf{b},.,(\mathbf{c}\times \mathbf{\hat{d}})=0 $
$ \Rightarrow \begin{vmatrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ a_1 & a_2 & a_3 \\ \end{vmatrix} =-1(-a_3-a_1)-1(-a_2) $
$ \therefore ,a_1+a_2+a_3=0\Rightarrow a_1-a_2+0a_3=0 $ , {from (ii)}
$ \therefore ,\frac{a_1}{0+1}=\frac{a_2}{1-0}=\frac{a_3}{-1-1}\Rightarrow \frac{a_1}{1}=\frac{a_2}{1}=\frac{a_3}{-2}=\lambda $ , (say)
$ \Rightarrow a_1=\lambda , $ $ a_2=\lambda , $ $ a_3=-2\lambda $
$ \therefore ,{{\lambda }^{2}}+{{\lambda }^{2}}+4{{\lambda }^{2}}=1 $ , {from (i)}
$ \Rightarrow 6{{\lambda }^{2}}=1\Rightarrow \lambda =\pm \frac{1}{\sqrt{6}}; $
$ \therefore ,\mathbf{\hat{d}}=\pm \frac{\mathbf{i}+\mathbf{j}-2\mathbf{k}}{\sqrt{6}} $ .
BETA
