Vector Algebra Question 343
Question: A unit vector which is coplanar to vector $ \mathbf{i}+\mathbf{j}+2k $ and $ \mathbf{i}+2\mathbf{j}+\mathbf{k} $ and perpendicular to $ \mathbf{i}+\mathbf{j}+\mathbf{k}, $ is
[IIT 1992; Kurukshetra CEE 2002]
Options:
A) $ \frac{\mathbf{i}-\mathbf{j}}{\sqrt{2}} $
B) $ \pm ,( \frac{\mathbf{j}-\mathbf{k}}{\sqrt{2}} ) $
C) $ \frac{\mathbf{k}-\mathbf{i}}{\sqrt{2}} $
D) $ \frac{\mathbf{i}+\mathbf{j}+\mathbf{k}}{\sqrt{3}} $
Show Answer
Answer:
Correct Answer: B
Solution:
- Let the vector be given as $ a\mathbf{i}+b\mathbf{j}+c\mathbf{k}. $ For this vector to be coplanar with $ \mathbf{i}+\mathbf{j}+2\mathbf{k} $ and $ \mathbf{i}+2\mathbf{j}+\mathbf{k}, $ we will have $ a\mathbf{i}+b\mathbf{j}+c\mathbf{k}=p(\mathbf{i}+\mathbf{j}+2\mathbf{k})+r(\mathbf{i}+2\mathbf{j}+\mathbf{k}) $ This gives, $ a=p+r $ …..(i) $ b=p+2r $ …..(ii) $ c=2p+r $ …..(iii) For the vector $ a\mathbf{i}+b\mathbf{j}+c\mathbf{k} $ to be perpendicular to $ \mathbf{i}+\mathbf{j}+\mathbf{k}, $ we will have $ (a\mathbf{i}+b\mathbf{j}+c\mathbf{k}),.,(\mathbf{i}+\mathbf{j}+\mathbf{k})=0 $
$ \Rightarrow a+b+c=0 $ ……(iv) Adding equation (i) to (iii), we get $ 4p+4r=a+b+c $
$ \Rightarrow 4(p+r)=0\Rightarrow p=-r $ Now with the help of (i), (ii) and (iii), we get $ a=0, $ $ b=r, $ $ c=p=-r $ Hence the required vector is $ r(\mathbf{j}-\mathbf{k}) $ To be its unit vector $ r^{2}+r^{2}=1\Rightarrow r=\pm \frac{1}{\sqrt{2}} $ Hence the required unit vector is, $ \pm \frac{1}{\sqrt{2}}(\mathbf{j}-\mathbf{k}) $ . Trick : Check for option $ \text{(a)},\frac{\mathbf{i}-\mathbf{j}}{\sqrt{2}} $ is a unit vector and perpendicular to $ \mathbf{i}+\mathbf{j}+\mathbf{k} $ . But $ \begin{vmatrix} \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} & 0 \\ 1 & 1 & 2 \\ 1 & 2 & 1 \\ \end{vmatrix} =-\frac{4}{\sqrt{2}}\ne 0 $ . So it is not coplanar with the given vector. Check for option $ \text{(b}),\pm ( \frac{\mathbf{j}-\mathbf{k}}{\sqrt{2}} ) $ is a unit vector and also perpendicular to $ \mathbf{i}+\mathbf{j}+\mathbf{k}, $ $ | ,\begin{matrix} 0 & \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} \\ 1 & 1 & 2 \\ 1 & 2 & 1 \\ \end{matrix}, |=0 $ . So, it is also coplanar with the given vectors.
BETA
