Vector Algebra Question 57
Question: A vector whose modulus is $ \sqrt{51} $ and makes the same angle with $ \mathbf{a}=\frac{\mathbf{i}-2\mathbf{j}+2\mathbf{k}}{3},\mathbf{b}=\frac{-,4\mathbf{i}-3\mathbf{k}}{5} $ and $ \mathbf{c}=\mathbf{j}, $ will be
[Roorkee 1987]
Options:
A) $ 5\mathbf{i}+5\mathbf{j}+\mathbf{k} $
B) $ 5\mathbf{i}+\mathbf{j}-5\mathbf{k} $
C) $ 5\mathbf{i}+\mathbf{j}+5\mathbf{k} $
D) $ \pm ,(5\mathbf{i}-\mathbf{j}-5\mathbf{k}) $
Show Answer
Answer:
Correct Answer: D
Solution:
- Let the required vector be $ \alpha =d_1\mathbf{i}+d_2\mathbf{j}+d_3\mathbf{k} $ , where $ d_1^{2}+d_2^{2}+d_3^{2}=51 $ , (given) …..(i) Now, each of the given vectors $ \mathbf{a},\mathbf{b},\mathbf{c} $ is a unit vector $ \cos \theta =\frac{\mathbf{d},.,\mathbf{a}}{|\mathbf{d}|,|\mathbf{a}|}=\frac{\mathbf{d},.,\mathbf{b}}{|\mathbf{d}|,|\mathbf{b}|}=\frac{\mathbf{d},.,\mathbf{c}}{|\mathbf{d}|,|\mathbf{c}|} $ or $ \mathbf{d},.,\mathbf{a}=\mathbf{d},.,\mathbf{b}=\mathbf{d},.,\mathbf{c} $ $ |\mathbf{d}|=\sqrt{51} $ cancels out and $ |\mathbf{a}|=|\mathbf{b}|,=,|\mathbf{c}|,=1 $ Hence, $ \frac{1}{3}(d_1-2d_2+2d_3)=\frac{1}{5}(-4d_1+0d_2-3d_3)=d_2 $
$ \Rightarrow d_1-5d_2+2d_3=0 $ and $ 4d_1+5d_2+3d_3=0 $ On solving, $ \frac{d_1}{5}=\frac{d_2}{-1}=\frac{d_3}{-5}=\lambda $ (say) Putting $ d_1,,d_2 $ and $ d_3 $ in (i), we get $ \lambda =\pm 1 $ Hence the required vectors are $ \pm (5\mathbf{i}-\mathbf{j}-5\mathbf{k}). $ Trick: Check it with the options.
BETA
