Vector Algebra Question 378
Question: A vector of length 3 perpendicular to each of the vectors $ 3,\mathbf{i}+\mathbf{j}-4,\mathbf{k} $ and $ 6,\mathbf{i}+5,\mathbf{j}-2,\mathbf{k} $ is
Options:
A) $ 2,\mathbf{i}-2,\mathbf{j}+\mathbf{k} $
B) $ -,2,\mathbf{i}+2,\mathbf{j}+\mathbf{k} $
C) $ 2,\mathbf{i}+2,\mathbf{j}-\mathbf{k} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- Let the vector is $ x\mathbf{i}+y\mathbf{j}+z\mathbf{k}. $ Now according to the conditions, $ \sqrt{x^{2}+y^{2}+z^{2}}=3\Rightarrow x^{2}+y^{2}+z^{2}=9 $ …..(i) $ 6x+5y-2z=0 $ …..(ii) and $ 3x+y-4z=0 $ …..(iii) $ [\because $ it is perpendicular to both vectors, hence by $ a_1b_1+a_2b_2+a_3b_3=0] $ On solving the equation (i), (ii) and (iii), we get $ x=2, $ $ y=-2 $ and $ z=1. $ Therefore, the required vector is $ 2\mathbf{i}-2\mathbf{j}+\mathbf{k}. $ Trick : By inspection, the vector $ 2\mathbf{i}-2\mathbf{j}+\mathbf{k} $ is of length 3 and also perpendicular to the given vectors.
BETA
