Vector Algebra Question 387
Question: A unit vector in the $ xy- $ plane which is perpendicular to $ 4\mathbf{i}-3\mathbf{j}+\mathbf{k} $ is
[RPET 1991]
Options:
A) $ \frac{\mathbf{i}+\mathbf{j}}{\sqrt{2}} $
B) $ \frac{1}{5}(3\mathbf{i}+4\mathbf{j}) $
C) $ \frac{1}{5},(3\mathbf{i}-4\mathbf{j}) $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- $ x^{2}+y^{2}=1 $ Let vector be $ x\mathbf{i}+y\mathbf{j}, $ then $ 4x-3y=0 $
$ \Rightarrow 4x=3y\Rightarrow x=\frac{3}{5},y=\frac{4}{5}, $ Hence the required vector is $ \frac{1}{5}(3\mathbf{i}+4\mathbf{j}). $
BETA
