Vector Algebra Question 257
Question: The vector a coplanar with the vectors i and j, perpendicular to the vector $ \mathbf{b}=4\mathbf{i}-3\mathbf{j}+5\mathbf{k} $ such that $ |\mathbf{a}|,=,|,\mathbf{b}| $ is
Options:
A) $ \sqrt{2},(3\mathbf{i}+4\mathbf{j}) $ or $ -\sqrt{2},(3\mathbf{i}+4\mathbf{j}) $
B) $ \sqrt{2},(4\mathbf{i}+3\mathbf{j}) $ or $ -\sqrt{2},(4\mathbf{i}+3\mathbf{j}) $
C) $ \sqrt{3},(4\mathbf{i}+5\mathbf{j}) $ or $ -\sqrt{3},(4\mathbf{i}+5\mathbf{j}) $
D) $ \sqrt{3},(5\mathbf{i}+4\mathbf{j}) $ or $ -\sqrt{3},(5\mathbf{i}+4\mathbf{j}) $
Show Answer
Answer:
Correct Answer: A
Solution:
- Let $ \mathbf{a}=x\mathbf{i}+y\mathbf{j}, $ then $ \mathbf{a},.,\mathbf{b}=0 $
$ \Rightarrow 4x-3y=0\Rightarrow \frac{x}{3}=\frac{y}{4}\Rightarrow x=3\lambda ,y=4\lambda ,\lambda \in R. $ Now $ |\mathbf{a}|,=,|\mathbf{b}|,\Rightarrow x^{2}+y^{2}=16+9+25 $ $ =9{{\lambda }^{2}}+16{{\lambda }^{2}}=50 $
$ \Rightarrow \lambda =\pm \sqrt{2} $
$ \Rightarrow x=\pm 3\sqrt{2},y=\pm 4\sqrt{2} $ Hence, $ \mathbf{a}=\pm \sqrt{2}(3\mathbf{i}+4\mathbf{j}) $ .
BETA
