Vector Algebra Question 295
Question: A unit vector a makes an angle $ \frac{\pi }{4} $ with z-axis. If $ \mathbf{a}+\mathbf{i}+\mathbf{j} $ is a unit vector, then a is equal to
[IIT 1988]
Options:
A) $ \frac{\mathbf{i}}{2}+\frac{\mathbf{j}}{2}+\frac{\mathbf{k}}{\sqrt{2}} $
B) $ \frac{\mathbf{i}}{2}+\frac{\mathbf{j}}{2}-\frac{\mathbf{k}}{\sqrt{2}} $
C) $ -\frac{\mathbf{i}}{2}-\frac{\mathbf{j}}{2}+\frac{\mathbf{k}}{\sqrt{2}} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- Let $ \mathbf{a}=l\mathbf{i}+m\mathbf{j}+n\mathbf{k}, $ where $ l^{2}+m^{2}+n^{2}=1. $ a makes an angle $ \frac{\pi }{4} $ with $ z- $ axis.
$ \therefore n=\frac{1}{\sqrt{2}}, $ $ l^{2}+m^{2}=\frac{1}{2} $ ?..(i)
$ \therefore \mathbf{a}=l,\mathbf{i}+m,\mathbf{j}+\frac{\mathbf{k}}{\sqrt{2}} $ $ \mathbf{a}+\mathbf{i}+\mathbf{j}=(l+1)\mathbf{i}+(m+1)\mathbf{j}+\frac{\mathbf{k}}{\sqrt{2}} $ Its magnitude is 1, hence $ {{(l+1)}^{2}}+{{(m+1)}^{2}}=\frac{1}{2} $ …..(ii) From (i) and (ii), $ 2lm=\frac{1}{2}\Rightarrow l=m=-\frac{1}{2} $ Hence $ \mathbf{a}=-\frac{\mathbf{i}}{2}-\frac{\mathbf{j}}{2}+\frac{\mathbf{k}}{\sqrt{2}} $ .
BETA
