Question

A unit vector in $xy$-plane that makes an angle of ${45}^{\circ }$ with the vector $\overrightarrow{i}+\overrightarrow{j}$ and an angle of ${60}^{\circ }$ with the vector $3\overrightarrow{i}+4\overrightarrow{j}$ is

• A. $\overrightarrow{i}$
• B. $\frac{(a+\overrightarrow{j})}{\sqrt{2}}$
• C. $\frac{(\overrightarrow{i}-\overrightarrow{j})}{\sqrt{2}}$
• D. none of these

Answer 1

Answer: (D)

none of these

Let the required unit vector in the $xy$-plane be
$r=x\overrightarrow{i}+y\overrightarrow{j}\therefore \left|r\right|=\sqrt{x^2+y^2}=1\Rightarrow x^2+y^2=1$ ...(1)
Since angle between $r$ and vector $\overrightarrow{i}+\overrightarrow{j}$ is ${45}^o$ and the angle between $r$ and vector $3\overrightarrow{i}-4\overrightarrow{j}$ is ${60}^o$,
$\therefore \cos {45}^o=\frac{(x\overrightarrow{i}+y\overrightarrow{j}).(\overrightarrow{i}+\overrightarrow{j})}{|x\overrightarrow{i}+y\overrightarrow{j}||\overrightarrow{i}+\overrightarrow{j}|}$ and $\cos {60}^o=\frac{(x\overrightarrow{i}+y\overrightarrow{j}).(3\overrightarrow{i}-4\overrightarrow{j})}{|x\overrightarrow{i}+y\overrightarrow{j}||3\overrightarrow{i}-4\overrightarrow{j}|}$
$\Rightarrow x+y=1$ ...(2)
and $3x-4y=\frac{5}{2}$ ...(3)
There exists no real value of $x$ and $y$ satisfying equation (1), (2) and (3).

Hence, correct option is D.