Question

A unit vector in the $xy$-plane that makes an angle of ${45}^0$ with the vector $\hat{i}+\hat{j}$ and an angle of ${60}^0$ with the vector $3\hat{i}+4\hat{j}$ is

• A. $\hat{i}$
• B. $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$
• C. $\frac{\hat{i}-\hat{j}}{\sqrt{2}}$
• D. None of these

Answer 1

Answer: (D)

None of these

Let the unit vector be $\overrightarrow{p}=a\hat{i}+b\hat{j}$, so that $a^2+b^2=1$

Given $(a\hat{i}+b\hat{j}).(\hat{i}+\hat{j})=|a\hat{i}+b\hat{j}||\hat{i}+\hat{j}|\cos {45}^0$

$\Rightarrow a+b=\sqrt{a^2+b^2}\sqrt{2}.\frac{1}{\sqrt{2}}(\because a^2+b^2=1)$ ...(1)

and $(a\hat{i}+b\hat{j})(3\hat{i}+4\hat{j})=1\times 5\times \cos {60}^0\Rightarrow 3a-4b=\frac{5}{2}$ ...(2)

Solving (1) and (2), we get

$a=\frac{13}{14},b=\frac{1}{14}$ but $a^2+b^2=1$ is not satisfied.