Question

If an unit vector $(a\hat{i}+b\hat{j})$ is perpendicular to $(\hat{i}-\hat{j})$ then the value a and b are

• A. $\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}$
• B. $\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}$
• C. $-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}$
• D. Can't be determined

Answer 1

Answer: (A), (C)

The correct options are
A $\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}$
C $-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}$

Two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ are perpendicular if $\overrightarrow{A}.\overrightarrow{B}=0\Rightarrow (a\hat{i}+b\hat{j}).(\hat{i}-\hat{j})=0$

$\Rightarrow a-b=0\Rightarrow a=b$
But $\sqrt{a^2+b^2}=1\Rightarrow a=\frac{1}{\sqrt{2}},b=\frac{1}{\sqrt{2}}$ or $a=-\frac{1}{\sqrt{2}},b=-\frac{1}{\sqrt{2}}$