Physics - Vectors Question with Solution | TestHub

Given ${\vec{v}}_1=a\hat{i}+b\hat{j}=\hat{n}$ and ${\vec{v}}_2=\hat{i}+\hat{j}$.

Since ${\vec{v}}_1$ is a unit vector, its magnitude is $1$.

$a^2+b^2=1$

Since ${\vec{v}}_1$ and ${\vec{v}}_2$ are perpendicular, their dot product is zero.

$(a\hat{i}+b\hat{j})\cdot (\hat{i}+\hat{j})=0$

$a+b=0\Rightarrow b=-a$

Substituting $b=-a$ in the magnitude equation:

$a^2+(-a{)}^2=1$

$2a^2=1\Rightarrow a=\pm \frac{1}{\sqrt{2}}$

For $a=\frac{1}{\sqrt{2}}$, $b=-\frac{1}{\sqrt{2}}$.

Answer: $\left(\frac{1}{\sqrt{2}}\right),\left(-\frac{1}{\sqrt{2}}\right)$