Question

If ${\overline{z}}_1=a\hat{i}+b\hat{j}$ and ${\overline{z}}_2=c\hat{i}+d\hat{j}$ are two vectors in $\hat{i}$ and $\hat{j}$ system where $\left|{\overline{z}}_1\right|=\left|{\overline{z}}_2\right|=r$ and ${\overline{z}}_1\cdot {\overline{z}}_2=0$, then ${\overline{\omega }}_1=a\hat{i}+c\hat{j}$ and ${\overline{\omega }}_2=b\hat{i}+d\hat{j}$ satisfy

• A. $\left|\overline{{\omega }_1}\right|=r$
• B. $\left|\overline{{\omega }_2}\right|\ne r$
• C. $\overline{{\omega }_1},\overline{{\omega }_2}=0$
• D. $Noneofthese$

Answer 1

The correct option is A $\left|\overline{{\omega }_1}\right|=r$

The first condition gives $a^2+b^2=r^2$ and $c^2+d^2=r^2$
$z_1\cdot z_2=0$ gives $ac+bd=0$

Substituting $b=\frac{-ac}{d}$ in the above equation, we get

$a^2+\frac{a^2{c}^2}{d^2}=r^2$

$\Rightarrow a^2\left(\frac{c^2+d^2}{d^2}\right)=r^2$

$\Rightarrow a=d$

$\Rightarrow w_1=\sqrt{a^2+c^2}=\sqrt{c^2+d^2}=r$