If $\overrightarrow{\mathrm{A}}=a_{1} \hat{\imath}+a_{2} \hat{\jmath}$ and…
If $\overrightarrow{\mathrm{A}}=a_{1} \hat{\imath}+a_{2} \hat{\jmath}$ and $\overrightarrow{\mathrm{B}}=b_{1} \hat{\imath}+b_{2} \hat{\jmath}$ are perpendicular to each other then
When two vectors are perpendicular, the cross product of the two vectors is zero
A. $B=0$
$a_1 b_1+a_2 b_{\overline{2}} 0$
$-a_1 b_1=a_2 b_2$
$\frac{\mathrm{a}_1}{\mathrm{~b}_2}=-\frac{\mathrm{a}_2}{\mathrm{~b}_1}$