Question

If the vectors $\overline{a}$ and $\overline{b}$ are perpendicular to each other, then a vector $\overline{v}$ in terms of $\overline{a}$ and $\overline{b}$ satisfying the equations $\overline{v}.\overline{a}=0,\overline{v}.\overline{b}=1$ and $[\overline{v}$ $\overline{a}$ $\overline{b}]=1$ is

• A. $\frac{\overline{b}}{|\overline{b}{|}^2}+\frac{\overline{a}\times \overline{b}}{|\overline{a}\times \overline{b}{|}^2}$
• B. $\frac{\overline{b}}{|\overline{b}|}+\frac{\overline{a}\times \overline{b}}{|\overline{a}\times \overline{b}{|}^2}$
• C. $\frac{\overline{b}}{|\overline{b}{|}^2}$+$\frac{\overline{a}\times \overline{b}}{|\overline{a}\times \overline{b}|}$
• D. None of these

Answer 1

The correct option is A $\frac{\overline{b}}{|\overline{b}{|}^2}+\frac{\overline{a}\times \overline{b}}{|\overline{a}\times \overline{b}{|}^2}$

Given $\overline{a},\overline{b}$ are perpendicular

$\overline{v}\cdot \overline{a}=0$

$\overline{r}\cdot \overline{b}=1$

$\left[\overline{r}\overline{a}\overline{b}\right]=1$

$\overline{v}\cdot \overline{a}=0$ means $\overline{v}$ is perpendicular to $\overline{a}$

This means $\overline{v}$ is parallel to $\overline{b}$ or $\overline{a}\times \overline{b}$

$\overline{v}=x\overline{b}+y(\overline{a}\times \overline{b})$

$\overline{v}\cdot \overline{b}=1=x{|b|}^2+y\left(0\right)$

$\overline{v}\cdot (\overline{a}\times \overline{b})=1=x\left(0\right)+y{|(\overline{a}\times \overline{b})|}^2$

$v=\frac{\overline{b}}{{|\overline{b}|}^2}+\frac{\overline{a}\times \overline{b}}{{|(\overline{a}\times \overline{b})|}^2}$