Question

If the vector $a$ and $b$ are perpendicular to each other, then a vector $v$ in terms of $a$ and $b$ satisfying the equation $v.a=0,v.b=1$ and $\left[vab\right]=1$ is

• A. $\frac{1}{{\left|b\right|}^2}b+\frac{1}{{|a\times b|}^2}a\times b$
• B. $\frac{b}{\left|b\right|}+\frac{a\times b}{{|a\times b|}^2}$
• C. $\frac{b}{{\left|b\right|}^2}+\frac{a\times b}{{|a\times b|}^2}$
• D. none of these

Answer 1

The correct option is A $\frac{1}{{\left|b\right|}^2}b+\frac{1}{{|a\times b|}^2}a\times b$

We have $a\perp b\Rightarrow a,b,a\times b$ are linearly independent.
$v$ can be expressed uniquely in terms of $a,b$ and $a\times b.$
Let $v=xa+yb+za\times b$ ...(i)
Given that, $:a.b=0,v.a=0,v.b=1,[v,a,b]=1$
$\therefore v.a=x{a}^2$ or $x{a}^2=0$ or $x=0$ ...(ii)
$v.b=xv.b+y{b}^2+zb.(a\times b)$
$\Rightarrow y{b}^2=1$ or $y=\frac{1}{b^2}$ ...(iii)
$v.a\times b=x.0+y.0+z{|a\times b|}^2\Rightarrow z{|a\times b|}^2$
$\Rightarrow z=\frac{1}{{|a\times b|}^2}$ ...(iv)
From (1),(2),(3) and (4), we get
$v=\frac{1}{|b{|}^2}b+\frac{1}{{|a\times b|}^2}a\times b.$