Question

If $a=i-j+k,a\cdot b=0,a\times b=c$, where $c^{-}=-2i-j+k$, then $b^{-}$ is equal to

• A. $(1,0,-1)$
• B. $(0,1,1)$
• C. $(-1,-1,0)$
• D. $(-1,0,1)$

Answer 1

Answer: (B)

$(0,1,1)$

Let $b=(x,y,z)$, then
$a\cdot b=0$
$\Rightarrow x+y+z=0....\left(i\right)$
Also, $a\times b=c$
$\Rightarrow \left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& 1\\ x& y& z\end{array}\right|=-2\hat{i}-\hat{j}+\hat{k}$
$\Rightarrow (-z-y)\hat{i}+(x-z)\hat{j}+(y+x)\hat{k}=-2\hat{i}-\hat{j}+\hat{k}$
$\Rightarrow -z-y=-2,x-z=-1,y+x=1....\left(ii\right)$
From Eqs. (i) and (ii), we get
$x=0,y=1,z=1$.