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

The correct option is B $(0,1,1)$

$\overrightarrow{a}=\hat{i}-\hat{j}+\hat{k},\overrightarrow{c}=-2\hat{i}-\hat{j}+\hat{k}$
Let $\overrightarrow{b}=x\hat{i}+y\hat{j}+z\hat{k}$
$\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{c}$
$\Rightarrow \left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& 1\\ x& y& z\end{array}\right|=-2\hat{i}-\hat{j}+\hat{k}$
$\Rightarrow \hat{i}(-z-y)-\hat{j}(z-x)+\hat{k}(y+x)=-2\hat{i}-\hat{j}+\hat{k}$
Therefore,
$-z-y=-2...\left(1\right)$
$z-x=1...\left(2\right)$
$x+y=1...\left(3\right)$
It is given, $\overrightarrow{a}\cdot \overrightarrow{b}=0$
$\Rightarrow x-y+z=\mathrm{0...}\left(4\right)$
Substracting $\left(1\right)$ from $\left(3\right)$
$x+z+2y=3\Rightarrow x+z=3-2y$
put in equation $\left(4\right)$
$3-2y-y=0\Rightarrow y=1$
From equation $\left(3\right)$, $x=0$
From equation $\left(2\right)$, $z=1$
Hence, $(x,y,z)\equiv (0,1,1)$