Question

If $a=(0,1,-1)$ and $c=(1,1,1)$ are given vectors, then find ${\left|b\right|}^2$, where $b$ satisfies $a\times b+c=0$ and $a\cdot b=3$.

• A. 6
• B. 12
• C. 18
• D. 3

Answer 1

The correct option is A 6

Let $b=xi+yj+zk.$
So $a\times b=(z+y)i-xj-xk$
$a\times b+c=0\Rightarrow z+y+1=0,-x+1=0\Rightarrow x=1$
$a\cdot b=3\Rightarrow y-z=3$
Solving these equations we have $y=1,z=-2.$
Thus $b=(1,1,-2).$ i.e.${\left|b\right|}^2=6$