Answer: i - j - k
Step-by-step explanation:
Taking the cross product between two vectors will give you a third vector that is orthogonal(perpendicular) to both vectors.
<1,1,0> x <1,0,1>
[tex]det(\left[\begin{array}{ccc}i&j&k\\1&1&0\\1&0&1\end{array}\right] )[/tex]
the determinate of the matrix: <1,-(1),-1>
or: i - j - k
