Answer
$(1-t)i+(t^3-t^2)k$
Yes, $a \times b$ is orthogonal to both $a$ and $b$.
Work Step by Step
$a= \lt t,1,1/t \gt$ and $b= \lt t^2,t^2,1 \gt$
$a \times b= \lt t,1,1/t \gt \times \lt t^2,t^2,1 \gt$
$a \times b=(1-t)i+(t^3-t^2)k$
To verify that it is orthogonal to $a$, we will compute:
$(a\times b).a=(1-t,0,t^3-t^2) \cdot (t,1,1/t)=0$
To verify that it is orthogonal to $b$, we will compute:
$(a\times b).b=(1-t,0,t^3-t^2) \cdot (t^2,t^2,1)=0$
Yes, $a \times b$ is orthogonal to both $a$ and $b$.
