Question

Find the vector, not with determinants, but by using properties of cross products.
k × (i − 5j)

Answer 1

what a waste of time. Well, you know that
|axb| = |a| |b| sinθ
and is ⊥ to both a and b
|k| = 1
|i-5j| = √26
Now, we know that k ⊥ i-5j, so θ=90°
The cross product will also lie in the x-y plane, and it is easy to show that it will be parallel to -5i+j (think of rotating i-5j). It looks like our magnitude is already ok,
So, k × (i − 5j) = -5i+j

You can verify using determinants if you like.