#cross product vectors
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Hey there Jessica!
This question is a multi-step question.
a) What is PQ? Point Q - Point P right?
So we have -6i + 3j or [-6, 3] as a vector.
b) QR = [9, 18]
QR dot PR is zero, so thus 90 degrees. This can be useful (but optional) because we can use the formula:
|a x b| = |a| * |b| sin90
|a x b| = |a| * |b|
c) PS = 3QR = [27, 54]
We are missing a vector PR to do cross product with PS. However:
PR = PQ + QR
= [-6, 3] + [9, 18]
= [3, 21]
So we have everything we need to do cross product:
PS = [27, 54], PQ = [3, 21]
QP = -PQ = [6, -3], QR = [9, 18] (here is 90°)
Area = (|PS x PQ| + |QP x QR|)/2
Your solution is correct in the end!