Finding Perpendicular Vector and Area of Triangle Using Cross Product

[meadow]

I am having trouble setting this problem up.
The problem says: Find a vector N that is perpendicular to the plane determined by the points P(0,1,0), Q(-1,1,2), R(2,1,-1), and find the area of triangle PQR.

I know that the cross product of two vectors is perpendicular to the plane of a and b, so do I just cross the three vectors. I tried PxQ and then (PxQ)xR, but I didn't get the correct answer.

Also, how would you find the area of the triangle? I tried finding the distance of PQ and PR, multipying and dividing by 2, but I still didn't get the correct answer.

Any help?

 

[LeonhardEuler]

I am having trouble setting this problem up.
The problem says: Find a vector N that is perpendicular to the plane determined by the points P(0,1,0), Q(-1,1,2), R(2,1,-1), and find the area of triangle PQR.

I know that the cross product of two vectors is perpendicular to the plane of a and b, so do I just cross the three vectors. I tried PxQ and then (PxQ)xR, but I didn't get the correct answer.

Also, how would you find the area of the triangle? I tried finding the distance of PQ and PR, multipying and dividing by 2, but I still didn't get the correct answer.

Any help?

P,Q, and R are points, not vectors. You could find vectors in this plane by looking at the vectors that begin at one of these points and end at another. $\vec{(P-Q)}$ and $\vec{(R-Q)}$ would do. Then take the cross product. As per the area of the triangle, do you remember that the magnitude of the cross product is the area of the parallelagram spanned by the two vectors? What is half of this parallelagram?

 

[meadow]

thank you

thank you.
i was able to do the rest of the problems:)

thanks again.