Calc III Help?

Lardalish

Ok, Im studying for my exam that is in..... 6 hours. Im going over my tests and I dont have the solution to this one and I got it really wrong so I was wondering if I could get some help on how to solve it.


Find the vector orthogonal to the plane through the points P(0,-2,0) Q(4,1,-2) R(5,3,1)

Ok, I got vectors PQ(4,3,-2) and PR(5,5,1) and then found the crossproduct -7i-6j+5k, so thats my normal vector that Ill be using. This is where Im stuck. Ive got the equation: a(dot)b=|a||b|cos(theta) but Im not sure how to solve that for the b vector. Maybe its just cause its late but I could use some help.

Also, right now Im drawing a blank for an easy way to find a vector orthogonal to those two, all I can think of is trial and error and I know thats not right.

This probly wont be on the exam but I would rather have my bases covered. Thanks guys!

Clipse

The normal vector is orthogonal to those two. (Assuming that by "those two" you meant PQ and PR.)

In general, AxB is orthogonal to A and B.

Kerbob97

Damn it! Do you have any idea how much alcohol I have had to drink to purge those foul thoughts of Calc III from my head!?!?


Seriously though, good luck with the final!

Lardalish

Clipse wrote: »

The normal vector is orthogonal to those two. (Assuming that by "those two" you meant PQ and PR.)

In general, AxB is orthogonal to A and B.

Well damnit.

Thanks for the help.



Also, thanks for the good wishes from Kerbob, I hope I do well. Oh god I hope I do well.

BlazeFire

Hopefully I'm not too late. It's been 8 months so I don't remember specifics. Vectors PQ and PR both lie in the same plane. It's easy to find what those vectors are, you just subtract one point from the other. Then, if you go PQ x PR, you'll get a vector that is normal to the plane PQ and PR are in. This resultant vector will therefore also be normal to the plane. Be careful though, there are two orthogonal vectors, one from each side of the plane.

Good luck.