Find a vector which is perpendicular to both AB and BC. I only managed to form two equations which is AB.v=0 a

Question

Find a vector which is perpendicular to both AB and BC.

I only managed to form two equations which is AB.v=0 and BC.v=0 where v is the unknown vector and AB and BC are vectors.
I need one more equation as i have three unknowns so what more can you extract out from the question?
one more information is that A,B and C are not collinear.

Information if you want to work out the problem:
OA= i + j + k
OB= 2i + 3j+ 4k
OC= -i - 4j - 7k

Answer 1

AB = OB - OA = 2i + 3j + 4k - (i + j + k) = i + 2j + 3k

BC = OC - OB = -i - 4j - 7k - (2i + 3j + 4k) = -3i - 7j - 11k

You can find a perpendicular vector using the cross-product.

(i + 2j + 3k) × (-3i - 7j - 11k) =

(2(-11) - 3(-7))i - (1(-11) - 3(-3))j + (1(-7) - (2(-3))k =

(-22 + 21)i - (-11 + 9)j + (-7 + 6)k =

-i + 2j - k

If I didn't mess up.

OK, I made one subtraction error and one determinant/cross-product error, both of which I've corrected. Northstar and I agree now, except for the direction of BC (and therefore the direction of the perpendicular). I think I'm right on that count.

Answer 2

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Answer 3

Take the cross product. The result will be a vector perpendicular to both AB and BC. First though, find AB and BC

AB = <2-1,3-1,4-1> = <1,2,3>
BC = <2+1,3+4,4+7> = <3,7,11>

AB X BC = i - 2j + k

Answer 4

Cross product is the best way to go.

You could also get a linear system with:

A dot C= 0
B dot C = 0
(1,1,1) dot C = c1+c2+c3

You can manipulate these to get a solution. But, cross product is the best way.