Vector question (scalar products)

[jaejoon89]

Find scalars s, t s.t. C - sA - tB is perpendicular to A and B

A = i + j + 2k
B = 2i - j + k
C = 2i - j + 4k

I took the cross product of A and B
(1, 1, 2) x (2, -1, 1) =
|i j k|
|1 1 2|
|2 -1 1|
= 3i + 3j - 3k

OK, that should be perpendicular to both vectors A and B I'm guessing...
But how do I still determine the scalars s and t -- there is only one equation!

 

[danago]

Ill let Q = C - sA - tB.

If Q is perpendicular to both A and B, and the vector (3,3,-3) is also perpendicular to both A and B (as you have calculated), then it should follow that Q is a scalar multiple of (3,3,-3) right?

i.e.
k(3,3,-3) = Q

Now you have a system of 3 equations in 3 variables (s,t,k).

____

Another way that you could solve it would be to realize that you need to find s,t such that Q.A = Q.B = 0 (a zero scalar product implies orthogonality), and form a system of two equations in two variables (s,t).