# Thread: Orthogonal Vectors in a plane

1. ## Orthogonal Vectors in a plane

Find two orthogonal vectors in the plane containing the points A(-5,3,-4) B(-4,-3,-1) and C(-2,5,3)

Ok so i have that
AB = <1,-6,3>
AC = <3,2,7>

AB X AC = n = determinant of the 3x3 matrix = -48i + 2j +20k

so i need u and v such that u . v = 0 and u . n = 0 and v . n =0
i have a feeling it should be easy but i can't figure it out?

2. ## Re: Orthogonal Vectors in a plane

Hey linalg123.

Hint: If you need two orthogonal vectors that lie in the plane I suggest you use one of the vectors (like AB), then given the plane normal take AB X N and get the second vector that is both orthogonal to AB and the Normal vector.

3. ## Re: Orthogonal Vectors in a plane

thanks i knew it was gonna be obvious!

4. ## Re: Orthogonal Vectors in a plane

As an alternative method, use the fact that any vector in the plane of the triangle can be expressed as some linear combination of (say) $\underline{AB} \text{ and } \underline{AC}$.

Suppose then that a suitable vector is $\alpha\underline{AB}+\beta\underline{AC}.$

For this to be perpendicular to $\underline{AB},$ its scalar product with $\underline{AB}$ has to equal zero.

The result from that allows you to write either $\alpha \text{ or } \beta$ in terms of the other.

Remember that your answer is not going to be unique, if you have a vector that is perpendicular to $\underline{AB},$ then any (non-zero) scalar multiple will also be perpendicular to $\underline{AB},$ so you don't need an actual value for $\alpha \text{ or } \beta$.