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.
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?
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.
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) .
Suppose then that a suitable vector is
For this to be perpendicular to its scalar product with has to equal zero.
The result from that allows you to write either in terms of the other.
Remember that your answer is not going to be unique, if you have a vector that is perpendicular to then any (non-zero) scalar multiple will also be perpendicular to so you don't need an actual value for .