Find a unit length vector which is parallel to the plane with Cartesian equation and perpendicular to the vector .
The plane with Cartesian equation has intercepts . Therefore, the plane is the set of points
So, the vector being parallel to implies is in the span of . Therefore,
for some . Then, perpendicular to implies
Finally, since has unit length we have
This then gives us
Tom said cross product is useful when we need to construct a vector perpendicular to 2 other vectors. Here one vector is the normal vector of the plane and the other one is given.
A is parallel to plane M: x – y + 5z = 1, so A is in the span of M: kS + mT and should be perpendicular to normal vector of M: (1, -1, 5). So let A = (a, b, c) must satisfy equations like:
a – b + 5c = 0 (from A perpendicular to M), a + 2b – 3c = 0 (from A perpendicular to i + 2j -3k) and
(A of length 1).
Can get the solution z^2=9/122 from which can get A can be two vectors as solution.