Find all unit vectors in the plane determined by u=(1,-2,1) & v=(2,1,0), that are perpendicular to (-1, 0, 1)?

Answer 1

The set of all such vectors are perpendicular both to the given vector w = <-1, 0, 1>, and also the normal vector of the plane since they lie in the plane.

If the vector w is perpendicular to the plane defined by u and v, that is, if it is the normal vector of the plane, then there are an infinite number of solutions. If it is not, then there are two solutions. By inspection I see that the solution set of vectors are:

±u / || u || = <1, -2, 1>/√6 and <-1, 2, -1>/√6
______________

But suppose we didn't notice that.

The normal vector n, of the plane is perpendicular to both u and v. Take the cross product.

n = u X v = <1, -2, 1> X <2, 1, 0> = <-1, 2, 5>

The desired vectors are perpendicular to both n and w. Take the cross product.

t = n X w = <-1, 2, 5> X <-1, 0, 1> = <2, -4, 2>

Any non-zero multiple of t is also perpendicular to n and w . Divide by 2.

t = <1, -2, 1> = u

If we had divided by -2, we would have gotten -u.

So the answer is:

±u / || u || = <1, -2, 1>/√6 and <-1, 2, -1>/√6

As we saw above.