How do you find two unit vectors which are orthogonal to the vector 3i-j+2k and parallel to the yz plane?

1 Answer
Jul 21, 2016

General solution is #(+-(2+c, 3 c, -3))/|((2+c, 3 c, -3))|#
Particular solutions for + sign, and c =0 and 1, are
#1/sqrt 13(2, 0, -3) and 1/sqrt 3(1, 1, -1)#, respectively.

Explanation:

A vector parallel to yz-plane is # p=j+ck=(0, 1, c)#, with c at your

choice..

The other given vector is #q=3 i-j+2 k=(3, -1, 2)#

Now, #r=+-(p X q)=+-((0, 1, c) X (3, -1, 2)#

#=+-(2+c, 3 c, -3)# is

orthogonal to both #p and q#..

Such orthogonal unit vectors, for + sign, and c =0 and 1, are

#1/sqrt 13(2, 0, -3) and 1/sqrt 3(1, 1, -1)#, respectively.

The general solution is

#(+-(2+c, 3 c, -3))/|((2+c, 3 c, -3))|#, where c i a parameter.