1. ## Vector question:

"Find a unit vector parallel to $i + rj + 2k$ and perpendicular to $2i + 2j - k$."

If I call the unit vector U(a,b,c) , then kU (k=some constant) is equal to the parallel vector. U dotted with the second vector should equal zero to be perpendicular to each other.

The big problem arises: there are so many equations and unknowns, that everything becomes impossible! Can anyone help me figure out the smartes way to solve this question?

2. Originally Posted by karldiesen
"Find a unit vector parallel to $i + rj + 2k$ and perpendicular to $2i + 2j - k$."

If I call the unit vector U(a,b,c) , then kU (k=some constant) is equal to the parallel vector. U dotted with the second vector should equal zero to be perpendicular to each other.

The big problem arises: there are so many equations and unknowns, that everything becomes impossible! Can anyone help me figure out the smartes way to solve this question?
First, a "unit vector parallel to $i+ rj+ 2k$" can be "perpendicular to $2i+ 2j- k$" only if $i+ rj+ 2k$ is perpendicular to $2i+ 2j- k$ itself. So first take the dot product of those two vectors and set it equal to 0 to find r. Then find the unit vector in that direction.

You could, of course, do it your way. Since you say you have "too many equations" and "too many unknowns", I suspect you are not using the fact that U= ai+ bj+ c, is parallel to i+ rj+ 2k and so must be of the form ci+ crj+ 2ck. You have only two unknowns and two equations: the fact that the vector is perpendicular to 2i+ 2j- k gives you 2c+ 2cr-c= 0 and that its length, $\sqrt{c^2+ c^2r^2+ 2c^2}= |c|\sqrt{1+ r^2+ 4}= 1$. In the first equation "c" immediately cancels leaving a simple equation for r. So that boils down to what I said above.