Unit vector perpendicular to two known vectors

[cjwalle]

Homework Statement

Find a unit vector which is perpendicular to both of the vectors a = 4i + 2j - 3k and b = 2i - 3j + k

c = xi + yj + zk

Homework Equations

a[tex]\bot[/tex]c [tex]\longrightarrow[/tex] a [tex]\bullet[/tex] c

The Attempt at a Solution

Okay, here's what I've done so far.

Take the dot-product of a and c, and b and c
a [tex]\bullet[/tex]b: 4x + 2y -3z = 0
b [tex]\bullet[/tex]b: 2x - 3y + z = 0

(1) 4x + 2y -3z = 0
(2) 2x - 3y + z = 0

I isolate z and get rid of x by multiplying (1) with -2 and (2) with 4, then add them:
(1) -8x -4y = -6z
+
(2) 8x -12y = 5z

-16y = 10z

y/z = 10/16

Which again means that:
y = 10m
z = 16m
where m is a constant and [tex]\neq[/tex] 0

And then I insert this into (1) to find x:
4x + 2(10m) - 3(16m) = 0
4x + 20m - 48m = 0
x = 7m

c = m(7i + 10j + 16k)

For the easiest possible solution, m = 1.
c = 7i + 10j + 16k
As far as I can tell, this is a perfectly valid answer.

However, the answer key has the answer:
(1/9[tex]\sqrt{5}[/tex])(7i + 10j + 16k)

While this does not contradict my solution, that is a far too weird m to have been chosen randomly. Does anyone see how they were thinking?

Thanks

 

[Fightfish]

The question asked for a unit vector ie a vector with magnitude of 1. Hence m must be chosen so that the magnitude of the vector is one.
Also, there is a much much easier way of solving the problem. The cross product of two vectors yields a third vector that is perpendicular to both original vectors. In this case, simply take the cross product with a with b, and adjust to obtain a unit vector.

 

[cjwalle]

The question asked for a unit vector ie a vector with magnitude of 1. Hence m must be chosen so that the magnitude of the vector is one.
Also, there is a much much easier way of solving the problem. The cross product of two vectors yields a third vector that is perpendicular to both original vectors. In this case, simply take the cross product with a with b, and adjust to obtain a unit vector.

*facepalm*

Yes, of course.

And this exercise is from the sub-chapter before cross products, so I figured I'd try to do it the way they wanted me to do it.

Thanks a lot :)


Edit: Just to complete the solution:

| c | = [tex]\sqrt{7^{2} + 10^{2} + 16^{2}}[/tex] = [tex]\sqrt{405}[/tex]
= [tex]\sqrt{81}[/tex][tex]\sqrt{5}[/tex] = 9[tex]\sqrt{5}[/tex]

Thus m = 1/ 9[tex]\sqrt{5}[/tex]