# Finding a vector perpendicular to two vectors

1. Sep 7, 2009

### Kevin2341

1. The problem statement, all variables and given/known data
"A vector perpendicular to both vectors V & W
Vector V = 3i + 2j - 2k
Vector W = 4i -3j + k

2. Relevant equations
Cross product equation

3. The attempt at a solution
The way our teacher tried to teach us was to do the cross product of vectors V x W
Which supposedly equals (area of parallelogram)n-hat (or carrot, what ever you call unit vectors).

But the only example my teacher did this was for an extremely simple set of vectors dealing only with the i and j components and no k component. So basically she went and did V x W for this extremely simple vector problem, and when she chose her perpendicular vector to both V and W, she took a shortcut and just choose k to be the perpendicular vector, which makes perfect sense because it is so simple, but it doesn't help us at all when it comes to trying to do more complex variations of this problem.

2. Sep 7, 2009

### smk037

Vector V = 3i + 2j - 2k
Vector W = 4i -3j + k

I think you can just multiply them as you would scalar numbers, just keeping in mind
i × j = k
j × k = i
k × i = j
i × i = j × j = k × k = 0

so you would get
(3i + 2j - 2k)(4i -3j + k)
12(0) + (3i)(-3j) + ... + (-2)(0)

I hope I'm remembering this correctly and not providing you with wrong information

3. Sep 8, 2009

### Kevin2341

Oh also for the work I have done so far...

I calculated the area of my parallelogram created by the two vectors to be ||V x W||, so its a scalar value, which makes sense. But where I am getting stuck is where to begin with finding the perpendicular vector, I can't just choose a vector going off in a straight component direction because the vectors sort of make a down-ward facing V shape which is at a slant (left side is 2 units higher than right side). So I know that this vector is shooting off in some direction. I can imagine it really easy in my head, its just being able to come up with an actual vector that I'm stuck with, because I have nothing to work with, at least that I know of.

4. Sep 8, 2009

### Kevin2341

Hmm, okay so the vector you came up with is 14i - 6j - 2k, which can be reduced to 7i - 3j - k

But as I remember, is a vector is perpendicular to another vector, their dot product = 0. Using this vector you just calculated, neither V or W = 0 when DP'ed with the perpendicular vector (lets call it P)

5. Sep 8, 2009

### ehild

Don't forget, the vector product is not commutative: a x b = - b x a. So

i × j = k, j × i = -k
j × k = i, k × j = -i
k × i = j, i × k = -j

ehild

6. Sep 8, 2009

### Kevin2341

Okay, I gotta ask, are you guys doing the Cross Product here? Because I haven't ever seen these things before. For cross products we use determinates

so you have
|..i......j......k.....|
|xi(1) yj(1) zk(1)|
|xi(2) yj(2) zk(2)|

= (yj(1)zk(2) - yj(2)zk(1))i - (xi(1)zk(2) - xi(2)zk(1))j - (xi(1)yj(2) - xi(2)zk(1))k

7. Sep 8, 2009

### VeeEight

Yes they are doing cross products. You can do them several ways (your way is correct too).