# Homework Help: Vector perpendicular to only 1 vector

1. Oct 27, 2012

### Ruzic

1. The problem statement, all variables and given/known data
The vector a = [2,-5,6] is given. Determine one vector that is perpendicular to a.

2. Relevant equations

Cross product or dot product

3. The attempt at a solution

I messed around with the dot product by making the right side 0 because cos 0 = 90 but i dont think i can do anything with just 1 equation. For the cross product i got [-5z - 6y, 6x - 2z, 2y +5x] and idk where to take it from there.

i thought maybe subbing this into a x b = |a||b|Sin(1) so

[-5z - 6y, 6x - 2z, 2y +5x] = (√65)(√x^2 + y^2 + z^2)

but idk if theres any point in doing that

If you know how to do this please help :)

2. Oct 27, 2012

### Staff: Mentor

Show us what you did. The dot product is the right way to go.

3. Oct 27, 2012

### Ruzic

i just got 2x - 5y +6z = 0

our teacher said we can make z any number we want then solve which works for finding a normal to a plane but if i do that here everything just cancels out

4. Oct 27, 2012

### Staff: Mentor

Since you have one equation in three unknowns, there are two free variables. Pick values for any two variables, and then solve for the remaining variable.

The values for x, y, and z will give you a vector that is perpendicular to <2, -5, 6>.

5. Oct 27, 2012

### aralbrec

Sometimes it helps to visualize what you are doing. Given one vector, can you find a perpendicular one?

Try this experiment. Take two pens, each will represent a vector. You are given the direction of one of them. This is pen number one, set it on the desk so it points vertically. Can you position the second pen to make it perpendicular to the first vertical one? Sure you can, lay the pen horizontally on the table. But you can rotate that pen on the table at any angle and it will still be perpendicular to the first one. There are an infinite number of solutions; your job is to find one of them.

Many possible solutions shows up in math as having too many unknowns for the number of equations you have. A previous poster stated you have one equation and three unknowns, which means you can freely choose two values and solve for the third. The number of free variables will tell you the dimensionality of your possible solutions. Two free variables means your solutions will be 2d (ie lie in a plane, just like your second pen).

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