Finding vector perpendicular to another vector

[TsAmE]

Homework Statement

How would you find a vector perpendicular to (2,-1,1)?

Homework Equations

None.

The Attempt at a Solution

I tried saying (x,y,z) x (2,-1,1) = 1 (since sin90 max at 1)

but have 3 unknowns, x, y and z so don't know how to get a numerical answer

 

[danago]

Homework Statement

How would you find a vector perpendicular to (2,-1,1)?

Homework Equations

None.

The Attempt at a Solution

I tried saying (x,y,z) x (2,-1,1) = 1 (since sin90 max at 1)

but have 3 unknowns, x, y and z so don't know how to get a numerical answer

When you cross two vectors, the result should also be a vector, so writing (x,y,z) x (2,-1,1) = 1 doesn't really make sense.

Consider using the dot product.

 

[HallsofIvy]

There exist an infinite number of vectors (an entire plane) perpendicular to a given vector in three dimensions. If you want to use the cross product, then it would be the magnitude of the cross product that would be 1. But, as danago says, it would be simplest to use the dot product. $(2, -1, 1)\cdot(x, y, z)= 2x- y+ z= 0$.

That gives a single equation in three unknown values. If you really just want a vector perpendicular to (2, -1, 1), choose any values you want for, say, x and y, and solve for the corresponding z.

If you want to be able to express all vectors perpendicular to (2, -1, 1), solve for z as a function of x and y and use x and y as parameters.