Find several normals to the vector represented by (1, 2, 2)^{T}.

- ( 0, 0, 0 )
- ( -2, 1, 0 )
- ( 0, -1, 1 )
- ( -4, 1, 1 )
- ( 0, 1, -1 )
- ... and infinitely many more

These are all normal to the given vector because their dot product with it is zero. These are not unit normals. But even if you were to insist on unit normals, there would still be an infinite number.

Notice the first answer (above). The dot product of the zero vector with the given vector is zero, so the zero vector must be orthogonal to the given vector. This is OK. Math books often use the fact that the zero vector is orthogonal to every vector (of the same type). We probably won't need this; but future math courses might. Of course, the length of the zero vector is zero, so it is not a unit normal.

From analytic geometry you may recall the formula for the slope of a line in two dimensions not parallel to the Y axis:

m = (change in y) / (change in x)

This is the same formula as for the tangent of the angle with the x axis. This formula is related to the reason that the dot product of orthogonal 2D vectors is zero.

- What is the slope of ( 2, 5 )
^{T}? (Call it m_{1}) - What is the slope of ( -5, 2)
^{T}? (Call it m_{2}) - What is the dot product?
- What is m
_{1}times m_{2}?