Show that if and only if A.B=0

Answer 1

To prove the statement "if and only if A.B = 0," we need to show two things:

1. If A.B = 0, then A and B are orthogonal.
2. If A and B are orthogonal, then A.B = 0.

Let's start with statement 1.

1. If A.B = 0, then A and B are orthogonal:

Orthogonal vectors are defined as vectors that have a dot product of zero.
So, if A.B = 0, it means that the dot product of A and B is zero, which implies that A and B are orthogonal.

Statement 1 is proven.

Now, let's move on to statement 2.

2. If A and B are orthogonal, then A.B = 0:

Suppose A and B are orthogonal. By definition, this means that their dot product is zero.

We can express the dot product A.B as:

A.B = |A| |B| cos(theta)

where |A| and |B| are the magnitudes of vectors A and B, respectively, and theta is the angle between them.

Since A and B are orthogonal, cos(theta) = 0, which means that the dot product A.B is equal to zero.

Statement 2 is proven.

Therefore, we have shown both directions of the "if and only if" statement, and we can conclude that A.B = 0 if and only if A and B are orthogonal.