Does $\mathbf{a}\,.\mathbf{a} = \mathbf{a}\,.\mathbf{b}$ mean that $\mathbf{a}=\mathbf{b}$?

Question

I want to intuitively say tht the answer is yes, but if it so happens that $|\mathbf{a}|=|\mathbf{b}|cos(\theta)$, where $\theta$ is the angle between the two vectors, then the equation will be satisfied without the two vectors being the same.

However, my friend keeps telling me that I'm wrong and that this would contradict the given result in our homework question anyway, which tells us that $\mathbf{a}\times\mathbf{b} = \mathbf{a}-\mathbf{b}$ and then asks us to prove $\mathbf{a}=\mathbf{b}$ (the equation in the question was obtained by dotting both sides with $\mathbf{a}$.

Which one of us is wrong, and why?

Answer 1

As a simple counter-example see

$$\begin{pmatrix}1\\2\\3\end{pmatrix}.\begin{pmatrix}1\\2\\3\end{pmatrix}=14 = \begin{pmatrix}1\\2\\3\end{pmatrix}.\begin{pmatrix}1\\5\\1\end{pmatrix}$$ and clearly

$$\begin{pmatrix}1\\2\\3\end{pmatrix} \not = \begin{pmatrix}1\\5\\1\end{pmatrix}.$$

Answer 2

If $a$ is the zero vector and $b$ is not zero you get a contradiction to your statement.

Answer 3

The equation $\mathbf{a} \cdot \mathbf{a} = \mathbf{a} \cdot \mathbf{b}$ is equivalent to $\mathbf{a} \cdot (\mathbf{a}-\mathbf{b})=0$, which is the same as saying that $\mathbf{a}$ is orthogonal to $\mathbf{a} - \mathbf{b}$. And this can of course happen without either $\mathbf{a}$ or $\mathbf{a} - \mathbf{b}$ being the zero vector.

So you cannot conclude that $\mathbf{a} = \mathbf{b}$ (even if you assume that $\mathbf{a} \neq \mathbf{0}$).