A question regarding dot product

Question

I'm struggling to prove that a (dot product) b < |a| |b| . I have to prove this algebraically and I can't use the equation of dot product involving the cosine.

I wrote each side in terms of the vector components and then squared each side.

Any help would be amazing!

Thanks

Answer 1

Let's think of the case that $\vec{a}$ and $\vec{b}$ are 2-dimensional vectors, for the sake of simplicity of notation. Let's write their components as $\vec{a} = (a_x, a_y)$ and $\vec{b} = (b_x, b_y)$. Proceed as you described in your question, and note that \begin{equation} a_y^2 b_x^2 - 2a_x a_y b_x b_y + a_x^2 b_y^2 = (b_x a_y -a_x b_y)^2 \geq 0. \end{equation}