Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

OACB is a parallelogram. In other words if $\left \|\mathbf{a}+k\mathbf{b} \right \|=1$ ($k\in\mathbb{R}$), prove that
$$\|\mathbf{a}\| \cdot \|\mathbf{b} \| \cdot \sin \theta \leq \|\mathbf{b} \| $$ where $\theta$ is the angle of the two vectors.

Any suggestion?

share|cite|improve this question

1 Answer 1

up vote 0 down vote accepted

If $\| a + kb \| = 1$ for all $k \in \mathbb{R}$, then in particular, you have $\| a + 0 b\| = \| a \| = 1$. The inequality follows immediately since $\sin x \leq |\sin x| \leq 1$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.