# Vectors a,b |a+kb|=1 .Show that |a||b|sinx<=b, x:angle of vectors a,b

1. Oct 19, 2011

### 3.1415926535

1. The problem statement, all variables and given/known data

Let there be two vectors $$\mathbf{OA},\mathbf{OB}\neq\mathbf{0}$$If $$\exists k\in \mathbb{R}$$ such as that $$\left \| \mathbf{OA} +k\mathbf{OB}\right \|=1$$ show that $$Area(OACB)\leq\left \| \mathbf{OB} \right \|$$ (OACB:parallelogram)

2. Relevant equations
None

3. The attempt at a solution

I proved that we need to show that $$\left \|\mathbf{a}\right \| \left \|\mathbf{b}\right \| \sin(\theta )\leq \left \|\mathbf{b} \right \|$$ where θ:angle of vectors a=ΟΑ,b=ΟΒ but after that I am stuck.
Any suggestions? Any hints on how I should proceed?

Last edited: Oct 19, 2011
2. Oct 19, 2011

### 3.1415926535

Nevermind, I solved it. Here is the solution

First of all,
$$\left \| \mathbf{a} +k\mathbf{b}\right \|=1\Leftrightarrow (\mathbf{a} +k\mathbf{b})^{2}=1\Leftrightarrow \mathbf{a}^{2} +k^{2}\mathbf{b}^{2}+2k\left \langle {\mathbf{a} ,\mathbf{b} } \right\rangle =1\Leftrightarrow\left \langle {\mathbf{a} ,\mathbf{b} } \right\rangle^{2}=\frac{(1-\mathbf{a}^{2} -k^{2}\mathbf{b}^{2} )^{2}}{4k^2} (1)$$

We need to show that
$$\left \|\mathbf{a}\right \| \left \|\mathbf{b}\right \| \sin(\theta )\leq \left \|\mathbf{b} \right \|\Leftrightarrow \left \|\mathbf{a}\right \|^{2} \left \|\mathbf{b}\right \|^{2} \sin(\theta )^{2}\leq \left \|\mathbf{b} \right \|^{2}\Leftrightarrow \left \|\mathbf{a}\right \|^{2} \left \|\mathbf{b}\right \|^{2}- \left \|\mathbf{a}\right \|^{2} \left \|\mathbf{b}\right \|^{2}\cos(\theta )^{2}\leq\left \|\mathbf{b} \right \|^{2}\ \Leftrightarrow\left \|\mathbf{a}\right \|^{2} \left \|\mathbf{b}\right \|^{2}-\left \|\mathbf{b} \right \|^{2}\leq\left\langle {\mathbf{a} ,\mathbf{b} } \right\rangle^{2}(2)$$

Finally,
$$(2)\overset{(1)}{\rightarrow}\mathbf{a}^{2} \mathbf{b}^{2}-\mathbf{b} ^{2}\leq\frac{(1-\mathbf{a}^{2}-k^{2}\mathbf{b}^{2}) ^{2}}{4k^2}\Leftrightarrow (1-\mathbf{a}^{2}+k^{2}\mathbf{b}^{2})^{2}\geq 0$$

which is true!