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

[3.1415926535]

Homework Statement

Let there be two vectors \mathbf{OA},\mathbf{OB}\neq\mathbf{0}If <br /> \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)

Homework Equations

None

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?

 

[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}\<br /> \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,
<br /> <br /> (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 <br />

which is true!