# If $\vec{w}$ is a vector such that $\vec{w}+(\vec{w}\times \vec{u})=\vec{v}$,then prove that $|(\vec{u}\times \vec{v}).\vec{w}|\leq\frac{1}{2}$

Let $\vec{u}$ and $\vec{v}$ be unit vectors.If $\vec{w}$ is a vector such that $\vec{w}+(\vec{w}\times \vec{u})=\vec{v}$,then prove that $|(\vec{u}\times \vec{v}).\vec{w}|\leq\frac{1}{2}$ and the equality holds if and only if $\vec{u}$ is perpendicular to $\vec{v}$.

I could not solve this question.I dont have any idea how to start with it.Please help me.Thanks.

• Are these vectors specifically in $\Bbb R^3$? – Arthur Nov 6 '15 at 11:11
• This is not specified in the problem. – Vinod Kumar Punia Nov 6 '15 at 11:18
• The cross product only works in $\Bbb R^3$, in all other cases the outer product of two vectors can not be identified again as a vector in the same space, thus $v=w+w×u$ would be impossible. – LutzL Nov 6 '15 at 11:20

Use $$(u×v)·w = (w×u)·v = (v-w)·v\\ \text{ or } = (w×u)·(w+(w×u))=\|w×u\|^2$$ Since $w$ and $w×u$ are orthogonal, one also gets $$1=\|v\|^2=\|w\|^2+\|w×u\|^2=2\|w\|^2-(u·w)^2\\ \implies \|w×u\|^2=1-\|w\|^2=1-\frac12(1+(u·w)^2)=\frac12-\frac12(u·w)^2$$