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.

  • $\begingroup$ Are these vectors specifically in $\Bbb R^3$? $\endgroup$ – Arthur Nov 6 '15 at 11:11
  • $\begingroup$ This is not specified in the problem. $\endgroup$ – Vinod Kumar Punia Nov 6 '15 at 11:18
  • $\begingroup$ 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. $\endgroup$ – 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 $$ 1=\|v\|^2=\|w\|^2+\|w×u\|^2=2\|w\|^2-(u·w)^2\\ \implies \|w×u\|^2=1-\|w\|^2=1-\frac12(1+(u·w)^2)=\frac12-\frac12(u·w)^2 $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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