## anonymous one year ago Show that there are no vectors "u" and "v" in R^3 such that ||u||=1, ||v||=2, and <v,u>=3.

1. thomas5267

$\langle u,v\rangle=\|u\|\|v\|\cos(\theta)\\ \text{Maximum of }\cos(\theta)\text{ is }1\\ \max(\langle u,v\rangle)=\|u\|\|v\|=2$

2. anonymous

^^ same answer I gave the last time he posted this yesterday

3. anonymous

alternatively we can work with components. i'll sketch an approach down below: consider $$u=(u_1,u_2,u_3),v=(v_1,v_2,v_3)$$ so $$\|u\|=1\implies u_1^2+u_2^2+u_3^2=1\\\|v\|=2\implies v_1^2+v_2^2+v_3^2=4$$ now we look at $$\langle u,v\rangle$$: $$\langle u,v\rangle=u_1v_1+u_2v_2+u_3v_3$$consider: $$(u_1v_1+u_2v_2+u_3v_3)^2=u_1^2v_1^2+u_2^2v_2^2+u_3^2v_3^2+2(u_1u_2v_1v_2+u_1u_3v_1v_3+u_2u_3v_2v_3)\\(u_1^2+u_2^2+u_3^2)(v_1^2+v_2^2+v_3^2)=u_1^2v_1^2+u_2^2v_2^2+u_3^2v_3^2\\\qquad\qquad\qquad\qquad\qquad\qquad\qquad+u_2^2(v_1^2+v_3^2)+u_1^2(v_2^2+v_3^2)+u_3^2(v_1^2+v_2^2)\\(u_1^2+u_2^2+u_3^2)(v_1^2+v_2^2+v_3^2)-(u_1v_1+u_2v_2+u_3v_3)^2=\dots\ge 0\\\implies (u_1^2+u_2^2+u_3^2)(v_1^2+v_2^2+v_3^2)\ge (u_1v_1+u_2v_2+u_3v_3)^2\\\implies \|u\|^2\|v\|^2\ge\langle u,v\rangle^2\\\implies \|u\|\|v\|\ge|\langle u,v\rangle|$$

4. anonymous

I did not mean to re-post this question I apologize