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chrisplusian
 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.
chrisplusian
 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.

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thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1\[ \langle u,v\rangle=\u\\v\\cos(\theta)\\ \text{Maximum of }\cos(\theta)\text{ is }1\\ \max(\langle u,v\rangle)=\u\\v\=2 \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0^^ same answer I gave the last time he posted this yesterday

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0alternatively 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: $$(u_1v_1+u_2v_2+u_3v_3)^2=u_1^2v_1^2+u_2^2v_2^2+u_3^2v_3^2+2(u_1u_2v_1v_2+u_1u_3v_1v_3+u_2u_3v_2v_3)\\(u_1^2+u_2^2+u_3^2)(v_1^2+v_2^2+v_3^2)=u_1^2v_1^2+u_2^2v_2^2+u_3^2v_3^2\\\qquad\qquad\qquad\qquad\qquad\qquad\qquad+u_2^2(v_1^2+v_3^2)+u_1^2(v_2^2+v_3^2)+u_3^2(v_1^2+v_2^2)\\(u_1^2+u_2^2+u_3^2)(v_1^2+v_2^2+v_3^2)(u_1v_1+u_2v_2+u_3v_3)^2=\dots\ge 0\\\implies (u_1^2+u_2^2+u_3^2)(v_1^2+v_2^2+v_3^2)\ge (u_1v_1+u_2v_2+u_3v_3)^2\\\implies \u\^2\v\^2\ge\langle u,v\rangle^2\\\implies \u\\v\\ge\langle u,v\rangle$$

chrisplusian
 one year ago
Best ResponseYou've already chosen the best response.0I did not mean to repost this question I apologize
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