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Fie vectorul ∨=3i+4j. Aflati a ∈ R stiind ca lungimea vectorului av=10.

Răspuns :

[tex]\vec{v}=3\vec{i}+4\vec{j}\\a\vec{v}=3a\vec{i}+4a\vec{j}\\|av|=\sqrt{(3a)^2+(4a)^2}=\sqrt{9a^2+16a^2}=\sqrt{25a^2}<=>10=\pm5a=>a_1=2;a_2=-2[/tex]