Question

Dados os vetores u=(1, a, -2a-1), v=(a, a-1, 1) e w=(a, -1, 1), determinar a de modo que u.v=(u+v).w

Answer 1

$\vec{u}.\vec{v}=(\vec{u}+\vec{v}).\vec{w}\\\\\\Para~facilitar~e~por~falta~de~espaco~adequado,~vamos~calcular~por~partes:\\\\\\\\\vec{u}.\vec{v}~=~(1~,~a~,~-2a-1)~.~(a~,~a-1~,~1)\\\\\\\vec{u}.\vec{v}~=~1~.~a~+~a~.~(a-1)~+~(-2a-1)~.~1\\\\\\\vec{u}.\vec{v}~=~a~+~a^2-a~-2a-1\\\\\\\boxed{\vec{u}.\vec{v}~=~a^2-2a-1}$

$\vec{u}~+~\vec{v}~=~(1~,~a~,~-2a-1)~+~(a~,~a-1~,~1)\\\\\\\vec{u}~+~\vec{v}~=~(1+a~,~a+(a-1)~,~(-2a-1)+1)\\\\\\\boxed{\vec{u}~+~\vec{v}~=~(1+a~,~2a-1~,~-2a)}\\\\\\\\(\vec{u}+\vec{v}).\vec{w}~=~(1+a~,~2a-1~,~-2a)~.~(a~,~-1~,~1)\\\\\\(\vec{u}+\vec{v}).\vec{w}~=~(1+a)~.~a~+~(2a-1)~.~(-1)~+~(-2a)~.~1\\\\\\(\vec{u}+\vec{v}).\vec{w}~=~a+a^2~-~2a+1~-~2a\\\\\\\boxed{(\vec{u}+\vec{v}).\vec{w}~=~a^2-3a+1}$

$\vec{u}.\vec{v}~=~(\vec{u}+\vec{v}).\vec{w}\\\\\\a^2-2a-1~=~a^2-3a+1\\\\\\a^2-a^2-2a+3a~=~1+1\\\\\\0+a~=~2\\\\\\\boxed{a~=~2}$

Answer 2

Resposta:

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