Para que valor de k os pontos A (k, -1, 5), B (7, 2, 1), C (-1, -3, -1) e D (1, 0, 3) são coplanares?
Possíveis Respostas
a) -2
b) 3
c) -1
d) 0
e) -3
Soluções para a tarefa
Respondido por
1
Olá
Resposta correta, letra E) k = -3
Dados:
A = (k, -1, 5)
B = (7, 2, 1)
C = (-1, -3, -1)
D = (1, 0, 3)
Criando os 3 vetores
BA = A-B = (k, -1, 5) - (7, 2, 1) = (k-7, -3, 4)
BA = (k-7, -3, 4)
BC = C-B = (-1, -3, -1) - (7, 2, 1) = (-8, -5, -2)
BC = (-8, -5, -2)
BD = D-B = (1, 0, 3) - (7, 2, 1) = (-6, -2, 2)
BD = (-6, -2, 2)
Calculando o produto misto entre os 3 vetores criados, e igualando a zero.
![\displaystyle (\vec{BA},\vec{BC},\vec{BD})~=~ \mathsf{\left[\begin{array}{ccc}k-7&-3&4\\-8&-5&-2\\-6&-2&2\end{array}\right] ~=~0}\\\\\\\\\mathsf{\underbrace{(\mathsf{-10k+70-36+64})}_{diag.~principal}-\underbrace{(\mathsf{48+4k-28+120})}_{diag.~secund\'aria}=0}\\\\\\\\\mathsf{-10k+98-4k-140=0}\\\\\mathsf{-14k-42=0}\\\\\mathsf{-14k=42}\\\\\\\mathsf{k= \frac{42}{-14} }\\\\\\\boxed{\mathsf{k=-3}}\qquad\qquad\qquad\Longrightarrow\qquad\mathsf{Letra~E)}](https://tex.z-dn.net/?f=%5Cdisplaystyle+%28%5Cvec%7BBA%7D%2C%5Cvec%7BBC%7D%2C%5Cvec%7BBD%7D%29%7E%3D%7E++%5Cmathsf%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dk-7%26amp%3B-3%26amp%3B4%5C%5C-8%26amp%3B-5%26amp%3B-2%5C%5C-6%26amp%3B-2%26amp%3B2%5Cend%7Barray%7D%5Cright%5D+%7E%3D%7E0%7D%5C%5C%5C%5C%5C%5C%5C%5C%5Cmathsf%7B%5Cunderbrace%7B%28%5Cmathsf%7B-10k%2B70-36%2B64%7D%29%7D_%7Bdiag.%7Eprincipal%7D-%5Cunderbrace%7B%28%5Cmathsf%7B48%2B4k-28%2B120%7D%29%7D_%7Bdiag.%7Esecund%5C%27aria%7D%3D0%7D%5C%5C%5C%5C%5C%5C%5C%5C%5Cmathsf%7B-10k%2B98-4k-140%3D0%7D%5C%5C%5C%5C%5Cmathsf%7B-14k-42%3D0%7D%5C%5C%5C%5C%5Cmathsf%7B-14k%3D42%7D%5C%5C%5C%5C%5C%5C%5Cmathsf%7Bk%3D+%5Cfrac%7B42%7D%7B-14%7D+%7D%5C%5C%5C%5C%5C%5C%5Cboxed%7B%5Cmathsf%7Bk%3D-3%7D%7D%5Cqquad%5Cqquad%5Cqquad%5CLongrightarrow%5Cqquad%5Cmathsf%7BLetra%7EE%29%7D "\displaystyle (\vec{BA},\vec{BC},\vec{BD})~=~ \mathsf{\left[\begin{array}{ccc}k-7&-3&4\\-8&-5&-2\\-6&-2&2\end{array}\right] ~=~0}\\\\\\\\\mathsf{\underbrace{(\mathsf{-10k+70-36+64})}_{diag.~principal}-\underbrace{(\mathsf{48+4k-28+120})}_{diag.~secund\'aria}=0}\\\\\\\\\mathsf{-10k+98-4k-140=0}\\\\\mathsf{-14k-42=0}\\\\\mathsf{-14k=42}\\\\\\\mathsf{k= \frac{42}{-14} }\\\\\\\boxed{\mathsf{k=-3}}\qquad\qquad\qquad\Longrightarrow\qquad\mathsf{Letra~E)}")
by AvengerCrawl
Resposta correta, letra E) k = -3
Dados:
A = (k, -1, 5)
B = (7, 2, 1)
C = (-1, -3, -1)
D = (1, 0, 3)
Criando os 3 vetores
BA = A-B = (k, -1, 5) - (7, 2, 1) = (k-7, -3, 4)
BA = (k-7, -3, 4)
BC = C-B = (-1, -3, -1) - (7, 2, 1) = (-8, -5, -2)
BC = (-8, -5, -2)
BD = D-B = (1, 0, 3) - (7, 2, 1) = (-6, -2, 2)
BD = (-6, -2, 2)
Calculando o produto misto entre os 3 vetores criados, e igualando a zero.
by AvengerCrawl
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