Question

Dados os pontos A(2,1, -1), B(3, 0, 1) e C(2, -1, -3), determinar o ponto D tal que AD= BC x AC

Answer 1

BC=(-1,-1,-4) e AC=(0,-2,-2)

AD=(-1,-1,-4) x (0,-2,-2)
AD=(0,2,8)

D=A+AD
D=(2,1,-1)+(0,2,8)
D=(2,3,7)

Answer 2

$\bmatrix A(2,1,-1)\\\\B(3,0,1)\\\\C(2,-1-3)\\\\D(x,y,z)(\end$

temos que
$\bmatrix AD= D-A = (x-2,y-1,z+1)\\\\BC=C-B=(-1,-1,-4)\\\\AC=C-A=(0,-2,-2)\end$

BC x AC = produto vetorial entre os vetores

$BC _XAC= \left[\begin{array}{ccc}\vec i&\vec j&\vec k\\-1&-1&-4\\0&-2&-2\end{array}\right] =-6\;\vec i -2\;\vec j +2\;\vec k =(-6,-2,2)$

temos então
$AD =BC_XAC\\\\(x-2,y-1,z+1) = (-6,-2,2)$

resolvendo
$x-2 = -6\\\\x=-4$

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$y-1 = -2\\\\y=-1$

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$z+1 = 2\\\\z=1$

então 
$\boxed{\boxed{D=(-4,-1,1)}}$