Question

dados os pontos A(1,-2,0), B(2,-1,-2) e C(4, 2, 1), calcular ABxBC e calcular os
ângulos

Answer 1

AB = B-A = (2;-1;-2) - (1;-2;-0) 
AB = ( 2-1 ; -1-(-2) ; -2-(-0)
AB = (1;1;-2)
módulo de AB 
$||AB||+ \sqrt{1^2+1^2+(-2)^2} = \sqrt{6}$

BC = C-B 
BC = (4;2;1)-(2;-1;-2)
BC= (4-2 ; 2-(-1) ; 1-(-2))
BC= (2;3;1)
módulo de BC
$||BC||= \sqrt{2^2+3^2+1^2}= \sqrt{14}$


angulo entre os dois vetores
$\boxed{\boxed{cos(\theta) = \frac{AB_XBC}{||AB||*||BC||} }}$
;;;;;;;;;;;;;;;;;

$cos(\theta) = \frac{(1;1;-2)_X(2;3;1)}{ \sqrt{6}* \sqrt{14} } \\\\cos(\theta)= \frac{(1*2)+(1*3)+(-2*1)}{ \sqrt{6*14} } \\\\cos(\theta) = \frac{2+3-2}{ \sqrt{84} } \\\\ \theta = arcos( \frac{3}{ \sqrt{84} } )\\\\\theta \approx 70,89^\circ$