Question

Вычислите угол между прямыми AB и CD, если A (√3;1;0), B(8;-2;4), С(0;2;0), D(√3,1;2√2).

Answer 1

(AB): (x-√3)/(8-√3) = (y-1)/(-2-1) = (z-0)/(4-0)
(x-√3)/(8-√3) = (y-1)/(-3) = z/4
m1 = 8-√3; n1 = -3; p1 = 4

(CD): (x-0)/(√3-0) = (y-2)/(1-2) = (z-0)/(2√2-0)
x/√3 = (y-2)/(-1) = z/(2√2)
m2 = √3; n2 = -1; p2 = 2√2

Косинус угла между прямыми:
$cos( \alpha )= \frac{m1*m2+n1*n2+p1*p2}{ \sqrt{m1^2+n1^2+p1^2}* \sqrt{m2^2+n2^2+p2^2} } = \frac{(8- \sqrt{3}) \sqrt{3}+(-3)(-1)+4*2 \sqrt{2} }{ \sqrt{(8- \sqrt{3} )^2+9+16}* \sqrt{3+1+8} } =$
$=\frac{8 \sqrt{3} - 3+3+8 \sqrt{2} }{ \sqrt{64-16 \sqrt{3} +3+25}* \sqrt{12} } =\frac{8 (\sqrt{3}+\sqrt{2}) }{ \sqrt{92-16 \sqrt{3}}* \sqrt{12} } =\frac{8 (\sqrt{3}+\sqrt{2}) }{ 2\sqrt{23-4 \sqrt{3}}* 2\sqrt{3} }=$
$\frac{2 (\sqrt{3}+\sqrt{2}) }{ \sqrt{69-12 \sqrt{3}}}=0,9062$