Trong không gian Oxyz, cho hai vecto u=(1;4;1)

Phương pháp giải:

Cho hai vecto \[\vec a\left( {{x_1};{\mkern 1mu} {\mkern 1mu} {y_1};{\mkern 1mu} {\mkern 1mu} {z_1}} \right),{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \vec b = \left( {{x_2};{\mkern 1mu} {\mkern 1mu} {y_2};{\mkern 1mu} {\mkern 1mu} {z_2}} \right).\] Khi đó α⁢ =∠⁢(a→;b→) có:

cos⁡α⁢ =a→.b→|a→|.|b→|=x1⁢x2+y1⁢y2+z1⁢z2x12+y12+z12.x22+y22+z22.

Giải chi tiết:

Cho hai vecto \[\vec u = \left( {1;{\mkern 1mu} {\mkern 1mu} 4;{\mkern 1mu} {\mkern 1mu} 1} \right)\]và \[\vec v = \left( { - 1;{\mkern 1mu} {\mkern 1mu} 1; - 3} \right)\]

\[ \Rightarrow \cos \left( {\vec u,{\mkern 1mu} {\mkern 1mu} \vec v} \right) = \frac{{1.\left( { - 1} \right) + 4.1 + 1.\left( { - 3} \right)}}{{\sqrt {{1^2} + {4^2} + {1^2}} .\sqrt {{{\left( { - 1} \right)}^2} + {1^2} + {{\left( { - 3} \right)}^2}} }} = 0\]

\[ \Rightarrow \angle \left( {\vec u,{\mkern 1mu} {\mkern 1mu} \vec v} \right) = {90^0}.\]