Đáp án:
`∆_1: 2x -y -10=0=> \vec{n_1} = (2;-1)`
`∆_2: x-3y -9=0=> \vec{n_2} = (1;-3)`
`cos(∆_1,∆_2) = \frac{| \vec{n}_1.\vec{n}_2|}{|\vec{n}_1 |.|\vec{n}_2|}`
`= \frac{ |2.(-1) +1.(-3)|}{\sqrt{ 2² +1²} .\sqrt{(-1)^2 + (-3)^2}}= (\sqrt{2})/2`
`=> (∆_1,∆_2) = 45°`
Giải thích các bước giải:
`∆_1: a_1 x + b_1y +c_1 =0`
`=> \vec{n}_1 = (a_1;b_1)`
`∆_2: a_2 x + b_2 y + c_2=0`
`=> \vec{n}_2 = (a_2;b_2)`
`=> cos (∆_1,∆_2) = \frac{| \vec{n}_1.\vec{n}_2|}{|\vec{n}_1 |.|\vec{n}_2|} = \frac{| a_1b_1 +a_2b_2|}{\sqrt{a_1² +a_2²}.\sqrt{b_1² +b_2²}}`
`=> (∆_1,∆_2)=??`
