Question

Tìm côsin góc giữa hai đường thẳng Δ₁ : \(2x+y-1=0\) vaf △₂ : \(\left\{{}\begin{matrix}x=2+t\\y=1-t\end{matrix}\right.\)

Answer 1

Đường thẳng \(\Delta_1\) nhận \(\overrightarrow{u_1}=\left(1;-2\right)\) là 1 vtcp

Đường thẳng \(\Delta_2\) nhận \(\left(1;-1\right)\) là 1 vtcp

\(\Rightarrow cos\left(\Delta_1;\Delta_2\right)=\dfrac{\left|\overrightarrow{u_1}.\overrightarrow{u_2}\right|}{\left|\overrightarrow{u_1}\right|.\left|\overrightarrow{u_2}\right|}=\dfrac{\left|1.1+\left(-2\right).\left(-1\right)\right|}{\sqrt{1^2+\left(-2\right)^2}.\sqrt{1^2+\left(-1\right)^2}}=\dfrac{3}{\sqrt{10}}\)