Cho hai đường thẳng \(d_{1}: m x+y=m+1\) và \(d_{2}: x+m y=2\).
Khi \(m=2\), góc giữa hai đường thẳng xấp xỉ bằng:
Hãy suy nghĩ và trả lời câu hỏi trước khi xem đáp án
Khi đó \(d_{1}: 2 x+y-3=0\) và \(d_{2}: x+2 y-2=0\)
Ta có vecto pháp tuyến của hai đường thẳng lần lượt là: \(\overrightarrow{n_{1}}=(2 ; 1), \overrightarrow{n_{2}}=(1 ; 2)\)
Suy ra \(\overrightarrow{n_{1}} \cdot \overrightarrow{n_{2}}=2.1+1.2=4 ;\left|\overrightarrow{n_{1}}\right|=\left|\overrightarrow{n_{2}}\right|=\sqrt{2^{2}+1^{2}}=\sqrt{5}\)
Suy ra \(\cos \left(d_{1}, d_{2}\right)=\frac{\left|\overrightarrow{n_{1}} \cdot \overrightarrow{n_{2}}\right|}{\left|\overrightarrow{n_{1}}\right| \cdot\left|\overrightarrow{n_{2}}\right|}=\frac{4}{\sqrt{5} \cdot \sqrt{5}}=\frac{4}{5} \Rightarrow\left(d_{1}, d_{2}\right) \approx 37^{\circ}\).
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