Câu 1:
Gọi góc giữa hai đường là A
\(\cos A=\dfrac{\left|3\cdot2+1\cdot3\right|}{\sqrt{3^2+1^2}\cdot\sqrt{2^2+3^2}}=\dfrac{9}{\sqrt{10\cdot13}}=\dfrac{9}{\sqrt{130}}\)
nên \(\widehat{A}\simeq37^052^0\)
Câu 2:
\(d\left(M;\Delta\right)=\dfrac{\left|3\cdot\left(-1\right)+\left(-4\right)\cdot1-3\right|}{\sqrt{3^2+\left(-4\right)^2}}=12\)