\(\cos\left(d1,d2\right)=\dfrac{\left|\left(m+1\right)\cdot m+\left(m-1\right)\cdot\left(-1\right)\right|}{\sqrt{\left(m+1\right)^2+\left(m-1\right)^2}\cdot\sqrt{m^2+\left(-1\right)^2}}\)
\(=\dfrac{\left|m^2+m-m+1\right|}{\sqrt{m^2+2m+1+m^2-2m+1}\cdot\sqrt{m^2+1}}=\dfrac{1}{2}\)
nên \(\left(d1,d2\right)=60^0\)