Question

向量 A=(1,-2,2,-3) 与B=(3,-1,5,-1）的夹角

Answer 1

问：向量 A=(1,-2,2,-3) 与B=(3,-1,5,-1）的夹角

答：1.求出A、B的模：$A = \sqrt{1^2+(-2)^2+2^2+(-3)^2} = \sqrt{14} \\B = \sqrt{3^2+(-1)^2+5^2+(-1)^2} = \sqrt{36}$2.计算A、B的内积：$A \cdot B = 1 \times 3+(-2) \times (-1)+2 \times 5+(-3) \times (-1) = 8$3.利用余弦定理计算夹角θ：$\cos \theta = \frac{A \cdot B}{\left|A\right|\left|B\right|} = \frac{8}{\sqrt{14} \sqrt{36}} = \frac{8}{6 \sqrt{14}}$4.由余弦值求夹角θ：$\tan \theta = \frac{A \cdot B}{\left|A\right|\left|B\right|} = \arccos{\cos{\theta}} = \arccos{(\frac{8}{6 \sqrt{14}})} \approx 65.9^{\circ}$