Space vectors 3D

The vectors u = (1; 3; -4), v = (0; 1; 1) are given. Find the size of these vectors, calculate the angle of the vectors, the distance between the vectors.

Correct result:

a =  5.099
b =  1.4142
A =  13.8979 °
h =  5.4772

Solution:

$$\begin{gathered} u=(1,3,-4) \\ v=(0,1,1) \\ a=\mathrm{\mid }u\mathrm{\mid }=\sqrt{u_x^2+u_y^2+u_z^2}=\sqrt{1^2+3^2+(-4{)}^2}=\sqrt{26}=5.099 \end{gathered}$$

$b=\mathrm{\mid }v\mathrm{\mid }=\sqrt{v_x^2+v_y^2+v_z^2}=\sqrt{0^2+1^2+1^2}=\sqrt{2}=1.4142$

$$\begin{gathered} c={\displaystyle \frac{u_x\cdot v_x+u_y+v_y+u_y\cdot v_y}{a\cdot b}}={\displaystyle \frac{1\cdot 0+3+1+3\cdot 1}{5.099\cdot 1.4142}}\doteq 0.9707 \\ A={\displaystyle \frac{180^{\circ }}{\pi }}\cdot \arccos (c)={\displaystyle \frac{180^{\circ }}{\pi }}\cdot \arccos (0.9707)=13.8979^{\circ }=13^{\circ }53^{\mathrm{\prime }}52\mathrm{"} \end{gathered}$$

$$\begin{gathered} w=u-v \\ w=(1,2,-5) \\ h=\mathrm{\mid }w\mathrm{\mid }=\sqrt{w_x^2+w_y^2+w_z^2}=\sqrt{1^2+2^2+(-5{)}^2}=\sqrt{30}=5.4772 \end{gathered}$$

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