Respuesta :
Answer:
The angle between the lines [tex]\frac{x-1}{3}= \frac{y-0.5}{2}=\frac{z-0}{1}[/tex] and [tex]\frac{x-0}{1}= \frac{y-0}{1}=\frac{z-0}{1}[/tex] is [tex]\sqrt{\frac{6}{7}}[/tex]
Step-by-step explanation:
The equation of a line with direction vector [tex]\vec{d}=(l,m.n)[/tex] that passes through the point [tex](x_{1},y_{1},z_{1})[/tex] is given by the formula
[tex]\frac{x-x_{1}}{l}= \frac{y-x_{1}}{m}=\frac{z-z_{1}}{n},[/tex] where l,m, and n are non-zero real numbers.
This is called the symmetric equations of the line.
The angle between two lines [tex]\frac{x-x_{1}}{l_{1} }= \frac{y-y_{1}}{m_{1} }=\frac{z-z_{1}}{n_{1}}[/tex] and [tex]\frac{x-x_{2}}{l_{2} }= \frac{y-y_{2}}{m_{2} }=\frac{z-z_{2}}{n_{2}}[/tex] equal the angle subtended by direction vectors, [tex]d_{1}[/tex] and [tex]d_{2}[/tex] of the lines
[tex]cos (\theta)=\frac{\vec{d_{1}}\cdot\vec{d_{2}}}{|\vec{d_{1}}|\cdot|\vec{d_{2}}|}=\frac{l_{1} \cdot\l_{2}+m_{1} \cdot\ m_{2}+n_{1} \cdot\ n_{2}}{\sqrt{l_{1}^{2}+m_{1}^{2}+n_{1}^{2}} \cdot \sqrt{l_{2}^{2}+m_{2}^{2}+n_{2}^{2}}}[/tex]
Given that
[tex]\frac{x-1}{3}= \frac{y-0.5}{2}=\frac{z-0}{1}[/tex] and [tex]\frac{x-0}{1}= \frac{y-0}{1}=\frac{z-0}{1}[/tex]
[tex]l_{1}=3, m_{1}=2,n_{1}=1\\ l_{2}=1, m_{2}=1,n_{2}=1[/tex]
We can use the formula above to find the cosine of the angle between the lines
[tex]cos(\theta)=\frac{3 \cdot 1+2 \cdot 1 +1 \cdot 1}{\sqrt{3^{2}+2^{2}+1^{2}} \cdot \sqrt{1^{2}+1^{2}+1^{2}}} = \sqrt{\frac{6}{7}}[/tex]
