Question

By vector methods, find the cosine of the angle between the lines (x - 1)/(3) = (y - 0.5)/(2) = z and x = y = z

Answer 1

Answer:

The angle between the lines $\frac{x-1}{3}= \frac{y-0.5}{2}=\frac{z-0}{1}$ and $\frac{x-0}{1}= \frac{y-0}{1}=\frac{z-0}{1}$ is $\sqrt{\frac{6}{7}}$

Step-by-step explanation:

The equation of a line with direction vector $\vec{d}=(l,m.n)$ that passes through the point $(x_{1},y_{1},z_{1})$ is given by the formula

$\frac{x-x_{1}}{l}= \frac{y-x_{1}}{m}=\frac{z-z_{1}}{n},$ where l,m, and n are non-zero real numbers.

This is called the symmetric equations of the line.

The angle between two lines $\frac{x-x_{1}}{l_{1} }= \frac{y-y_{1}}{m_{1} }=\frac{z-z_{1}}{n_{1}}$ and $\frac{x-x_{2}}{l_{2} }= \frac{y-y_{2}}{m_{2} }=\frac{z-z_{2}}{n_{2}}$ equal the angle subtended by direction vectors, $d_{1}$ and $d_{2}$ of the lines

$cos (\theta)=\frac{\vec{d_{1}}\cdot\vec{d_{2}}}{|\vec{d_{1}}|\cdot|\vec{d_{2}}|}=\frac{l_{1} \cdot\l_{2}+m_{1} \cdot\ m_{2}+n_{1} \cdot\ n_{2}}{\sqrt{l_{1}^{2}+m_{1}^{2}+n_{1}^{2}} \cdot \sqrt{l_{2}^{2}+m_{2}^{2}+n_{2}^{2}}}$

Given that

$\frac{x-1}{3}= \frac{y-0.5}{2}=\frac{z-0}{1}$ and $\frac{x-0}{1}= \frac{y-0}{1}=\frac{z-0}{1}$

$l_{1}=3, m_{1}=2,n_{1}=1\\ l_{2}=1, m_{2}=1,n_{2}=1$

We can use the formula above to find the cosine of the angle between the lines

$cos(\theta)=\frac{3 \cdot 1+2 \cdot 1 +1 \cdot 1}{\sqrt{3^{2}+2^{2}+1^{2}} \cdot \sqrt{1^{2}+1^{2}+1^{2}}} = \sqrt{\frac{6}{7}}$