Question

Find the angle between the lines 2x = 3y = -z and 6x = -y = -4z​

Answer 1

Step-by-step explanation:

2x-3y-6=0 has slope 2/3

x+5y-20=0 has slope -1/5

so, you want

arctan(-1/5) = -11.3°

arctan(2/3) = 33.7°

so, the acute angle between the lines is 45°

Answer 2

Consider the line,

$\longrightarrow L_1:2x=3y=-z$

$\longrightarrow L_1:\dfrac{x-0}{\left(\dfrac{1}{2}\right)}=\dfrac{y-0}{\left(\dfrac{1}{3}\right)}=\dfrac{z-0}{-1}$

So this line in vector form will be, $\left[\left<x,\ y,\ z\right>=x\,\hat i+y\,\hat j+z\,\hat k\right]$

$\longrightarrow L_1:\vec{r_1}=\left<0,\ 0,\ 0\right>+\lambda_1\left<\dfrac{1}{2},\ \dfrac{1}{3},\ -1\right>$

where $\left<\dfrac{1}{2},\ \dfrac{1}{3},\ -1\right>$ is a direction ratio of $L_1.$

$\longrightarrow\vec{b_1}=\left<\dfrac{1}{2},\ \dfrac{1}{3},\ -1\right>$

Consider the line,

$\longrightarrow L_2:6x=-y=-4z$

$\longrightarrow L_2:\dfrac{x-0}{\left(\dfrac{1}{6}\right)}=\dfrac{y-0}{-1}=\dfrac{z-0}{\left(-\dfrac{1}{4}\right)}$

So this line in vector form will be,

$\longrightarrow L_2:\vec{r_2}=\left<0,\ 0,\ 0\right>+\lambda_2\left<\dfrac{1}{6},\ -1,\ -\dfrac{1}{4}\right>$

where $\left<\dfrac{1}{6},\ -1,\ -\dfrac{1}{4}\right>$ is a direction ratio of $L_2.$

$\longrightarrow\vec{b_2}=\left<\dfrac{1}{6},\ -1,\ -\dfrac{1}{4}\right>$

Taking dot product of the direction ratios,

$\longrightarrow \vec{b_1}\cdot\vec{b_2}=\left<\dfrac{1}{2},\ \dfrac{1}{3},\ -1\right>\cdot\left<\dfrac{1}{6},\ -1,\ -\dfrac{1}{4}\right>$

$\longrightarrow \vec{b_1}\cdot\vec{b_2}=\dfrac{1}{2}\cdot\dfrac{1}{6}+\dfrac{1}{3}\cdot(-1)+(-1)\cdot\left(-\dfrac{1}{4}\right)$

$\longrightarrow \vec{b_1}\cdot\vec{b_2}=\dfrac{1}{12}-\dfrac{1}{3}+\dfrac{1}{4}$

$\longrightarrow \vec{b_1}\cdot\vec{b_2}=0$

This implies the direction ratios are perpendicular to each other, so are the lines $L_1$ and $L_2.$

Hence the angle between the lines is 90°.