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

Question

Find the angle between the lines $2\text{x}=3\text{y}=-\text{z}$ and $6\text{x}=-\text{y}=-4\text{z}.$

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Answer

The equations of the given lines can be re-writen as
$\frac{\text{x}}{3}=\frac{\text{y}}{2}=\frac{\text{z}}{-6}$ and $\frac{\text{x}}{2}=\frac{\text{y}}{-12}=\frac{\text{z}}{-3}$
We know that angle between the lines $\frac{\text{x}-\text{x}_1}{\text{a}_1}=\frac{\text{y}-\text{y}_1}{\text{b}_1}=\frac{\text{z}-\text{z}_1}{\text{c}_1}$ and $\frac{\text{x}-\text{x}_2}{\text{a}_2}=\frac{\text{y}-\text{y}_2}{\text{b}_2}=\frac{\text{z}-\text{z}_2}{\text{c}_2}$ is given by $\cos\theta=\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}^2_1+\text{b}^2_1+\text{c}^2_1}\sqrt{\text{a}^2_2+\text{b}^2_2+\text{c}^2_2}}.$
Let $\theta$ be the angle between the given lines.
$\therefore\cos\theta=\frac{3\times2+2\times(-12)+(-6)\times(-3)}{\sqrt{3^2+2^2+(-6)^2}\sqrt{2^2+(-12)^2+(-3)^2}}$
$=\frac{6-24+18}{\sqrt{49}\sqrt{157}}$
$=0$
$\Rightarrow\theta=\frac{\pi}{2}$
Thus, the angle between the given lines is $\frac{\pi}{2}.$

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