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

Question

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

✓

Answer

We have
2x = 3y = -z
6x = -y = -4z
The given lines can be re-written as
$\frac{\text{x}}{\frac{1}{2}}=\frac{\text{y}}{\frac{1}{3}}=\frac{\text{z}}{-1}$ and $\frac{\text{x}}{\frac{1}{6}}=\frac{\text{y}}{-1}=\frac{\text{z}}{-\frac{1}{4}}$
$\Rightarrow\frac{\text{x}}{3}=\frac{\text{y}}{2}=\frac{\text{z}}{-6}$ and $\frac{\text{x}}{2}=\frac{\text{y}}{-12}=\frac{\text{z}}{-3}$
These lines are parallel to vectors $\vec{\text{b}}_1=3\hat{\text{i}}+2\hat{\text{j}}-6\hat{\text{k}}$ and $\vec{\text{b}}_2=2\hat{\text{i}}-12\hat{\text{j}}-3\hat{\text{k}}$
Let $\theta$ be the angle between these lines.
Now,
$\cos\theta=\frac{\vec{\text{b}}_1.\vec{\text{b}}_2}{\big|\vec{\text{b}}_1\big|\big|\vec{\text{b}}_2\big|}$
$=\frac{\big(3\hat{\text{i}}+2\hat{\text{j}}-6\hat{\text{k}}\big).\big(2\hat{\text{i}}-12\hat{\text{j}}-3\hat{\text{k}}\big)}{\sqrt{3^2+2^2+(-6)^2}\sqrt{2^2+(-12)^2+(-3)^2}}$
$=\frac{6-24+18}{\sqrt{9+4+36}\sqrt{4+144+9}}$
$=0$
$\Rightarrow\theta=\frac{\pi}{2}$

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