Question

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

 

Answer 1

We have , 
2x = 3y = −z
6x = −y = −4z

The given lines can be re-written as

$$\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{-1}\text{ and }\frac{x}{\frac{1}{6}}=\frac{y}{-1}=\frac{z}{-\frac{1}{4}}$$

$$\Rightarrow \frac{x}{3}=\frac{y}{2}=\frac{z}{-6}\text{ and }\frac{x}{2}=\frac{y}{-12}=\frac{z}{-3}$$

These lines are parallel to vectors

$$\overrightarrow{b_1}=3\hat{i}+2\hat{j}-6\hat{k}\text{ and }\overrightarrow{b_2}=2\hat{i}-12\text{j}-3\hat{k}$$

Let  $$\theta$$ be the angle between these lines.
Now,

$$\cos \theta =\frac{\overrightarrow{b_1}.\overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|}$$

$$=\frac{(3\hat{i}+2\hat{j}-6\hat{k}).(2\hat{i}-12\hat{j}-3\hat{k})}{\sqrt{3^2+2^2+{(-6)}^2}\sqrt{2^2+{(-12)}^2+{(-3)}^2}}$$

$$=\frac{6-24+18}{\sqrt{9+4+36}\sqrt{4+144+9}}$$

$$=0$$

$$\Rightarrow \theta =\frac{\pi }{2}$$