Question

Show that the lines $\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}\mathrm{and}\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ are perpendicular to each other.

Answer 1

We have

$\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}\mathrm{and}\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$

These equations can be re-written as

$$\begin{gathered} \frac{x-5}{7}=\frac{y-\left(-2\right)}{-5}=\frac{z-0}{1}...\left(1\right) \\ \frac{x-0}{1}=\frac{y-0}{2}=\frac{z-0}{3}...\left(2\right) \end{gathered}$$

$$\begin{gathered} \therefore \overrightarrow{m_1}=\mathrm{Vector}\mathrm{parallel}\mathrm{to}\mathrm{line}\left(1\right)=7\hat{i}-5\hat{j}+\hat{k} \\ \overrightarrow{m_2}=\mathrm{Vector}\mathrm{parallel}\mathrm{to}\mathrm{line}\left(2\right)=\hat{i}+2\hat{j}+3\hat{k} \end{gathered}$$

Now,

$$\begin{gathered} \overrightarrow{m_1}.\overrightarrow{m_2}=\left(7\hat{i}-5\hat{j}+\hat{k}\right).\left(\hat{i}+2\hat{j}+3\hat{k}\right) \\ =7-10+3 \\ =0 \end{gathered}$$

Hence, the given two lines are perpendicular to each other.