Question

Find the direction cosines of the line  \[\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3} .\]  Also, reduce it to vector form. 

Answer 1

The cartesian equation of the given line is 

\[\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3}\]

It can be re-written as 

\[\frac{x - 4}{- 2} = \frac{y - 0}{6} = \frac{z - 1}{- 3}\]

This shows that the given line passes through the point (4,0,1) and its direction ratios are proportional to -2,6,-3 

So, its direction cosines are 

\[\frac{- 2}{\sqrt{\left( - 2 \right)^2 + 6^2 + \left( - 3 \right)^2}}, \frac{6}{\sqrt{\left( - 2 \right)^2 + 6^2 + \left( - 3 \right)^2}}, \frac{- 3}{\sqrt{\left( - 2 \right)^2 + 6^2 + \left( - 3 \right)^2}}\]

\[ = \frac{- 2}{7}, \frac{6}{7}, \frac{- 3}{7} \] 
Thus, the given line passes through the point having position vector  \[\overrightarrow{a} = 4 \hat{i} + \hat{k} \]  and is parallel to the vector \[\overrightarrow{b} = - 2 \hat{i} + 6 \hat{j} - 3 \hat{k}\]

We know that the vector equation of a line passing through a point with position vector `vec a` and parallel to the vector `vec b` is   \[\overrightarrow{r} = \overrightarrow{a} + \lambda \overrightarrow{b}\]  Here, 

\[\overrightarrow{a} = 4 \hat{i} + \hat{k} \]

\[ \overrightarrow{b} = - 2 \hat{i} + 6 \hat{j} - 3 \hat{k} \]

Vector equation of the required line is 

\[\overrightarrow{r} = \left( 4 \hat{i} + 0 \hat{j}+ \hat{k} \right) + \lambda \left( - 2 \hat{i} + 6 \hat{j} - 3 \hat{k} \right) \]

\[\text{ Here } , \lambda \text{ is a parameter } . \]