Question

The cartesian equations of a line are \[\frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2} .\]  Find a vector equation for the line.

Answer 1

The cartesian equation of the given line is \[\frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2}\] 

It can be re-written as 

\[\frac{x - 5}{3} = \frac{y - \left( - 4 \right)}{7} = \frac{z - 6}{2}\] 

Thus, the given line passes through the point having position vector   \[\vec{a} = 5 \hat{i} - 4 \hat{j} + 6 \hat{k}\] and is parallel to the vector  \[\vec{b} = 3 \hat{i}  + 7 \hat{j}+ 2 \hat{k} \] 

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  \[\vec{r} = \vec{a} + \lambda \vec{b}\] 

Vector equation of the required line is 

\[\vec{r} = \left( 5 \hat{i} - 4 \hat{j} + 6 \hat{k} \right) + \lambda \left( 3 \hat{i} + 7 \hat{j} + 2 \hat{k} \right)\]

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