If the Cartesian equation of a line `\frac{3-x}{5} = \frac{y+4}{7} = \frac{2z-6}{4}`,then write the vector equation for line.

 If the Cartesian equation of a line `\frac{3-x}{5} = \frac{y+4}{7} = \frac{2z-6}{4}`,then write the vector equation for line.

Solution : 

Given that 

`\frac{3-x}{5} = \frac{y+4}{7} = \frac{2z-6}{4}`

`\frac{-(x-3)}{5} = \frac{y+4}{7} = \frac{2(z-3)}{4}`

`\frac{x-3}{-5} = \frac{y+4}{7} = \frac{z-3}{2}`

so position vector 

`\vec{a}= 3\hat{i}-4\hat{j}+3\hat{k}`

direction vector 

`\vec{b}= -5\hat{i}+7\hat{j}+2\hat{k}`

vector equation 

`\vec{r} = \vec{a} + \lambda  \vec{b}`

`\vec{r} = (3\hat{i}-4\hat{j}+3\hat{k}) + \lambda (5\hat{i}+7\hat{j}+2\hat{k})`