Question

Write the equation $\frac{3-x}{1}=\frac{2-y}{3}=\frac{1-z}{4}$ in the vector form.

Answer 1

Consider the equation

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

$\begin{array}{c}\frac{-\left(x-3\right)}{1}=\frac{-\left(y-2\right)}{3}=\frac{-\left(z-1\right)}{4}\\ \\ \frac{\left(x-3\right)}{-1}=\frac{\left(y-2\right)}{-3}=\frac{\left(z-1\right)}{-4}\end{array}$

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

Comparing the equation with

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_{}}{c}$

we get,

$a=1,b=3c=4$

$x_1=3,y_1=2,z_1=1$

Here,

$\begin{array}{c}\overrightarrow{b}=\hat{i}+3\hat{j}+4\hat{k}\\ \\ \overrightarrow{a}=3\hat{i}+2\hat{j}+\hat{k}\end{array}$

So, vector equation of the line is

$\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$

$=\left(3\hat{i}+2\hat{j}+\hat{k}\right)+\lambda \left(\hat{i}+3\hat{j}+4\hat{k}\right)$