Question

The Cartesian equation of a line is:
$\dfrac{x-5}{3}=\dfrac{y+4}{7}=\dfrac{7-6}{2}$
Find out the vector form of the line.

Answer 1

Hint: At first find out the position vector of a point on the line $\dfrac{x-5}{3}=\dfrac{y+4}{7}=\dfrac{7-6}{2}$
Find out the direction vector of the line. Then find out the vector form of the line.

Complete step-by-step answer:

To find the equation of a line in a two dimensional space, we need to know a point through which the given line passes. In addition, we have to find out the slope of the given line.
Similarly, in a three dimensional space we can obtain the equation of a line if we know a point that the line passes through and a direction vector of that line.
The equation of a line with direction vector $\overrightarrow{b}=\left( {{b}_{1}},{{b}_{2}},{{b}_{3}} \right)$ and passes through the point $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ is given by the formula:
$\dfrac{x-{{x}_{1}}}{{{b}_{1}}}=\dfrac{y-{{y}_{1}}}{{{b}_{2}}}=\dfrac{z-{{z}_{1}}}{{{b}_{3}}}......(1)$
Equation (1) is the general form of the Cartesian equation of a line.
Here the Cartesian equation of a line is given by:
$\dfrac{x-5}{3}=\dfrac{y+4}{7}=\dfrac{7-6}{2}......(2)$
Now if we compare (2) with (1), we will get that:
The given line passes through the point $(5,-4,6)$.
So the position vector of this point is $\overrightarrow{a}=5\hat{i}-4\hat{j}+6\hat{k}$
Also, the direction ratios of the given line are 3, 7 and 2.
This means that the line is in the direction of vector:
$\overrightarrow{b}=3\hat{i}+7\hat{j}+2\hat{k}$
We know that the line through the position vector $\overrightarrow{a}$ and in the direction of the vector $\overrightarrow{b}$ is given by the equation:
$\overrightarrow{r}=\overrightarrow{a}+\alpha \overrightarrow{b}$, where $\alpha $ is a real number.
$\Rightarrow \overrightarrow{r}=\left( 5\hat{i}-4\hat{j}+6\hat{k} \right)+\alpha \left( 3\hat{i}+7\hat{j}+2\hat{k} \right)$
This is the vector form of the given equation.

Note: We can leave the answer in the form:
$\overrightarrow{r}=\left( 5\hat{i}-4\hat{j}+6\hat{k} \right)+\alpha \left( 3\hat{i}+7\hat{j}+2\hat{k} \right)$
Or we can say the vector form is:
$\overrightarrow{r}=\left( 5+3\alpha \right)\hat{i}+\left( -4+7\alpha \right)\hat{j}+\left( 7+6\alpha \right)\hat{k}$