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Question: The Cartesian equation of a line is: \(\dfrac{x-5}{3}=\dfrac{y+4}{7}=\dfrac{7-6}{2}\) Find out ...

The Cartesian equation of a line is:
x−53=y+47=7−62\dfrac{x-5}{3}=\dfrac{y+4}{7}=\dfrac{7-6}{2}
Find out the vector form of the line.

Explanation

Solution

Hint: At first find out the position vector of a point on the line x−53=y+47=7−62\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 b→=(b1,b2,b3)\overrightarrow{b}=\left( {{b}_{1}},{{b}_{2}},{{b}_{3}} \right) and passes through the point (x1,y1,z1)\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right) is given by the formula:
x−x1b1=y−y1b2=z−z1b3......(1)\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:
x−53=y+47=7−62......(2)\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)(5,-4,6).
So the position vector of this point is a→=5i^−4j^+6k^\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:
b→=3i^+7j^+2k^\overrightarrow{b}=3\hat{i}+7\hat{j}+2\hat{k}
We know that the line through the position vector a→\overrightarrow{a} and in the direction of the vector b→\overrightarrow{b} is given by the equation:
r→=a→+αb→\overrightarrow{r}=\overrightarrow{a}+\alpha \overrightarrow{b}, where α\alpha is a real number.
⇒r→=(5i^−4j^+6k^)+α(3i^+7j^+2k^)\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:
r→=(5i^−4j^+6k^)+α(3i^+7j^+2k^)\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:
r→=(5+3α)i^+(−4+7α)j^+(7+6α)k^\overrightarrow{r}=\left( 5+3\alpha \right)\hat{i}+\left( -4+7\alpha \right)\hat{j}+\left( 7+6\alpha \right)\hat{k}