2.3 Cartesian form of a line passing through a given point and parallel to a given vector

2.1 Vectorial form of a line passing through a given point and parallel to a given vector
2.2 Vectorial form of a line passing through two given points
2.3 Cartesian form of a line passing through a given point and parallel to a given vector
2.4 Cartesian form of a line passing through two given points

3.1 Condition for perpendicularity
3.2 Condition for parallelism
3.3 Co-planarity of two lines
3.4 Shortest distance between two lines
3.5 Distance between two skew lines
3.6 Distance between parallel lines
3.7 Perpendicular distance of a point from a line
3.8 Image of a point in a straight line

4.1 Equation of plane in normal form
4.2 Equation of a plane perpendicular to a given vector and passing through a given point
4.3 Equation of plane passing through a given point and parallel to the two given vectors
4.4 Equation of plane passing through three non-collinear points
4.5 Intercept form of plane
4.6 Equation of plane passing through the intersection of two given planes

5.1 Angle between two planes
5.2 Angle between a line and a plane
5.3 Distance of a point from a plane
5.4 Distance between two parallel planes

Let us consider the co-ordinates of any arbitrary point on line $P$ be $(x,y,z)$, the co-ordinates of given point $A$ through which line passes be $({x_1},{y_1},{z_1})$ and the direction ratios of parallel vector be $a$, $b$, $c$.

Therefore,

$$\vec r = x\hat i + y\hat j + z\hat k{\text{ ;}}\;{\text{ }}\vec a = {x_1}\hat i + {y_1}\hat j + {z_1}\hat k{\text{ }}\;;{\text{ }}\vec b = a\hat i + b\hat j + c\hat k$$

Now, substitute these values in the vectorial form of line i.e., $$\overrightarrow r = \overrightarrow a + \lambda \overrightarrow b $$ We get

$$\begin{equation} \begin{aligned} \left( {x\widehat i + y\widehat j + z\widehat k} \right) = \left( {{x_1}\widehat i + {y_1}\widehat j + {z_1}\widehat k} \right) + \lambda \left( {a\widehat i + b\widehat j + c\widehat k} \right) \\ \left( {x\widehat i + y\widehat j + z\widehat k} \right) = \left( {{x_1} + \lambda a} \right)\widehat i + \left( {{y_1} + \lambda b} \right)\widehat i + \left( {{z_1} + \lambda c} \right)\widehat i \\\end{aligned} \end{equation} $$

On comparing the co-efficients, we get $$x = {x_1} + \lambda a;{\text{ }}y = {y_1} + \lambda b;{\text{ }}z = {z_1} + \lambda c$$

which are the parametric equations of the line.

By eliminating the parameter $\lambda $, we get, $$\frac{{x - {x_1}}}{a} = \frac{{y - {y_1}}}{b} = \frac{{z - {z_1}}}{c}$$ which is the cartesian form of the line.

Note: If $l$, $m$ and $n$ are the direction cosines of the line, then the equation of line becomes $$\frac{{x - {x_1}}}{l} = \frac{{y - {y_1}}}{m} = \frac{{z - {z_1}}}{n}$$

Question 6. If the cartesian equation of line is $6x-2=3y+1=2z-2$. Find its direction cosines.

Solution: In order to find the direction cosines from the cartesian equation of line, we have to arrange the given equation in the standard form i.e., $$\frac{{x - {x_1}}}{l} = \frac{{y - {y_1}}}{m} = \frac{{z - {z_1}}}{n}$$ Therefore, we can write the given equation as $$\begin{equation} \begin{aligned} \Rightarrow 6\left( {x - \frac{1}{3}} \right) = 3\left( {y + \frac{1}{3}} \right) = 2\left( {z - 1} \right) \\ \Rightarrow \frac{{x - \frac{1}{3}}}{{1/6}} = \frac{{y + \frac{1}{3}}}{{1/3}} = \frac{{z - 1}}{{1/2}} \\ \Rightarrow \frac{{x - \frac{1}{3}}}{1} = \frac{{y + \frac{1}{3}}}{2} = \frac{{z - 1}}{3} \\\end{aligned} \end{equation} $$ On comparing with the standard form, we get $${x_1} = \frac{1}{3}{\text{ }}{y_1} = - \frac{1}{3}{\text{ }}{z_1} = 1$$ which means the line passes through the point having co-ordinates $\left( {\frac{1}{3}, - \frac{1}{3},1} \right)$ and $$l = 1{\text{ }}m = 2{\text{ }}n = 3$$ which means the direction cosines are $(1,2,3)$.