Question

Find the vector and the cartesian equations of the lines that pass through the origin and(5,-2,3).

Answer 1

The required line passes through the origin.

Therefore, its position vector is given by, $\vec a$=$\vec 0$…………....(1)

The direction ratios of the line through the origin and (5,-2,3) are (5-0)=5, (-2-0)=-2, (3-0)=3
The lines are parallel to the vector given by the equation, b→5$\hat i$-2$\hat j$+3$\hat k$

The equation of the line in vector form through a point with position vector a and parallel to $\vec b$ is,
$\vec r = \vec a+\lambda \vec b$, λ∈R
⇒ $\vec r=\vec 0+\lambda(5\vec i-2\vec j+3\vec k)$
⇒ $\vec r=\lambda(5\vec i-2\vec j+3\vec k)$

The equation of the line through the point (x1,y1,z1) and direction ratios a,b,c is given by,
$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$

Therefore, the equation of the required line in the cartesian form is
$\frac{x-0}{5}=\frac{y-0}{-2}=\frac{z-0}{3}$
⇒$\frac{x}{5}=\frac{y}{-2}=\frac{z}{3}$