## Cartesian form of a 3D line

Discussions for the Core part of the syllabus. Algebra, Functions and equations, Circular functions and trigonometry, Vectors, Statistics and probability, Calculus. IB Maths HL Revision Notes

### Cartesian form of a 3D line

Vectors, Cartesian equation form of a line 3D, IB Maths HL

How can we find the Cartesian equation form of the line passing through the point $(2,-1,8)$ and is parallel to the vector $b= \begin{pmatrix} 4 \\ -2 \\ 7 \end{pmatrix}$ ?

Thanks
lora

Posts: 0
Joined: Wed Apr 10, 2013 7:36 pm

### Re: Cartesian form of a 3D line

The Cartesian form for the equation of a straight line is :

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

where $a= \begin{pmatrix} a_{1} \\ a_{2} \\ a_{3} \end{pmatrix}$ is a position vector (a point on the line)

and $b= \begin{pmatrix} b_{1} \\ b_{2} \\ b_{3} \end{pmatrix}$ is a direction vector ( a vector parallel to the line)

Concerning your question we have that the Cartesian equation form is given by the following formula:

$\frac{x-2}{4}= \frac{y-(-1)}{-2}= \frac{z-8}{7}$

Hope these help!
lily

Posts: 0
Joined: Mon Jan 28, 2013 8:04 pm

### Re: Cartesian form of a 3D line

Thanks Lily!!
lora

Posts: 0
Joined: Wed Apr 10, 2013 7:36 pm