## Parametric 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

### Parametric form of a 3D line

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

How can we find the parametric form of the equations of the line passing through the point $(3,-5,2)$ and is parallel to the vector $b= \begin{pmatrix} 1 \\ 7 \\ -6 \end{pmatrix}$ ?

Thanks
lora

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Joined: Wed Apr 10, 2013 7:36 pm

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

Vectors, Parametric equation form of a line - IB Maths HL

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

$x=a_{1} + \lambda b_{1}$

$y=a_{2} + \lambda b_{2}$

$z=a_{3} + \lambda b_{3}$

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)

$\lambda$ is a real number parameter.

Therefore for your question we have that the parametric equation form is

$x=3 + 1\lambda$

$y=-5 +7 \lambda$

$y=2 -6 \lambda$

Hope these help!!
lily

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Joined: Mon Jan 28, 2013 8:04 pm

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

Thanks Lily!!
lora

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