IB Maths HL, SL, Studies: IB Mathematics HL – Vectors, Cartesian equation form of a line 3D

Thursday, March 28, 2013

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

How can we find the Cartesian equation form of the line passing through the point (1,4,9) and is parallel to the vector $b= \begin{pmatrix} 5 \\ 1 \\ -8 \end{pmatrix}$ ?

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:

$\frac{x-1}{5}= \frac{y-4}{1}= \frac{z-9}{-8}$

IB Maths HL, SL, Studies

Thursday, March 28, 2013

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

No comments:

Post a Comment

About Me