Hi there. In this math post, I cover the vector equation of a line.
Math images/text mostly from QuickLaTeX.com.
Before giving the vector equation of a line, I cover the parts of the equation first.
We have a position vector of a known point on the line. This is
There is a direction vector which tells how many units to move for x, y and z from the position vector.
A scalar parameter t is used that multiples the direction vector by any real number.
All of the above components gives the position vector r = (x, y, z) of any point on the line.
Note that negative values of the parameter t makes the direction vector go in the opposite direction.
Source 2: Nelson Calculus & Vectors Grade 12 Textbook for Ontario Math
Picture and example below from Jack's Math Youtube video.
The above vector equation can be also represented by:
This is also:
Find the vector equation of a line that passes through the point A(0, 2) with a direction vector of d = (4, 1).
We have the pieces for the vector equation formula. The position vector is the point A(0, 2). and the direction vector is d = (4, 1). Substitute accordingly.
I add the x co-ordinate pieces together to get 0 + 4t = 4t. Adding the y components gives 2 + t.
Find the vector equation of a line passing through point A(1, 2, -3) with a direction vector d = (0, 2, -3).
This example is similar to the first example. We have three dimensions now.
What if the direction vector is not given but you have two points? The two given points can be used to obtain a direction vector.
Find the equation of a line passing through C(3, -1, 7) and D(1, 2, 4).
The direction vector can be obtained by doing C minus D or D minus C. I do D minus C as I want to go from point C to point D.
Use point C as the position vector and the newly obtained direction vector for the vector equation formula.
Thank you for reading.
