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Lines and Planes – Creating a Perpendicular Line (Example 2)
The equation of a line in general is given by:
\[r = {r_O} + tv\]
\(r_O\) is the initial point, t is time, and v is the vector direction.
Example question: Find the line passing through (1,0,6) and perpendicular to the plane x + 3y + z = 5
When a plane is written in the notation shown above, the coefficients of x, y and z represent the “norm” of a plane, which is a vector that points perpendicular to the plane.
So we can just use the norm of that plane as a vector pointing in a perpendicular direction.
In this case the norm vector from the plane is <1,3,1>, which we multiply by t in our r(t) equation shown previously.
Use the point given as our initial point.
\[r = \left\langle {1,0,6} \right\rangle + \left\langle {t,3t,t} \right\rangle \]
\[r = \left\langle {1 + t,3t,6 + t} \right\rangle \]
Try another example, where we make a plane using three given points and the cross-product!