noooffence said:
i know the direction vector is <1,-2,-5>
That's right, <1,-2,-5> is perpendicular to the plane, and will be parallel to whichever line you chose.
noooffence said:
and i got a random point <5,8,11>.
To test whether this is a point in the plane, try substituting its components, <x,y,z>, into the equation x-2y-5z=6.
noooffence said:
i also know that for an equation that is perpendicular to the plane a.b = 0
but i don't know how to use the dot product to find the vector of that equation...
A parametric equation for a line looks like this:
\textbf{R}(t)=\textbf{R}(0)+t\textbf{V}
where \textbf{R}(t) for all values of t are position vectors representing each of the points on the line, \textbf{R}(0) is the position vector for one particular point on the line where your parameter t happens to be equal to 0, and \textbf{V} is a vector parallel to the line. Oh, and the parameter t takes the value of each of the real numbers.
One way to describe a plane is in terms of which vectors are perpendicular to a certain vector, \textbf{N}.
\textbf{N} \cdot (\textbf{R}-\textbf{R}_0)=0
\textbf{N} \cdot \textbf{R}= \textbf{N} \cdot \textbf{R}_0
Where \textbf{R}_0 is a constant position vector indicating some particular point in the plane, and position vectors of the form \textbf{R} stand for each of the other points in the plane. In your case,
\textbf{N} \cdot \textbf{R}= 6
with \textbf{R} = <x,y,z> and \textbf{N} = <1,-2,-5>, as you worked out.
