let P be a point NOT on line L that passes through points Q and R. [tex] \vec{A} = QR[/tex] [tex]\vec{B} = QP [/tex] prove that distance from point P to anywhere on line L is [tex] d = |\vec{A} x \vec{B}| divided by |\vec{A}| [/tex] so, i've tried doing the cross product after assigning variables for the A and B components. I ended up with a very tedious long multiplication of several variables, and I was wondering if there is an easier way to prove this formula.
Draw a picture of what is going on and note that |AxB| is the area of the parallelogram generated by A and B. It's also equal to |A||B|Sin(t) where t is the angle between A and B.
Of course the shortest distance from P to a line is along the line through P perpendicular to the line. You might start by finding the equation of a line through P perpendicular to [itex]\vec{QR}[/itex].