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Vector distance formula

by linuspauling
Tags: distance, formula, vector
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linuspauling
#1
Jul25-07, 02:59 AM
P: 11
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.
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ObsessiveMathsFreak
#2
Jul25-07, 06:04 AM
P: 406
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.
HallsofIvy
#3
Jul25-07, 02:10 PM
Math
Emeritus
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Thanks
PF Gold
P: 39,322
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].


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