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Vector distance formula 
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#1
Jul2507, 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. 


#2
Jul2507, 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 ABSin(t) where t is the angle between A and B.



#3
Jul2507, 02:10 PM

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P: 39,682

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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