Vector Perpendicular: Finding Point P from A & B

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SUMMARY

The discussion focuses on finding the position vector of point P such that OP is perpendicular to the line segment AB defined by the vectors A (i - 5j - 7k) and B (10i + 10j + 5k). The direction vector of the line AB is given as (9i + 15j + 12k). To achieve perpendicularity, the dot product of the direction vector and the vector OP must equal zero. The solution involves determining appropriate values for the components of the vector OP that satisfy this condition.

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thomas49th
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Homework Statement



The equation for the line that goes through vector points A and B is

i - 5j - 7k + t(9i + 15j + 12k)

A is defined: i - 5j - 7k
B is defined: 10i + 10j + 5k

Find the position vector of point P such that OP is perpendicular to AB

The Attempt at a Solution



well a.b = 0 to be perpendicular

but how does that help...

Im lost

Any ideas. My test is at 9am tomorrow (GMT). Can someone just explain the whole method for me.


Thanks
 
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(9i + 15j + 12k) is your direction vector

so you need <a,b,c> direction vector such that
<a,b,c> . (9, 15, 12,) = 0

pick any values for a,b,c

so the line perpendicular is r = (0,0,0) + t <a,b,c>

now, find intersection
 
thomas49th said:
The equation for the line that goes through vector points A and B is

i - 5j - 7k + t(9i + 15j + 12k)

A is defined: i - 5j - 7k
B is defined: 10i + 10j + 5k

Find the position vector of point P such that OP is perpendicular to AB

well a.b = 0 to be perpendicular

Hi thomas49th! :smile:

(btw, B isn't 10i + 10j + 5k, is it? :wink:)

I assume P is to be on AB?

ok, then OP must be i - 5j - 7k + t(9i + 15j + 12k)
for some value of t.

And, as you say, AB.OP = 0.

So … ? :smile:

I got to go to bed now … :zzz:
 

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