Vector Perpendicular: Finding Point P from A & B

thomas49th
Messages
645
Reaction score
0

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
 
Physics news on Phys.org
(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:
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top