How to Find C Coordinates if OA + BA + CA is Perpendicular to OB

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Michael_Light
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Homework Statement



Given A(1,7) and B(8,3). C lies on OB where O is the origin. If vector OA + BA + CA is perpendicular to vector OB, find the coordinates of C.


Homework Equations





The Attempt at a Solution



The answer i get is C=(112/73 , 42/73) but the answer given is (-71/2 , 195/7). What answer do you get? :smile:
 
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I agree, I think you may have written something wrong there.

The easiest way is using the dot product, as (OA+OB+OC) . (OB) = 0, then you'd get two equations, one for each C co-ordinate, that'd be easy to solve.

However I don't get the answer you say you were given.
 
Eight said:
I agree, I think you may have written something wrong there.

The easiest way is using the dot product, as (OA+OB+OC) . (OB) = 0, then you'd get two equations, one for each C co-ordinate, that'd be easy to solve.

However I don't get the answer you say you were given.

Why (OA+OB+OC) . (OB) = 0?

What's wrong with the question? Here is how i do.. probably someone can check it out for me...

C lies on OB, so C = k(8,3)

BA = OA-OB = (-7,4)

CA= OA-OC = (1-8k, 7-3k)

OA + BA + CA = (-5-8k, 18-3k)

OA + BA + CA is perpendicular to vector OB, so

OA + BA + CA . OB = 0

(-5-8k, 18-3k) . (8,3)=0

-8(5+8k) + 3(18-3k) = 0

k= 14/73

So C = k(8,3) = 14/73 . (8,3)

Which step have i go wrong?
 
Michael_Light said:
What's wrong with the question?
The question is okay. I was expecting a more common question, a little simpler. :smile:
 
The book's answer can't be correct. OB is in the first quadrant, so no point on it has an x-coordinate with a negative value. I reckon you could easily plot this exercise on graph paper, and determine the point C with sufficient accuracy to confirm (or not) your answer.

I looked at this, and found C as (232/73, 87/73) , just fractionally more than double what you calculated. I haven't checked my working though.