# Calc III, Determing the coordinates of P

mr_coffee
Hello everyone. I'm stuck on the last part of this problem. It says, Draw two distinct nonzero postion vectors a = <x1,y1> and b = <x2,y2> such that the angle between them is less than pi/2. Draw the line L perpendicular with the line determined by the bector a such that l passes through the terminal point of b. Let P = (s,t) be the point where L intersects a. Determine the length of the line segment connecting the origin to the point P. Determine the coordinates of P. I drew a picture and I found the L = mx+b. The line determined by a = y1/x1 * x. m = -x1/y1 through (x2,y2). So now I have 2 lines, I need to find the intercept of these 2 lines to determine the coordinates of P (s,t). Any ideas on how i can do this? I'm also confused on what my 2 lines are exactly. Thanks.

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Homework Helper
thanks for the reply but I'm lost.
Quote:

You can work out the slope of OA (=m, say) using the coords of A and you know that BP is perpindicular to OA, so you can write the slope of BP in terms of m. You can also work out the slope of BP using the coords of P and B. Now solve for λ, and you can then work out the coords of P and the length of OP.

Where is the point OA coming from? by OA do u mean the vector O standing for origin and A for the a vector? So your saying set BP, B is a vector <x2,y2> and P is a point with coords (s,t) = m. Am i allowed to do this? BP = (y2-t)/(x2-s); or are u not talking about m = (y2-y1)/(x2-x1)?

You can think of the problem as one in vectors or one in coordinate geometry.

OA is a vector starting at the origin O and ending at the point A. The vector OA has the position vector a<x1,y1>.
OB is a vector starting at the origin O and ending at the point B. The vector OB has the position vector b<x2,y2>.
OP is a vector starting at the origin O and ending at the point P. The vector OP has the position vector p<s,t>.

Or,

treat it as a problem in coordinate geometry.
With an origin O and the points A(x1,y1), B(x2,y2), P(s,t).
Let m be the slope of OA.

You have already worked this out as m = (y1 - 0)/(x1 - 0) = y1/x1

So, the slope of the line OA is m = y1/x1
Then the eqn of the line OA is y = mx, or
y = (y1/x1)x ---------------------------------------------------(1)
=========

BP is perpindicular to OA, so you can write the slope of BP in terms of m.
So, slope of BP = m' = -1/m = -x1/y1, which you have.
Since you have the coords of B as (x2,y2), and the slope of BP, then you can write the eqn of BP as,
(y - y2)/(x - x2) = m'
y = m'(x-x2) + y2
y = -(x1/y1)(x-x2) + y2 ----------------------------------------(2)
==================

Eqn (1) is the eqn of the line OA.
Eqn (2) is the eqn of the line BP.

You can find out where these lines intersect by equating (1) with (2) and solving for x and y, which will be s and t respectively, the coords of P.

mr_coffee
Thanks a lot for your explanation,it was very helpful! but I'm stuck I think. When you said set equation 1 and 2 together and solve for x and y, did you mean x1 and y1 or did u mean x and y? When I set the two equations eqqual to each other, y isn't there, y1 is though. I tried solving for x and got a really messy equation. Here is my work:
http://img143.imageshack.us/img143/5492/gsfdg5pa.jpg [Broken]
Thanks.

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Homework Helper
mr_coffee said:
Thanks a lot for your explanation,it was very helpful! but I'm stuck I think. When you said set equation 1 and 2 together and solve for x and y, did you mean x1 and y1 or did u mean x and y? When I set the two equations eqqual to each other, y isn't there, y1 is though. I tried solving for x and got a really messy equation. Here is my work:
http://img143.imageshack.us/img143/5492/gsfdg5pa.jpg [Broken]
Thanks.
You're almost there.

Yes, you have to solve for x and y. You will then get expressions for x and y in terms of x1, y1, x2, y2. The final expression you got can be rearranged to give,

$$x = x_1\frac{(x_1x_2 + y_1y_2)}{(x_1^2 + y_1^2)}$$

Substitute this value for x into eqn (1) and get the value of y.

These values of x and y are the x- and y-coords of the intersection of the line-eqns (1) and (2). And these coords are the coords of the point P. ie x = s and y = t.

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mr_coffee
Awesome, thanks so much for the help! :) I was wondering how u got $$x = x_1\frac{(x_1x_2 + y_1y_2)}{(x_1^2 + y_1^2)}$$ I posted how far I got with rearranging but I didn't see how you did that exactly. http://img143.imageshack.us/img143/650/pointsss7gn.jpg [Broken]

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Homework Helper
mr_coffee said:
Awesome, thanks so much for the help! :) I was wondering how u got $$x = x_1\frac{(x_1x_2 + y_1y_2)}{(x_1^2 + y_1^2)}$$ I posted how far I got with rearranging but I didn't see how you did that exactly. http://img143.imageshack.us/img143/650/pointsss7gn.jpg [Broken]
Staring from here,

$$\left (\frac{y_1}{x_1}\right )x + \frac{x_1(x)}{y_1} = \frac{x_1x_2}{y_1} + y_2$$

$$x\left (\frac{y_1^2 + x_1^2}{x_1y_1}\right ) =\frac{x_1x_2 + y_1y_2}{y_1}$$

$$x = x_1\left (\frac{x_1x_2 + y_1y_2}{x_1^2 + y_1^2}\right )$$

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mr_coffee
awesome, thanks a lot for the help!