Calc III, Determing the coordinates of P

In summary, the conversation discusses the process of finding the coordinates and length of a point P and line L based on the vectors a and b and their intersection. The participants discuss using slopes and equations to find the coordinates of P, with the final solution being x = x1*(x1x2 + y1y2)/(x1^2 + y1^2).
  • #1
mr_coffee
1,629
1
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.
http://s7.imagehosting.us/uploadpoint/imagehosting_upload_storage/nouser_654/T0_-1_654150.jpg
 
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  • #2
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)?
I missed your response on the other thread, sorry.

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.
 
  • #3
Thanks a lot for your explanation,it was very helpful! :smile: 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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  • #4
mr_coffee said:
Thanks a lot for your explanation,it was very helpful! :smile: 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,

[tex]x = x_1\frac{(x_1x_2 + y_1y_2)}{(x_1^2 + y_1^2)}[/tex]

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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  • #5
Awesome, thanks so much for the help! :) I was wondering how u got [tex]x = x_1\frac{(x_1x_2 + y_1y_2)}{(x_1^2 + y_1^2)}[/tex] 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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  • #6
mr_coffee said:
Awesome, thanks so much for the help! :) I was wondering how u got [tex]x = x_1\frac{(x_1x_2 + y_1y_2)}{(x_1^2 + y_1^2)}[/tex] 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,

[tex]\left (\frac{y_1}{x_1}\right )x + \frac{x_1(x)}{y_1} = \frac{x_1x_2}{y_1} + y_2[/tex]

[tex]x\left (\frac{y_1^2 + x_1^2}{x_1y_1}\right ) =\frac{x_1x_2 + y_1y_2}{y_1}[/tex]

[tex]x = x_1\left (\frac{x_1x_2 + y_1y_2}{x_1^2 + y_1^2}\right )[/tex]
 
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  • #7
awesome, thanks a lot for the help! :biggrin:
 

1. What is Calc III and why is it important?

Calc III, short for Calculus III, is a college-level mathematics course that builds upon the concepts learned in Calculus I and II. It focuses on multivariable calculus, which includes topics such as vectors, partial derivatives, multiple integrals, and vector calculus. This subject is important for students pursuing degrees in fields such as engineering, physics, and mathematics, as well as for those interested in advanced applied mathematics.

2. What are the steps for determining the coordinates of point P?

To determine the coordinates of point P, you will need to follow these steps:

  • Identify the given information, such as the equation or graph that contains point P.
  • Find the coordinates of any known points on the same line or curve as P.
  • Use equations or algebraic manipulation to solve for the coordinates of P.
  • Check your answer by plugging in the coordinates of P into the original equation or graph.

3. How do you find the coordinates of P using vectors?

When working with vectors, the coordinates of point P can be found by using vector operations such as addition, subtraction, and scalar multiplication. If P lies on a line, you can use the direction vector of the line and the coordinates of another point on the line to find the coordinates of P. If P lies on a curve, you will need to use parametric equations or vector-valued functions to find the coordinates of P.

4. Can you determine the coordinates of P without knowing its equation or graph?

Yes, it is possible to determine the coordinates of P without knowing its equation or graph. This can be done by using a system of equations or by using geometric properties such as symmetry or perpendicularity. It may also be helpful to use a graphing calculator or other technology to visualize the problem and find the coordinates of P.

5. What are some common mistakes to avoid when determining the coordinates of P?

Some common mistakes to avoid when determining the coordinates of P include:

  • Forgetting to check your answer by plugging it into the original equation or graph.
  • Not using the correct units or coordinate system.
  • Missing a negative sign or making a sign error when solving equations.
  • Not considering all possible solutions or cases when solving for P.

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