Calc III, Determing the coordinates of P

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? Thanks. I'm confused on what my 2 lines are exactly. Here is a picture, don't look at the left part, that is incorrect but the drawing itself is right. //w00t

Here's the approach I would take:

1. Equation of line corresponding to x1,y1:
y = (y1/x1)x, or y = m1x

2. Equation of line corresponding to x2,y2:
y = (y2/x2)x, or y = m2x

3. Equation of perpendicular line (using point/slope form):
y-y2 = (-x1/y1)(x-x2). Rearranging, y = (-x1/y1)x + (y2 + x2x1/y1), or y = m3x + b, where m3 = -x1/y1 and b = y2 + x2x1/y1

4. To find the x value of the intersection between line 1 and line 3 (I'll call x3), set equations (1) and (3) equal to each other and solve for x:
m1x3 = m3x3 + b, or (m1-m3)x3 = b, or x3 = b/(m1-m3)

5. To find the y value of the intersection between line 1 and line 3 (I'll call y3), take the value of x from (4), substitute into equation 1, and solve for y

6. Now you have the values of x3, y3 in terms of x1,y1,x2,y2. The length of the line segments is sqrt(x3*x3 + y3*y3).

VBAhack

P.S. This is Calc III? It's pure algebra and geometry

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