Trig, Calc, Equations of lines...?

Question

Okay, imagine a 3D plane. With X, Y, Z coordinates. You have a line drawn from a given point to a given point. For instance, line A starting at 0, 0, 0, to 50, 50, 50. Or line B starting at -20, 50, -10, to 14, -7, 22.75. Or line C, 10, -10, 0, to 10, -50, 0. I need an equation, or method to use where when you are given the start XYZ, and the end XYZ, and you can determine what the XYZ coordinates will be any distance along the line. For instance, on line C, you would say, what is the XYZ for the point 5 units along the line. And the equation when 5 is plugged in returns 10, -15, 0. Because that's the point 5 units away from the line's origin. Lines A and B however are pointing in directions where all three Axis values have to be calculated because the line is not pointing along one single axis. Thanks

Answer 1

Consider your points as P = (p1, p2, p3) and Q = (q1, q2, q3).
The if you want to know the points a distance d from P along the line PQ, they are given by:
P +/- d(Q-P)/||Q-P||

Q-P means the vector: (q1-p1, q2-p2, q3-p3)
||Q-P|| means the magnitude of that vector - square each of the components, sum, and take a square root: If R = (r1, r2, r3) then ||R|| = sqrt(r1^2 + r2^2 + r3^3)

So in your example C, P = (10, -10, 0), Q = (10, -50, 0) so that Q-P = (0, -40, 0) and ||Q-P|| = 40. So the two points distance d from P will be:
(10, -10, 0) + d(0, -40, 0)/40 =
(10, -10, 0) + d(0, -1, 0)

Answer 2

The equation of a line in three dimensions can be given with vectors. Let's take your line B for example. Two points on that line are:

P(-20,50,-10) and Q(14,-7,22.75)

We can define a vector v:
v = Q - P = <14+20,-7-50,22.75+10> = <34,-57,32.75>

The magnitude of v is ||v||. So v / ||v|| is a unit vector in the direction of the line.

Let
P = vector from the origin to point P.
t = a constant that ranges over all the real numbers.

Then the equation of line B is:

B = P + t*v / ||v|| = <-20,50,-10> + (t / ||v||)<34,-57,32,75>

As t increases or decreases by 1 you will move a distance of 1 along line B.
____________________

||v|| = √[(14+20)² + (-7 - 50)² + (22.75 + 10)²]
||v|| = √(34² + 57² + 32.75²) = √5477.5625 ≈ 74.010557

Answer 3

f'(x) = 2cos(x)(-sin(x)) = -sin(2x) f'(pi/3) = -sin(2pi/3)=-sqrt(3)/2 so the equation of tangent line is y = -sqrt(3)/2 (x-pi/3) + a million/4 (cos(pi/3) = a million/2, cos^2(pi/3) = a million/4)