Question 1201855: from my 'Vector Equations of Planes' lesson
1.
the plane with the equation (vector) r = (1,2,3)+ s (1,2,5) + t(1,-1,3) where s, t E R inrersects the y and z-axes at the points A and B. Determine the equation of the line thru the points A and B.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Answer:
Either
< x,y,z > = < 0,3,0 > + t*< 0,-3,-2 >
Or
< x,y,z > = < 0,3-3t,-2t >
where t is any real number.
Explanation:
Each point in 3-space, aka 3d space, is of the form (x,y,z).
Points on the y axis have the x and z coordinates equal to zero.
The y coordinate could be zero, or could be any other real number.
So we have points of the form (0,y,0).
Since point A is the y-intercept, let's have the point be labeled (0,A,0).
We also have (0,0,B) which represents the z intercept located at point B.
Now let's rewrite the equation of the plane.
r = (1,2,3)+ s(1,2,5) + t(1,-1,3)
r = (1,2,3)+ (s,2s,5s) + (t,-t,3t)
r = (1+s+t,2+2s-t,3+5s+3t)
(x,y,z) = (1+s+t,2+2s-t,3+5s+3t)
That last equation breaks down into these three equations
x = 1+s+t
y = 2+2s-t
z = 3+5s+3t
If we knew what s and t were, then we'd determine a location (x,y,z)
Doing this a whole bunch of times for infinitely many r and s values will generate a flat plane that goes on forever in all directions.
Plug in the coordinates of the y intercept.
0 = 1+s+t
A = 2+2s-t
0 = 3+5s+3t
Solve the subsystem
0 = 1+s+t
0 = 3+5s+3t
and you should get the solution
s = 0, t = -1
I skipped the steps to solving, which I'll leave for the student to do.
Let me know if you get stuck here.
So,
A = 2+2s-t
A = 2+2*0-(-1)
A = 3
Meaning the point A is located at (0,3,0) as the y intercept of this plane.
Now plug in the coordinates of the z intercept.
0 = 1+s+t
0 = 2+2s-t
B = 3+5s+3t
Solve this subsystem
0 = 1+s+t,
0 = 2+2s-t
to get
s = -1, t = 0
Therefore,
B = 3+5s+3t
B = 3+5*(-1)+3*0
B = -2
The z intercept is located at (0,0,-2)
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We now need to find the equation of the line through A(0,3,0) and B(0,0,-2)
Subtract the two sets of coordinates to determine the vector that points from A to B
B-A = (0,0,-2) - (0,3,0)
B-A = (0-0,0-3,-2-0)
B-A = (0,-3,-2)
To go from A to B, we do three things:- Move 0 units along the x axis.
- Move 3 units along the negative y axis.
- Move 2 units along the negative z axis.
This vector <0,-3,-2> gives us the direction in which to move from point to point along this line.
It is analogous to the slope in 2D settings.
Let's say we started at A(0,3,0) which is the y intercept.
Add on the direction vector mentioned earlier, but scaled up by some parameter t.
< x,y,z > = some vector in 3 space
< x,y,z > = startPoint + t*(direction vector)
< x,y,z > = < 0,3,0 > + t*< 0,-3,-2 >
< x,y,z > = < 0,3,0 > + < t*0,t*(-3),t*(-2) >
< x,y,z > = < 0,3,0 > + <0,-3t,-2t>
< x,y,z > = < 0+0,3+(-3t),0+(-2t) >
< x,y,z > = < 0,3-3t,-2t >
< x,y,z > = < 0,3-3t,-2t > where t is any real number.
Let's see what happens when t = 0
(x,y,z) = (0,3-3t,-2t)
(x,y,z) = (0,3-3*0,-2*0)
(x,y,z) = (0,3,0)
We're at the point A(0,3,0) which is the y intercept.
Now try t = 1.
This represents moving forward 1 unit in time.
(x,y,z) = (0,3-3t,-2t)
(x,y,z) = (0,3-3*1,-2*1)
(x,y,z) = (0,0,-2)
We've arrived at the z intercept (0,0,-2) which is point B.
This confirms we have the correct equation of the line.
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