Vectors- 4 Gr. 12 Questions - Equations of Line and Planes

[crazycat23]

Homework Statement

The point A(2,4,5) is reflected in the line with equation r = (0,0,1) + s(4,2,1) , sER, to give the point A'. Determine the co-ordinates of A'.

Homework Equations

1. The equation of the line (parametric equations)
2. A (2,4,5)
Dot Product, and distance from a point to a line

The Attempt at a Solution

I first began by finding the distance between r and A, equating it to r and A'. With A' being (a,b,c), and substituting this into the parametric equations of the line, i got the following:

824 = 5a^2 + 17b^2 + 20c^2 - 4bc - 8ac - 16ab

I'm unsure where to proceed from here, or even if this equation is correct.
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Homework Statement

For the lines L1 and L2, determine the co-ordinates of points on these lines that produce the minimal distance between L1 and L2.

Homework Equations

L1: r = (1,-2, 5) + s(0,1,-1), sER
L2: r = (1,-1,-2) + t(1,0,-1), tER
P1 (lies on L1) = (a,b,c)
P2 (lies on L2) = (d,e,f)

The Attempt at a Solution

I know that the distance from each point to it's opposite line is equal, and I also broke up each line equation into parameters and substituted its respective point. I arrived at this answer however, and looking back, I can't figure out my thought process as to how I got to it, so I wouldn't trust it. lol

distance = (2t^2 +2s^2 +14t - 14s - 2st + 49)^1/2

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Homework Statement

Point A(1,0,4) is reflected in the plane with equation x - y + z - 1 = 0.
Determine the co-ordinates of the image point.

Homework Equations

The plane equation
A(1,0,4) and it's reflection B(a,b,c)

The Attempt at a Solution

Again, distances are equal. I have two distrance from a point to a plane formulas equated to each other. I end up with this:

4 = l a - b + c - 1 l

I'm not exactly sure where to proceed from here.
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Homework Statement

A perpendicular line is drawn from point X(3,2,-5) to the plane 4x - 5y + z - 9 = 0 and meets the plane at point M. Determine the co-ordinates of point M.

Homework Equations

The points X, M and the plane equation

The Attempt at a Solution

Since the distance from a point to a plane is perpendicular, I found the distance from X to the plane as 12/(42^1/2).

I converted the plane equation into a vector equation, and then broke that into parametric equations. Here were mine: x = s + m, y = -s, z = 9 - 9s + 4m, with s and m being the parameters.

Using the distance between two points formula, and equating that to my distance, substituting the parameters, I receive this equation:

7a^2 + 7b^2 + 7c^2 - 42a - 28b + 70c + 242 = 0

Again, I don't know where to proceed
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Sorry I have so many questions. I hope you can help me, especially with my problem with finding equations and then getting stumped!

 

[HallsofIvy]

Construct the plane perpendicular to the given line, containing the given point.

That's easy: r = (0,0,1) + s(4,2,1) has "direction vector" <4, 2, 1> and that is perpendicular to any plane perpendicular to the line. In particular, the perpendicular plane that contains (2,4,5) has equation 4(x- 2)+ 2(y- 4)+ 1(z- 5)= 0 or 4x+ 2y+ z= 21.

Now determine where the line crosses that plane: where 4(4s)+ 2(2s)+ 1(1+ s)= 21.

Once you have found that point, construct the line through the two points and use the fact that point you want must be on that line, the same distance on the opposite sides of the point of intersetion.