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## 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!