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### Vectors and Linear Combinations

```
Date: 01/25/99 at 21:02:53
From: Stacy Doud
Subject: Plane equations

I cannot explain the following statement. If vector v0 = (x0, y0, z0)
and vector v1 = (x1, y1, z1) both satisfy the linear equation ax + by
+ cz = d, then so does mv0 + nv1, provided that m + n = 1.

Is this because v0 and v1 are solutions to the linear equation, and
the addition of these two vectors must have scalar multipliers that
equal 1, so that the addition of mv0 + nv1 is also a solution?

Another part of the question states that all combinations of v0 and
v1 satisfy the linear equation when the right side is zero. Why is
that true?
```

```
Date: 01/26/99 at 12:57:13
From: Doctor Peterson
Subject: Re: Plane equations

Hi, Stacy. Let's look at what these statements mean, restricting
ourselves to a plane to make my drawings simpler. A general line
(d non-zero) will look like this:

\|
\
|\
| \
|  \ v0
|   *
|  / \
| /   \
|/     \
-------------------+-------\-----------
| \      \
|   \     \
|     \    \
|       \   \
|         \  \
|           \ \
|             \\ v1
|               *
|                \

The sum mv0 + nv1, with m + n = 1 is a specific kind of linear
combination: a weighted average of the two vectors. Here's one way to
see that only these combinations will be on the line:

We are given two points on the line. Any point on the line will be the
sum of v0 and some multiple of (v1 - v0), which is a vector in the
direction of the line. So the general vector on the line will be

v = v0 + n(v1 - v0) = v0 + nv1 - nv0 = (1-n)v0 + nv1

Do you see how that gives you what you are looking for?

A line with d = 0 looks like this:

\        |
\       |
\      |
\ v0  |
*    |
\   |
\  |
\ |
\|
-------------------+------------------
|\
| \
|  \
|   \
|    \
|     \
|      \ v1
|       *
|        \

have to restrict m and n as we did before. Notice that v0 and v1 are
no longer independent. Each is a multiple of the other.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 01/26/99 at 14:35:56
From: Stacy Doud
Subject: Re: plane equations

Dr. Peterson,

Thank you for the prompt response. And yes, that helps me tremendously.
I have a better understanding and can do the problem now.

Thank you,
Stacy Doud
```
Associated Topics:
High School Linear Equations

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