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 | * | \ Think about this the same way, and you should see why you no longer 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
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