# Suppose V1,v2,v3 Is An Orthogonal Set Of Vectors In R5. Let W Be In Span(v1,v2,v3) Such That ...

Okay in this linear algebra problem, we want to suppose that V 1 V 2, V 3 is an orthogonal set of vectors in R 5. And we want to let W be in span of v1 v2 v3, such that v1 dot v1 equals 12 v2 dot, v2 equals 2, V, 3, V, 3 equals 4 and W dot v1 equals negative 12 and W dot v2 equals 8 and WV 3 equals negative 16 the way the computer they're doing this on a computer would want you to put it in will actually just be the answer regardless. But most people use computers.

These days it's going to be W this Where. So a lower case equals unknown to the one plus z2, plus b3. Okay.

So you look at this I, don't know if you look at this, it may confuse you very much I'm, not going to go into the details of the theory behind everything because you should never pretty good understand it for taking linear algebra of this kind of things right up. When you look at this, you should immediately recognize that you've got dot products in you say, you got v1 dot v1, which is the magnitude of v1 squared based on the. Properties of dot products and so forth.

So if you go to properties of the dot products, and you go to number three is usually what it would be listed as in your book or so you would have you would have a dot B plus C equals a dot B equals a dot C. And because of that property, you can recognize that we have a V dot B in aw dot B. And we can use this in the same fashion that we would use this. And so this is negative 12. And this is 12.

So we would take I'm working with the first one here. We. Would take v1 dot v1 equals 12.

Now we want to make this equal to 12. And so we would just move the negative sign over so equals negative W dot v1. And so we can get rid of the 12 part here and just say, v1 dot, v1 equals a negative W dot v1. And since v1 dot v1 equals a ww1 based on this property here we can divide them by each other and get v1 dot v1 over negative. W dot, v1 equals negative 1.

And so your answer for W is negative 1. And this was equal. Please 1 1 1 times, v1 and same for the rest of them. You just simply divide them out. And so he divided by 2 is going to be before and negative 16/14 means it's negative for him. So just simple properties of the dot product you just apply those.