Problem

In the figure, the tip of $ c $ and the tail of $…

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Problem 6 Easy Difficulty

Copy the vectors in the figure and use them to draw the following vectors.
(a) $ a + b $
(b) $ a - b $
(c) $ \frac{1}{2} a $
(d) $ -3b $
(e) $ a + 2b $
(f) $ 2b - a $

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Video Transcript

So in order for us to add these vectors, let's first look at what the vectors are. We're gonna call the red Vector. We'll have a red vector A and that's going to look like I will have it be longer. So this is going to be a red vector A. And then we're going to have a blue vector. And that's going to be be. So now what we have is be vector, and we have our a vector. So using this will be able to create all of the vectors that we're looking for. So the first one is just a plus B. That's pretty simple. All we have to do is take this factor here. Then we take this factor here, and we want to put it its tail at the head of the A vector. And then our answer for a plus B is going to be the pink factor. So this will be a plus B. So now let's do that with the next problem. Only this time, rather than a plus B, it's going to end up being a minus B. So for a minus, B, um, it's going to be the same thing as a plus, a negative B. So what we mean by that is we'll take our again well duplicated. And then rather than adding B to it, we're going to add a negative B. So it's going to look like this because essentially what we did as we flip this around, so that's gonna be our minus B. And then, as you can figure, this right here will be a plus, a negative B, which is the same thing as a minus. B next victor. We'll bring these down, so we have them with us always. So next we want to do one half a. So that would be the same thing as taking this a vector and just shrinking it by a factor of two. So if we just take it and drink it by a factor of two, that right there is are one half a. It's just the magnitude scaled down by half. Now we want to look at negative three B, so the first thing we want to do for negative three b is we first want to flip it around so it's facing this way and then we want to expand that. So it goes three times the length and this is not going to be perfect. But if we stretch it out enough and it looks right, that right there will be our negative three B. Then we have a plus to be so a plus two b is going to look like a what we have right here and then to get the to be we want this to be twice the length. So here's something fun that we can do to make it easier to view. So we'll do one B to be so that right there is a plus to be we could draw it as one large vector or to individual be vectors. The same thing is true, and this right here will be a plus to be. Then, lastly, we want to B minus a so to do, to be minus a Let's just copy these guys right here. Since that's our to be And then, um, to do the minus A. We're gonna want the A to go the opposite direction. So we're gonna take this, duplicate it, and it would be as if it's going the opposite direction. So in order to do that, we're gonna flip it like this. So now that's our negative. A. So now we can add to be plus, um, to be plus that negative a value. So, um, that right there would go here, and we want to go. Um, we're going to move that here. So that way, this a can now get dragged right here. So that's two B minus A. And that's going to look like this vector right here to be plus a negative a vector, which is the same thing as to be minus a.

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Problem

In the figure, the tip of $ c $ and the tail of $…

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Watch More Solved Questions in Chapter 12

Video Transcript

James Stewart

Calculus: Early Transcendentals

Lily An

Caleb Elmore

Kristen Karbon

Joseph Lentino

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Lily An

Caleb Elmore

Kristen Karbon

Joseph Lentino

James StewartCalculus: Early Transcendentals

Lily An

Caleb Elmore

Kristen Karbon

Joseph Lentino

James Stewart

Calculus: Early Transcendentals