For the vectors A⃗ and B⃗ in Fig. E 1.28, use a scale drawing to find the magnitude and directio

step 1 For this question, we're asked to reference figure E1 .28. And we're supposed to use drawings to

For the vectors A⃗ and B⃗ in Fig. E 1.28, use a scale...

Key Concepts

Magnitude and Direction

Magnitude and direction are the two key properties that define a vector. The magnitude is a measure of the vector's length or size, while the direction indicates the angle or orientation of the vector relative to a reference axis. These properties are crucial in understanding and representing vectors in both two-dimensional and three-dimensional spaces.

Negative Vectors

A negative vector is obtained by reversing the direction of the original vector while maintaining its magnitude. This concept is important for operations like vector subtraction, as subtracting a vector is equivalent to adding its negative. It is also used to indicate oppositely directed forces or displacements in various physical contexts.

Scale Drawing

A scale drawing is a graphical method used to represent vectors accurately by adhering to a specific scale factor. It allows for the visualization and measurement of vectors' magnitudes and directions on paper, which can then be utilized to approximate the resultant vector from vector addition or subtraction. This method is especially useful when numerical computation is less feasible or to provide intuitive insight into vector relationships.

Vector Addition

Vector addition is the process of combining two or more vectors to produce a resultant vector. It is typically performed using the head-to-tail method, where the tail of one vector is placed at the head of the preceding vector, and the resultant vector is drawn from the tail of the first vector to the head of the last vector. This concept is fundamental in physics and mathematics as it describes how different vector quantities interact.

Vector Subtraction

Vector subtraction involves finding the difference between two vectors, which is equivalent to adding one vector to the negative of the other. This is often performed using the same geometric methods as vector addition, but with a focus on reversing the direction of the vector being subtracted. The result represents the relative displacement between the two vectors.

Transcript

00:01 For this question, we're asked to reference figure e1 .28.

00:05 And we're supposed to use drawings to figure out the vector sum for a plus b for part a and the vector difference of a minus b for part b.

00:15 And then using those answers, find the magnitude and direction of minus a minus b for c and b for part d.

00:25 Okay.

00:26 So we're going to start out with just writing down the magnitude of vector a and vector b here.

00:31 Which doesn't include direction.

00:33 But for part a and b, we're going to draw images here.

00:37 So first, let's just start off by simply drawing the x and y axis.

00:43 Okay.

00:43 Now i'm going to use red to draw our vectors that we're summing together.

00:49 So the first one here, let's go ahead and write it up top.

00:52 This one is for a vector plus b vector.

00:58 And we're using that figure, given in the question, to reference these.

01:01 And we're adding them together.

01:03 So first we have vector a.

01:05 Vector a is 8 meters, and it is straight downed on the y -axis.

01:11 So there's vector a.

01:15 If you look at the diagram, vector b makes an angle of 30 degrees away from the y -axis, and it's going to come straight out from vector a if we're adding them vectorally.

01:24 So it kind of goes up like this.

01:30 There's vector b, and it makes an angle here of 30 degrees.

01:40 Now, what we want to find is the resulting vector from the sum, which would be right there, and that's vector r, as well as the angle here that it makes theta with that x -axis.

01:57 So we want to figure out the angle with the x -axis.

01:59 Okay.

02:00 So in order to do this, we're going to use some trickery here.

02:05 I'm going to use some green dash lines here to kind of draw it down.

02:08 So you have a couple of right angles i get made.

02:11 There's a right angle there.

02:13 And then there's also a right angle over here.

02:17 So what you'll realize is that the rx value, so the value of r, the vector r projected on the x axis, is the same as b at x.

02:27 Because what we want to try to find is that magnitude r.

02:29 And i'll just draw it over to the side so we can kind of see what i'm talking about here.

02:32 So you have this magnitude r that we're trying to find.

02:35 But r must have some kind of x component and some kind of y component.

02:38 It, right? so if you can, this would be r of y and this would be r of x.

02:43 So if you can find r of y and r of x, you can find r using pythagorean theorem and you can also find theta, right? okay, so we're going to use those right angles that i drew over here to find r of y and r of x.

02:58 So again, like i said, the first thing we're going to notice is that r of x is equal to b of x.

03:06 So the projection of on the x -axis is going to be the exact same as the projection of b on the x -axis because there's no x value of a right the y value of r is going to have components of b value of w or y value of b and y value of a because they both have y values but a is only on the y -axis so r of x is going to be equal to b of x and uh using our trigon metric identities b of x you'll see that it makes a a a larger right angle here let's draw i guess just not to confuse let's draw this one in blue you see this right angle here that b makes if you could follow that green line all the way down so here's a right angle this angle then here theta has to be equal to 30 minus or 90 minus 30 so this has to be equal to 60 degrees right so then the b projection or b of x projection is going to be b times the cosine of 60 degrees.

04:07 So b times the cosine of 60 degrees comes out to equal 7 .5 meters.

04:12 So now we have r of x.

04:15 Okay.

04:16 Well now to get r of y, r of y is going to be the b value of y, right? because that would be the entire right side of that blue dash green dotted line that's made from the b right triangle.

04:36 So it's going to be the b value of y minus a of y, because a of y goes negative there.

04:41 So this would be minus a of y.

04:43 Well, the b value of y is b times the sign of that angle 60 degrees, right, minus a of y, but a of y is just all of a, right? it's just going to be that eight meters because the y or a is only in the y direction.

05:05 Okay.

05:06 So it's just going to be minus the magnitude of a because that's all, everything is, everything for a is in the y direction.

05:13 So carrying this operation out, we find that this is equal to 4 .99 meters.

05:21 So now that we know r of x and r of y, we can find r magnitude from pythagorean theorem.

05:28 We start a new page here, so we ran out of room.

05:32 That's going to be r of x squared plus r of y squared.

05:41 Plugging those values in, we find a magnitude of r of approximately nine meters.

05:48 Next, we need to define that angle in the graph, the angle theta, right? okay, well, now that we know r of x and r of y, we can just use our trigonometric identities.

06:00 So the angle theta is equal to the inverse tangent of the ratio r of y to r of x.

06:08 And that's because in the trigonometric identities, tangent theta is equal to opposite over adjacent.

06:14 Opposite would be r of y, adjacent would be r of x.

06:17 So you just have to take the inverse of, or inverse tangent of both sides to get theta by itself.

06:22 So inverse tangent of r of y over r of x.

06:24 And we find that this is equal to approximately 34 degrees.

06:29 So we can box that in as our angle that's made with the x axis.

06:33 Okay.

06:34 Now for part b, we're asked to do the same thing, but this time we're going to use a minus b instead of a plus b.

06:49 So again, referencing the same figure to see a and b.

06:52 Let's go ahead and draw out.

06:55 Actually, let's move over a little bit.