Proof Prove that 𝐮 ×𝐯=𝐮𝐯 if 𝐮 and 𝐯 are orthogonal. |...

Name: Problem 61, Chapter 10: Precalculus with Limits : A Graphing Approach
Uploaded: 2020-12-17T19:52:53-08:00
Duration: 17 min 52 s
Channel: Cullen Miller
Description: Proof Prove that 𝐮 ×𝐯=𝐮𝐯 if 𝐮 and 𝐯 are orthogonal.

Key Concepts

Cross Product

The cross product of two vectors in three-dimensional space is a binary operation that results in another vector which is orthogonal to both of the original vectors. It is defined in terms of the magnitudes of the original vectors and the sine of the angle between them, making it closely related to geometric concepts such as area and orientation.

Norm (Magnitude) of a Vector

The norm or magnitude of a vector is a measure of its length in space. It is computed using the square root of the sum of the squares of its components, and it plays a crucial role in understanding vector operations, including the cross product and the calculation of areas or distances.

Orthogonality

Two vectors are said to be orthogonal if their dot product is zero, meaning they are at right angles to each other. This property simplifies many vector calculations, as it implies that the sine of the angle between the vectors is one, thus making certain expressions involving the cross product simpler.

Trigonometric Relationship in the Cross Product

The cross product involves the sine of the angle between two vectors, encapsulated in the formula ||u × v|| = ||u|| ||v|| sin(?). When the vectors are orthogonal, the angle ? is 90 degrees, which leads to sin(?) = 1. This trigonometric fact directly leads to the simplification of the cross product's magnitude to the product of the vectors’ magnitudes.

Transcript

00:01 Hello, hope you're doing well.

00:03 So this problem asks us to prove that this statement right here is true if u and v are orthogonal, which means they form a 90 degree angle between them.

00:14 So this is u and v, there's a 90 degree angle.

00:17 So first, before we get started, we need to keep in mind that the cosine of the angle between two vectors, say u and v is equal to the dot product of u and v.

00:31 Over the length of the u vector times the length of the v vector.

00:38 So that's one thing to keep in mind.

00:40 We also need to keep in mind our trigonometric identities.

00:43 So remember that sine squared theta plus cosine squared of theta is equal to 1.

00:51 Then lastly, we need to remember that these two lines on either side of our vector mean that we're finding the length of our vector.

00:58 So to find the length of the vector u, we would take the square root, of the sum of the squares of each of the vector components.

01:05 So let's say u is equal to the vector u1, u2, u3, then the length of u would be squared of u1 squared plus u2 squared plus u3 squared.

01:19 All right, so keeping that in mind, let's go ahead and prove this.

01:23 So what we're going to do is we're going to start out with the statement the length of u times the length of v times sine theta.

01:31 So you just want to see what is that equal to.

01:37 So remembering our trigonometric identity, sine squared plus cosine square is equal to 1.

01:42 So this means that if we subtract a cosine squared theta from both sides and take the square root, we end up with sine theta is equal to the square root of 1 minus cosine squared theta.

01:58 Okay, so substituting in this statement on the left here is equal to length of u.

02:04 Times the length of v times the square root of 1 minus cosine square theta.

02:14 Okay, so now that we have that, we can substitute what we got, this cosine theta is equal to this right here.

02:24 So we can substitute that in for cosine theta here.

02:28 So continuing this, this is equal to length of u times the length of the vector v, times the square root of 1 minus that it's, we're squaring the cosine, so we have u.

02:44 .v squared over the length of u squared times the length of v squared.

02:54 Okay, so that's what this is equal to.

02:56 So continuing this, we can take this length of u times length of v, and if we square it, so this length of u times length of v is equal to the square root of length of u squared times length of v squared.

03:15 So when we do that, if we replace this here with this expression, we've got two square roots multiplied by each other, so we can just combine them.

03:25 We can put this inside this square root.

03:28 So we're left with essentially square root of length of u squared, length of v squared times 1 minus u .d .v squared over length of u squared times the length of v squared.

03:46 So we can do that because this right here is equivalent to this.

03:49 We're just combining the square roots.

03:51 So now we can distribute this, use the distributed property, multiply this by the 1, multiply this by this expression.

04:00 So we end up with the square root of u squared v squared, multiplied by 1, that gives us still u, length of u squared, length of v squared.

04:11 Then when you multiply this by the second term, it cancels out with the bottom.

04:16 So you end up with minus u.

04:18 .

04:21 Okay, so then continuing to simplify this, we know we got from before that to the length of u is equal to square root of u1 squared plus u2 squared plus u3 squared.

04:34 That's this right there.

04:36 You can just substitute that in for u is equal to the square root of.

04:47 So if u is equal to this, square to u1 squared plus u2 squared plus u3 squared, that means u squared is just going to be equal to that quantity squared.

04:58 So you're just getting rid of the square roots.

05:00 You end up with u1 squared plus u2 squared plus u3 squared.

05:06 So writing that in we have u1 squared plus u2 squared plus u3 squared.

05:13 We'll splying this, we do the same thing with our v vector.

05:16 It's the same rule applies.

05:17 So v1 squared plus v2 squared plus v3 squared.

05:24 Then got minus, minus, and this is all under the square root, the dot product of u.

05:29 .

05:31 So u.

05:36 Let's see, yeah, so u.

05:37 .v squared, u.

05:40 .v is equal to you multiply the x components together and add them to the product of the y components, add them to the product of the z component.

05:48 So you have the dot product is equal to u1 v1 plus u2 v2, i'm sorry, u2 v2 plus u3 v3.

06:00 That's all squared.

06:03 Okay, so you can simplify this.

06:07 You end up with, so essentially, so yeah, you got u1 squared v1 squared, multiplying everything out plus u1 squared v2 squared plus u1 square v3 squared, plus u2 squared, plus u2 squared v3 squared, plus u2 squared, plus u3 squared, plus u3 squared, plus u3 squared, plus u3 squared, plus u3 squared, plus u3 squared, plus u3 squared, and then you subtract this quantity.

06:48 So taking that squared, it's minus u1 squared, v1 squared, and it's minus u1, u2, u2, minus u1, u3, u3, v1, u3.

07:06 You have minus u2 squared, u2 squared, you two, you one, u3, u3 .s, u3, u2 squared, you have minus u2 squared, u2 squared, you two, v1, v2, minus u2, u2, u3, v2, v3.

07:26 We've got minus u1, u3, u3 v1, v3, minus u2, u3, v2, v3, minus u3 squared, squared.

07:44 So it's going to be kind of a pain to simplify all this, but let's see what we can do.

07:49 So u1 squared v1 squared, these cancel out.

07:51 I'll just do those in blue just to make it clear which ones we've canceled out.

07:57 So you end up with u1v1 squared minus u1 v1 squared.

07:59 Those cancel out.

08:01 Let's see what else.

08:05 We have u2 v2 squared minus u2 v2 squared, then u3 v3 squared minus u3 v3 squared.

08:16 Okay, so let's combine like terms is equal to the square root.

08:24 We have u1 squared v2 squared, plus u1 squared v3 squared, plus u2 squared, plus u2 squared, and let's see where we can combine electricity.

08:41 Yeah, so plus u2 squared v1 squared, plus u2 squared, plus u3 squared, plus u3 squared, plus u3 squared, plus u3 squared, plus u3 squared, plus u3 squared, plus u3 squared, v2 squared.

08:55 Okay, so those are all our terms with the squares in them.

08:59 Then we're going to combine like terms with these that have four numbers in them.

09:06 So four different quantities in them.

09:08 So we've got two of these.

09:11 So we have minus 2, u1, u2, v1, v2, minus 2, u1, u3, minus 2, u1, u3, minus 2, u1, u3, minus 2, u2, u2, u2, and 3 .3, 2, u2, u2, u3.

09:28 V2, v3.

09:31 Okay, so let's think about this...