The cross product 𝐮 ×𝐯 of the vectors 𝐮=u1 𝐢+u2 𝐣+u3 𝐤 and 𝐯=v1 𝐢+v2 𝐣+v3 𝐤 is 𝐮 ×𝐯 =| 𝐢

The cross product 𝐮 ×𝐯 of the vectors 𝐮=u1 𝐢+u2 𝐣+u3 𝐤 and...

The cross product $\mathbf{u} \times \mathbf{v}$ of the vectors $\mathbf{u}=u_{1} \mathbf{i}+u_{2} \mathbf{j}+u_{3} \mathbf{k}$ and $\mathbf{v}=v_{1} \mathbf{i}+v_{2} \mathbf{j}+v_{3} \mathbf{k}$ is $$ \begin{aligned} \mathbf{u} \times \mathbf{v} &=\left|\begin{array}{ccc}{\mathbf{i}} & {\mathbf{j}} & {\mathbf{k}} \\ {u_{1}} & {u_{2}} & {u_{3}} \\ {v_{1}} & {v_{2}} & {v_{3}}\end{array}\right| \\ &=\left(u_{2} v_{3}-u_{3} v_{2}\right) \mathbf{i}+\left(u_{3} v_{1}-u_{1} v_{3}\right) \mathbf{j}+\left(u_{1} v_{2}-u_{2} v_{1}\right) \mathbf{k} \end{aligned} $$ Use this definition in Exercises $65-68$ Assuming the theorem about angles between vectors (Section 6.2 ) holds for three-dimensional vectors, prove that $\mathbf{u} \times \mathbf{v}$ is perpendicular to both u and v if they are nonzero.

The cross product $\mathbf{u} \times \mathbf{v}$ of the vectors $\mathbf{u}=u_{1} \mathbf{i}+u_{2} \mathbf{j}+u_{3} \mathbf{k}$ and $\mathbf{v}=v_{1} \mathbf{i}+v_{2} \mathbf{j}+v_{3} \mathbf{k}$ is
$$
\begin{aligned} \mathbf{u} \times \mathbf{v} &=\left|\begin{array}{ccc}{\mathbf{i}} & {\mathbf{j}} & {\mathbf{k}} \\ {u_{1}} & {u_{2}} & {u_{3}} \\ {v_{1}} & {v_{2}} & {v_{3}}\end{array}\right| \\ &=\left(u_{2} v_{3}-u_{3} v_{2}\right) \mathbf{i}+\left(u_{3} v_{1}-u_{1} v_{3}\right) \mathbf{j}+\left(u_{1} v_{2}-u_{2} v_{1}\right) \mathbf{k} \end{aligned}
$$
Use this definition in Exercises $65-68$
Assuming the theorem about angles between vectors (Section 6.2 ) holds for three-dimensional vectors, prove that $\mathbf{u} \times \mathbf{v}$ is perpendicular to both u and v if they are nonzero.

Key Concepts

Cross Product

The cross product is an operation on two three-dimensional vectors that produces another vector perpendicular to the plane containing the original vectors. It is defined using the determinant of a 3×3 matrix, where the first row consists of the unit vectors and the following rows consist of the components of the given vectors. This operation not only determines the direction of the resultant vector but also its magnitude, which is equal to the area of the parallelogram generated by the two original vectors.

Determinant in Vector Operations

The determinant of a 3×3 matrix is used to compute the cross product, providing a systematic way to calculate the components of the resultant vector. By expanding the determinant along the first row, we obtain expressions for the i, j, and k components that combine the elements of the two vectors. This method encapsulates both the algebraic and geometric interpretations required to establish the perpendicularity of the cross product to the original vectors.

Dot Product and Orthogonality

The dot product of two vectors is a scalar value that, among other properties, is used to determine the angle between the vectors. Two vectors are orthogonal (i.e., perpendicular) if and only if their dot product is zero. In the context of the cross product, showing that the dot product of the cross product with either of the original vectors equals zero confirms that the cross product is perpendicular to both vectors.

Angle Between Vectors

The theorem regarding the angle between two vectors states that the cosine of the angle is equal to the dot product of the vectors divided by the product of their magnitudes. This theorem is critical when proving that the cross product yields a vector perpendicular to the original vectors, as it relies on the fact that orthogonal vectors have a cosine of zero.

Transcript

00:05 All right, we're supposed to prove that u cross the v is perpendicular to both u and v, assuming u and b are non -zero.

00:11 It also in the book says recall the theorem from 6 .2.

00:15 When it says that, what we want to use is specifically the theorem that says that two vectors are orthogonal, if and only if their dot product is zero.

00:21 That's the theorem we care about using.

00:23 All right, so i'm going to let u equals some non -zero vector, so a, b, c, and v equals some other non -zero vector, d, e, f.

00:36 When i say non -zero, it'd be okay if, like, say, c was equal to zero or b was equal zero.

00:40 We just can't have all being equal to zero, because if every component is equal to zero, then you just get the zero vector crossed with something.

00:48 It's still going to be the zero vector.

00:50 But we are, you could, like, have one of these be zero.

00:53 Anyway, we have our two non -zero vectors.

00:57 And so you crossed with fee is equal to.

01:22 And i'm just taking a little bit shorthood here because it tells us that we can use that theorem which tells us what the vector is i guess i should technically a theorem doesn't skip straight to the vector so i'll write it out a little bit more but not be lazy okay so this is equal to bf minus e c i hat minus a f minus d c j hat plus a e minus d c b k hat, which is just the vector bf, e, c, comma, d, c, minus a, f, because we got to distribute that negative through a, e, minus d, b.

02:18 If you write it as a vector rather than it's ijk form.

02:24 Okay.

02:26 Now, so here's my product, and we want to prove that's perpendicular both u and v.

02:32 So we're going to do two cases.

02:33 So we have case one.

02:38 It actually rather than saying case one.

02:40 I'll say, all right, i'll say, perpendicular to you.

02:50 So for this, i'm just going to take the vector.

02:55 For notation, simplicity, to abbreviation, i'm just going to call this vector.

03:01 Well, i'm actually why am i doing that? i'm going to call it r.

03:09 Well, actually, i don't even need to do that.

03:10 I'm making notation when i don't need to.