Notice what this means for the direction of
$\text{v}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{u}.$ If we apply the right-hand rule to
$\text{v}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{u},$ we start with our fingers pointed in the direction of
$\text{v},$ then curl our fingers toward the vector
$\text{u}.$ In this case, the thumb points in the opposite direction of
$\text{u}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{v}.$ (Try it!)
Anticommutativity of the cross product
Let
$\text{u}=\u27e80,2,1\u27e9$ and
$\text{v}=\u27e83,\mathrm{-1},0\u27e9.$ Calculate
$\text{u}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{v}$ and
$\text{v}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{u}$ and graph them.
We see that, in this case,
$\text{u}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{v}=\text{\u2212}\left(\text{v}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{u}\right)$ (
[link] ). We prove this in general later in this section.
Suppose vectors
$\text{u}$ and
$\text{v}$ lie in the
xy -plane (the
z -component of each vector is zero). Now suppose the
x - and
y -components of
$\text{u}$ and the
y -component of
$\text{v}$ are all positive, whereas the
x -component of
$\text{v}$ is negative. Assuming the coordinate axes are oriented in the usual positions, in which direction does
$\text{u}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{v}$ point?
The cross products of the standard unit vectors
$\text{i},\text{j},$ and
$\text{k}$ can be useful for simplifying some calculations, so let’s consider these cross products. A straightforward application of the definition shows that
(The cross product of two vectors is a vector, so each of these products results in the zero vector, not the scalar
$0.)$ It’s up to you to verify the calculations on your own.
Furthermore, because the cross product of two vectors is orthogonal to each of these vectors, we know that the cross product of
$\text{i}$ and
$\text{j}$ is parallel to
$\text{k}.$ Similarly, the vector product of
$\text{i}$ and
$\text{k}$ is parallel to
$\text{j},$ and the vector product of
$\text{j}$ and
$\text{k}$ is parallel to
$\text{i}.$ We can use the right-hand rule to determine the direction of each product. Then we have
We know that
$\text{j}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{k}=\text{i}.$ Therefore,
$\text{i}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\left(\text{j}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{k}\right)=\text{i}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{i}=0.$
As we have seen, the dot product is often called the
scalar product because it results in a scalar. The cross product results in a vector, so it is sometimes called the
vector product . These operations are both versions of vector multiplication, but they have very different properties and applications. Let’s explore some properties of the cross product. We prove only a few of them. Proofs of the other properties are left as exercises.
Properties of the cross product
Let
$\text{u},\text{v},$ and
$\text{w}$ be vectors in space, and let
$c$ be a scalar.
For property
$\text{i}.,$ we want to show
$\text{u}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{v}=\text{\u2212}\left(\text{v}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{u}\right).$ We have
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