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The cross product of vectors and is vector ,

that is perpendicular to the plane determined by vectors and , and where

To get the cross product, we have to calculate this determinant.

It is funny that determinants will only appear at a later stage, but in spite of that, we will try to calculate this now.

We will expand the determinant along the first row, but don't worry, everything will be simple. Well, here it is:

We will figure out these soon. It is intentional that we have a minus here, instead of a plus.

All of these would be much easier to grasp if we knew the expansion rules included in the chapter about determinants. If someone decides to check out what those rules are, we can safely assume that nobody can prevent them from doing so.

Now let's get down to business.

Let's see a specific example.

Here is a specific example:

Now let's get down to business.

Here is the cross product:

Let's see two exercises for a typical geometry application of the cross product.

One exercise is to find the equation of a line in the plane where two given points are on the line. We don't need the cross product for this.

The other one involves finding the equation of a plane in space where three given points are on the plane. We do need the cross product for this.

Find the equation of the line in the plane, where point is on the line, and the line is perpendicular to line

Find the equation of the plane, where is on the plane that is perpendicular to the line described by:

The normal vector of line is

We can make use of this vector if

we rotate it by , because then

it will be the normal vector of the line we are trying to define.

To rotate a vector in the plane by ,

we swap its coordinates,

and multiply one of them by .

We have the normal vector, so

the equation of the line is:

Let's see what we can do over here.

The normal vector of the plane happens to be the direction vector of the line.

Here comes the equation of the plane:

And finally, another typical exercise.

Find the equation of a line in the plane, where points and are on the line.

Find the equation of the plane in space, where points , , and are on the plane.

If is a point on the plane and

is the normal vector of the plane, the equation of the line will be:

If is a point on the plane and

is the normal vector of the plane, the equation of the plane will be:

We have plenty of points, but we don't have a single normal vector,

so we have to make one.

Let's rotate this by , and that gives us the normal vector.

To rotate a vector in the plane by ,

we swap its coordinates,

and multiply one of them by .

The equation of the line:

Here, at the plane, the cross product will yield the normal vector.

The equation of the plane:

And then it is done.

# The cross product, and its use

07
Let's see this
Calculus 3 episode