Let us continue with more examples.

Consider a line in -space

Choose a point on the line, call it Our goal is to explain any other point on this line. Obviously, the yellow vector is Now if ia any other point on the line, then the vector must be parallel to the line. We show this vector in red. Observe that for the vector which has its terminal point on the line at the point

(1)

This gives us a hint that if we have any vector parallel to the line, with multiplying by the right parameter, we can get the vector In fact

(2)

This way, any vector with the initial point at the origin and the terminal point on the line can be explained. Observe that this information explains every point on the line. This form of equation of a line is called the vector form. One can obtain

(3)

which is called the parametric form. Solving for ,

(4)

which is called the symmetric form of equation of a line.

Consider a plane in -space.

Choose a point on this plane. Now we would like to explain any other point on this plane. Clearly, for any other point on the plane, the vector is on the plane. In the event that we have a vector perpendicular to the plane, we can write

(5)

This choice of a perpendicular vector to the plane is called a normal to the plane and in the picture is shown in yellow. Obviously, a different choice of a perpendicular will not result on a different plane since the point is fixed on the plane to begin with. One can write the above plane equation in the form

(6)

or

(7)

The right side of the above equation is a number.

Suppose and are two vectors in -space. We define cross product of and by

(8)

We find cross product of a vector with itself

(9)

This is always true and note that the result of the cross product of two vectors is another vector. Many text books use the notion of determinant to compute the cross product of two vectors; here it is:

(10)

which essentially provides what we used as a definition. You can use this while computation but in this set of notes, we avoid it when possible.

A set of properties of cross product are as followed. Suppose is a real number and are vectors in -space:

1.

2.

3.

We want to show

(11)

Observe that on the left side

(12)

Now on the right side of the given expression (a seems to be an identity but we don’t know it yet!)

(13)

Therefore, the given statement is true. We are now done. Now consider the angle between and Another interesting relationship, is due to the fact that

(14)

and so

(15)

which provides

(16)

Geometrically, we see that this is area of the shaded part that is a parallelogram with sides being vectors and Recall that area of a parallelogram is obtained by multiplying base with hight.

Suppose and are vectors. We would like to find the angle between and We know

(17)

Computing the numerator we get

(18)

So the two vectors and are perpendicular to each other.

We want to write equation of a line that is parallel to the the vector and passes the point . We start by the vector form

(19)

The parametric form of the equation is

(20)

The symmetric form of the equation is

(21)

Suppose we are given two lines

(22)

We note that the line is parallel to the vector and is parallel to the vector . Since

(23)

the two vectors are parallel to each other. So the two lines are parallel to one another.

Suppose we are given the two lines

(24)

Of course the two lines are not parallel since there is no such that

(25)

As we have seen before, the two vectors and determine the direction of each line. Note that for the two vectors to be parallel to each other, we must have

(26)

The first equation says, while the last equation claims If you believe, then these two vectors are not parallel to each other. Now, do we know if the lines intersect each other? Let us find out. If the lines intersect, then there is a point that satisfies both equations. Assume the two lines intersect; define the point of intersection. Then from the first line

(27)

and from the second line

(28)

Since is on both lines

(29)

We have three equations and two unknowns. From the second equation, we get

(30)

Put this into the first equation

(31)

This means . If these values for and satisfy the last equation, we have found parameters which give us the point ; otherwise, there are no such parameters and in result no such point The last equation dictates

(32)

This is not true, so our original assumption is not true. The two lines are not parallel and don’t intersect.

Suppose we are given a line and a point We would like to find the distance between these two. Let us list the given information first followed by a strategy.

1. The vector can be used as a direction vector for the line.

2. The point is on the line.

3. Using the given points and we can construct a vector, say .

4. Projection vector of onto the vector on the line is computed by

(33)

5. We have many options. One way is that if we go along the line from the point by the projection vector that we just constructed, we find the closest point on the line to the point outside of the line This way, we can find the distance between the two points and we are done.

(34)

Now the distance between and is

(35)

And we are done.

Suppose we are given a plane and the point . We would like to find the distance between the point and the given plane. Let us again list the information and a strategy:

1. The point is on the plane.

2. The vector is normal to the plane.

3. We can use a point on the plane and a point outside to make a vector, .

4. If we project this vector onto the normal vector to the plane, we can find its length and done.

(36)

and magnitude of this vector is

(37)

In the event that we have to find the distance between two planes, the process is as easy (just choose a point on a plane and find the distance from this point to the other plane.)

Suppose we are given two planes and We know if the two planes are not parallel to each other, then they intersect in a line. We are interested to find a parametric equation for such line. Again, there are many ways to do this. For this exercise, we compare the normals to the planes and to see that the two planes are not parallel to each other. To construct an equation of any line in space, we need a point on the line and a direction vector. Finding a point is easy; put to get So the point is on both planes. To find a direction vector for the line, observe that cross product of the vectors normal to each plane provides a new vector normal to both vectors. With the same token, this new vector is in the direction of the intersection line. So

(38)

The line

(39)

is the line of intersection.

Please see a few additional problems for this section.