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# Prove that the intersection of a line and plane which are not parallel contains exactly one point

Prove that the intersection of a line and a plane such that the line is not parallel to the plane contains one and only one point.

Proof. Denote the line by and the plane by . Let be the set of points

Since is not parallel to w know that its direction vector is not in the span of and . Further, by definition of a plane, we know the vectors and are linearly independent. Hence, are linearly independent. Then, any point in the intersection must be a solution to the system of equations

By the linear independence of we know this system has exactly one solution . Hence, contains exactly one point

# Find the parametric equation for a line through a point and parallel to two planes

We say that a line is parallel to a plane if the direction vector of the line is parallel to the plane. Let be the line containing the point and parallel to the planes

Find a vector parametric equation for .

The normal vectors of the planes are and . So, the direction vector of will be perpendicular to both of these,

From the first equation we have . Plugging this into the second equation we obtain , which then gives us . Since is arbitrary, we take to obtain a direction vector . Therefore, the vector parametric equation for the line is

# Prove that there is exactly one plane through a line and a point not on the line

Let be a line and a point not on . Prove that there is exactly one plane through containing every point of .

Proof. Existence follows from a previous exercise (Section 13.8, Exercise #12) by letting be any two distinct points on , then there is a plane containing and every point on the line through and .
Now, to prove there is only one such plane we apply Theorem 13.10 to the points to conclude the plane that contains is unique. But, if we choose any other two points on , the plane must still contain and (since it contains every point on ); hence, it is the same plane . Thus, there is exactly one plane containing and

# Find a Cartesian equation for a plane through a point containing a given line

Let be the line through the point and parallel to the vector and let be a point not on . Find a Cartesian equation for the plane which passes through and entirely contains .

The line is the set of points

Then, the plane is the set of points

Then, to get the Cartesian equation, we have

The first two equations give and so . This implies , and is arbitrary. So, the plane is described by the equation .

# Prove that the line through two distinct points on a plane is entirely contained in the plane

Let be a plane containing two distinct points and . Prove that every point on the line through and is in .

Proof. Fist, the line through and is given by

Then, let be any point on not on . The plane is given by

So, for any point let and . Then we have

# Determine whether a line is parallel to given planes

We say that a line in the direction of a vector is parallel to a plane if is parallel to . Consider the line through the point and parallel to the vector . Determine whether is parallel to the following planes.

1. The plane through the point and spanned by and .
2. The plane through the points .
3. The plane determined by the Cartesian equation .

1. This asks if is in the span of , i.e., does there exist such that

From the second equation we have . Then, from the first, which implies

But then,

Thus, there is no solution, so the line is not parallel to the plane.

2. The plane through the points is the set of points

For to be in the span of we must have such that

From the first equation we have . Then from the second we have which implies

But then,

Hence, is not parallel to .

3. The plane with Cartesian equation is the set of points

The points are all in . So,

Thus, we ask if is in the span of . This requires that there exist such that

But, this fails since the second and third equations implies and . But then

Hence, this line is not parallel to the plane.

# Prove that there is exactly one point in the intersection of a given line and given plane

Consider the line through the point and parallel to the vector . Then, let be the plane through the point and spanned by the vectors and . Prove that there is exactly one point in , and determine the point.

Proof. First, we have that and are the sets of points

Then, the points in are those which satisfy

This gives us the Cartesian equation

The points on are those such that

Thus, if then we have

Hence,

is the unique point in

# Prove that the intersection of two non-parallel lines is either empty or contains exactly one point

Let and be two lines in which are not parallel. Prove that the intersection is either empty or contains exactly one point.

Proof. Since and are not parallel we know . Now, suppose there exist two distinct points . This means there exist real numbers such that

Since are distinct points we also know and . Then we have

But then

for some . Thus, the lines are parallel, contradicting our assumption that the lines are not parallel. Hence, the intersection contains at most one point

# Prove that two parallel lines are either equal or have empty intersection

Let and be two parallel lines in . Prove that or .

Proof. Assume . Suppose that the intersection is not empty, so that there exists a point and . Then, we know there exist such that

But then

But, by Theorem 13.2, this means , contradicting our assumption. Hence, either or the intersection is empty

# Prove some properties about a line and a point not on the line

Let be a point not on the line in .

1. Consider the function

Prove is a quadratic polynomial in and that there is a unique value of , say , at which this polynomial takes on its minimum.

2. Prove that is orthogonal to .

1. Proof. First, we compute

But, are all just real scalars (by definition of the dot product); therefore,

for scalars given by

Then,

which implies has a unique minimum at

2. Proof. We compute the dot product,

Then we have

Therefore,