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# Find the Cartesian equation of a plane passing through a point and parallel to a given plane

Let be a plane which passes through the point and is parallel to the plane given by the equation . Find the Cartesian equation of . Further, find the distance between the two planes.

Let denote the plane given by the equation . Since and are parallel, we know they share a common normal vector. Therefore, the Cartesian equation of is of the form

Since is on , we have

Hence, the Cartesian equation of is

The distance between and is then

# Find geometric properties of a plane with given Cartesian equation

Consider a plane with the Cartesian equation

Find each of the following:

1. A normal vector to the plane with length 1.
2. The intercepts of the plane.
3. The distance from the plane to the origin.
4. The point on the plane, , nearest to the origin.

1. From the Cartesian equation we know that a normal vector to the plane is . Making this of unit length, we have a unit normal given by

2. Since the Cartesian equation gives us we have the -intercept is , the -intercept is and the -intercept is .
3. The distance from the plane to the origin is given by

4. The point on the plane which is nearest to the origin is in the direction of the normal vector, and at a distance of , therefore,

# Give a Cartesian for planes through given points spanned by given vectors

Consider the vectors

1. Find a nonzero vector perpendicular to both and .
2. Find a Cartesian equation for the plane through which is spanned by and .
3. Find a Cartesian equation for the plane through which is spanned by and .

1. Since and are independent, we can take

2. From part (a) we have is perpendicular to both and , so a Cartesian equation for the plane is given by

Further, since the point is on the plane, we must have . Hence, the Cartesian equation for the plane is

3. Again, we have a Cartesian equation for the plane given by

Since is on the plane we must have

Hence, the Cartesian equation for the plane is given by

# Give a vector based proof of Heron’s formula for computing the area of a triangle

Let denote the area of a triangle with sides of lengths . Heron’s formula states that

We prove this formula using vectors via the following steps. Assume the triangle has vertices , and with

1. Using the identities

prove the formula

2. Simplify the formula in part (a) to obtain the formula

and use this to deduce Heron’s formula.

1. Proof. We know the area of the triangle (in terms of the vectors and ) is

Therefore,

2. Proof. We simplify the formula in part (a),

# Prove a formula for the perpendicular distance between a point and a line

1. Assume that . Prove that the perpendicular distance from to the line passing through the points and is given by the formula

2. Compute the distance in the case

1. Proof. We know the area of the parallelogram determined by and is given by

But this is exactly twice the area of the triangle with base and height the perpendicular line from to the line through and . Hence, is the perpendicular distance from to the line through and , and

2. When we have

# Prove a vector formula for the volume of a tetrahedron

1. Consider a tetrahedron with vertices . Prove that the volume of the tetrahedron is given by the formula

2. Compute the volume in the case that

1. Proof. We know the volume of a tetrahedron is given by (where denotes the altitude of the tetrahedron). We know (page 490 of Apostol) that the volume of the parallelepiped with base formed by vector and height formed by vector is given by . In this case we have that the base of the tetrahedron is formed by the vectors and , and the height is formed by the vector . Further, we know that the area of the base described by the vectors and is one half that of the parallelepiped whose base is given by vectors and (since the base of the parallelepiped described by vectors and is a rectangle, and the base of the tetrahedron is the triangle formed by cutting this rectangle along the diagonal). Therefore we have

2. Using the formula in part (a) with the given values of we have

# Compute the cross product of given vectors in terms of the unit coordinate vectors

Let such that

In terms of the unit coordinate vectors compute the cross product

Using part (a) of the previous exercise and equation (13.10) on page 490 of Apostol () we compute

Then we use the other given relations

Which all implies

# Prove some vector identities using the “cab minus bac” formula

In the previous exercise (Section 13.14, Exercise #9) we proved the “cab minus bac” formula:

Using this formula prove the following identities:

1. .
2. .
3. if and only if .
4. .

1. Proof. Using the cab minus bac formula with in place of , in place of , and in place of we have

2. Proof. Applying the cab minus back formula to each of the three terms in the sum we have

So, putting these together we have

3. Proof. From cab minus bac we have

Furthermore, since , we can apply bac minus cab to get

Therefore,

4. Proof. From a previous exercise (Section 13.14, Exercise #7(d)) we know the identity . In this case we have in place of , in place of and in place of . This gives us

# Prove the “cab minus bac” formula

The “cab minus bac” formula is the vector identity

Let and . Prove that

This is the “cab minus bac” formula in the case . Prove similar formulas for the special cases and . Put these three results together to prove the formula in general.

Proof. For the case we have

Similarly, for and we have

So, if is any vector in then we have

# Find vectors which satisfy given relations

1. Find all of the vectors which satisfy

2. Find the shortest length vector which satisfies the relation in part (a).

1. We can compute,

So, can take any value, and then we must have for the relation to be satisfied. Therefore, any vector

satisfies the relations.

2. So, we know the vectors satisfying the relation are of the form . This means we want to minimize

This is minimal when (since for any other value of ). Then we want to find the value of which minimizes . Taking the derivative and setting it equal to 0 we have

Hence, the vector of minimal length which satisfies the given relation is