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# 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

# Prove an identity of cross products and the unit coordinate vectors

Prove that we have the identity:

Proof. Let and compute,

# Prove a formula for the cross product of vectors in terms of the unit coordinate vectors

Prove that we have the formula

Proof. Let and . Then we compute,

# Compute the volume of a parallelpiped determined by given vectors

Compute the volume of the parallelpiped determined by the vectors .

The volume is given by the scalar triple product

# Find all values of t so that (1,t,1), (t,1,0), (0,1,t) are dependent

Find all such that the vectors

are linearly dependent.

The vectors are linearly dependent if and only if the scalar triple product

So, we compute

So, the vectors are dependent for all such that . Thus the vectors are dependent for

# Compute the scalar triple product of given vectors

For each of the following sets of vectors compute the scalar triple product.

1. .
2. .
3. .

1. We compute

2. We compute

3. We compute

# Prove statements about a vector satisfying a given vector equation

Consider unit length, orthogonal vectors , and a vector such that

Prove the following.

1. and are orthogonal and the length of is .
2. The vectors form a basis for .
3. .
4. .

1. Proof. We compute,

since and since and are orthogonal by assumption. Thus, and are orthogonal. Next,

since by hypothesis and . Hence, from the vector equation we have

2. Proof. Since and are orthogonal (part (a)), we know the vectors are independent. Thus, they form a basis for since any three independent vectors in are a basis
3. Proof. We compute, the vector is given by

Then the three coordinates of this cross product are given by

Expanding these out we obtain the coordinates

Since we know and since we know . So, simplifying the expressions above, for each of the coordinates we have

Hence, we indeed have

4. Proof. We compute

# Prove some facts about vectors in R3

1. Prove that are all on the same line if and only if

2. For two vectors with , prove that the line through and is the set of all vectors such that .

1. Proof. Assume all lie on a line , and let . Then

So, we have

Thus, is the vector

Conversely, assume . From Theorem 13.12(g) we know this is the case if and only if the vectors and are linearly dependent. Hence, there are nonzero such that

Since is nonzero we may divide by , and we obtain

But this means is on the line passing through and

2. Proof. If , then we know there is a unique line passing through and , say

Thus, any point is given by

Therefore, we have