Vectors: Proving the value of a•(b x c)

[Sandro Romualdez]

Homework Statement

Given that vector a = (1, 2, -5), b = (-12, 41, 75) and c = a + 2b, explain why (without doing any calculations whatsoever) the value of a•(b x c) = 0

Homework Equations

No specific equations, as the question asks for the value without making any calculations. This problem mostly takes a knowledge of the properties of vectors, vector addition, the dot product, and the cross product.

The Attempt at a Solution

Using calculations, I could probably manipulate the equation and calculate the values/magnitudes of the vectors and stuff, but since the problem asks for a solution without calculations, it would be pretty futile to try..

I know that for the dot product between two vectors to equal zero, the vectors must be perpendicular. The cross product is the vector that is perpendicular to both vectors b and c, but in this case (3D/Three space/R3), aren't there an infinite number of vectors that can be perpendicular to b and c? In that case how would I be able to explain why a•(b x c) is equal to zero?

 

[SammyS]

Homework Statement

Given that vector a = (1, 2, -5), b = (-12, 41, 75) and c = a + 2b, explain why (without doing any calculations whatsoever) the value of a•(b x c) = 0

Homework Equations

No specific equations, as the question asks for the value without making any calculations. This problem mostly takes a knowledge of the properties of vectors, vector addition, the dot product, and the cross product.

The Attempt at a Solution

Using calculations, I could probably manipulate the equation and calculate the values/magnitudes of the vectors and stuff, but since the problem asks for a solution without calculations, it would be pretty futile to try..

I know that for the dot product between two vectors to equal zero, the vectors must be perpendicular. The cross product is the vector that is perpendicular to both vectors b and c, but in this case (3D/Three space/R3), aren't there an infinite number of vectors that can be perpendicular to b and c? In that case how would I be able to explain why a•(b x c) is equal to zero?

Hello @Sandro Romualdez . Welcome to PF.Does the distributive law hold for either (or both) forms of the product of two vectors?

 

[Mark44]

Using calculations, I could probably manipulate the equation and calculate the values/magnitudes of the vectors and stuff, but since the problem asks for a solution without calculations, it would be pretty futile to try..

Right, especially because the problem says to not do any calculations at all.

I know that for the dot product between two vectors to equal zero, the vectors must be perpendicular. The cross product is the vector that is perpendicular to both vectors b and c, but in this case (3D/Three space/R3), aren't there an infinite number of vectors that can be perpendicular to b and c? In that case how would I be able to explain why a•(b x c) is equal to zero?

@SammyS's hint about the distributive properties for both the dot product and cross product is a good one. A key item in this problem is that c = a + 2b.

The problem directions don't prohibit you from doing algebraic manipulations with the variables -- basically, you can't just plug in the given vectors to calculate the specific dot or cross products.

 

[Sandro Romualdez]

Hello @Sandro Romualdez . Welcome to PF.Does the distributive law hold for either (or both) forms of the product of two vectors?

Thank you, I've been a member of PF for awhile, but for some reason my old account got deleted or something.

What we've discussed in class is that the distributive property of the dot product is that a•(b+c) = a•b + c•a. I'm assuming that can also be applied with the cross product instead of addition? Meaning that a•(b x c) is a•b x c•a = 0. I know that for the cross product to equal zero, the vectors are parallel. Does that show that the vectors are coplanar maybe?

 

[Sandro Romualdez]

@SammyS's hint about the distributive properties for both the dot product and cross product is a good one. A key item in this problem is that c = a + 2b.

With this knowledge, would I be able to state that since c can be expressed as a linear combination of vectors a and b, all three vectors lie on the same plane? Following that, since the cross product is perpendicular to all three vectors, the dot product of a and (b x c) would be zero, following the definition of perpendicular vectors.

Is this a logical/sufficient solution?

 

[Mark44]

With this knowledge, would I be able to state that since c can be expressed as a linear combination of vectors a and b, all three vectors lie on the same plane? Following that, since the cross product is perpendicular to all three vectors, the dot product of a and (b x c) would be zero, following the definition of perpendicular vectors.

Is this a logical/sufficient solution?

This is logical, but it might or might not be sufficient. Personally, I would explain a is perpendicular to b x c, using the given information about c.

 

[Stavros Kiri]

Note: the numbers for a, b don't even matter

 

[SammyS]

What we've discussed in class is that the distributive property of the dot product is that a•(b+c) = a•b + c•a. I'm assuming that can also be applied with the cross product instead of addition? Meaning that a•(b x c) is a•b x c•a = 0. I know that for the cross product to equal zero, the vectors are parallel. Does that show that the vectors are coplanar maybe?

The highlighted assumption in the quote is false.

My intention was that you express vector, $\ \vec c\,,\$ as $\ \vec a +2\vec b \,,\$ then take the cross product $\ \vec b\times ( \,\vec a +2\vec b \,) \,.\$

Use the distributive law on that.

 

[tnich]

Homework Statement

Given that vector a = (1, 2, -5), b = (-12, 41, 75) and c = a + 2b, explain why (without doing any calculations whatsoever) the value of a•(b x c) = 0

Homework Equations

No specific equations, as the question asks for the value without making any calculations. This problem mostly takes a knowledge of the properties of vectors, vector addition, the dot product, and the cross product.

The Attempt at a Solution

Using calculations, I could probably manipulate the equation and calculate the values/magnitudes of the vectors and stuff, but since the problem asks for a solution without calculations, it would be pretty futile to try..

I know that for the dot product between two vectors to equal zero, the vectors must be perpendicular. The cross product is the vector that is perpendicular to both vectors b and c, but in this case (3D/Three space/R3), aren't there an infinite number of vectors that can be perpendicular to b and c? In that case how would I be able to explain why a•(b x c) is equal to zero?

There is a neat trick for this problem. Are you familiar with the method of calculating a cross product like this?
For the cross product of $\vec b$ and $\vec c$, form a matrix:
$M=\begin{pmatrix} 1 & 1 & 1\\ b_x & b_y & b_z\\ c_x & c_y & c_z \end{pmatrix}$

If you find the determinant of this matrix using expansion by minors, you get:
$det(M) = 1(b_y c_z - b_z c_y) + 1(b_z c_x - b_x c_y) + 1(b_x c_y - b_y c_x)$

You will notice that the terms in the sum are exactly the elements of $\vec b \times \vec c$.

Now try the same thing with matrix
$N = \begin{pmatrix} a_x & a_y & a_z\\ b_x & b_y & b_z\\ c_x & c_y & c_z \end{pmatrix}$

Finding the determinant in the same way, do you get something related to your problem?
How can you use your knowledge of matrices to get your desired result?

 

[Mark44]

There is a neat trick for this problem. Are you familiar with the method of calculating a cross product like this?

The OP is expressly forbidden from doing calculations of the sort I think you are leading up to.

 

[Sandro Romualdez]

My intention was that you express vector, $\ \vec c\,,\$ as $\ \vec a +2\vec b \,,\$ then take the cross product $\ \vec b\times ( \,\vec a +2\vec b \,) \,.\$

Use the distributive law on that.

Using the distributive law, I can manipulate the equation to be:
a • [(b x a) + (b x 2b)]

I can further use the distributive law on this as well? Making it
[a•(b x a)] + [a•(b x 2b)]

Using my knowledge of the dot and cross product, the cross product of b and a will be perpendicular to both vectors. This means that the first term will equal zero, as the dot product of perpendicular vectors is zero. Logically, the second term is also zero as (b x 2b) also results in a vector perpendicular to a. Therefore, since both terms are zero, a•(b x c) = 0.

Is this logical and sufficient? And if so, would there be a more concise way to state this?

 

[Sandro Romualdez]

There is a neat trick for this problem.

Thanks for your help, and I understand how the determinant can help with this problem. The only limitation from me using this solution though is the fact that I can't use calculations, as Mark44 stated.

 

[Sandro Romualdez]

Note: the numbers for a, b don't even matter

Thanks. The numbers are quite big anyways so I assumed that they weren't too significant, and maybe served as a deterrent from actually calculating

 

[tnich]

Thanks for your help, and I understand how the determinant can help with this problem. The only limitation from me using this solution though is the fact that I can't use calculations, as Mark44 stated.

You don't know to do calculations for this solution, you only need to know properties of matrices and their determinants. Under what condition would the determinant be zero?

 

[Stavros Kiri]

You don't know to do calculations for this solution, you only need to know properties of matrices and their determinants. Under what condition would the determinant be zero?

It's actually a lot simpler than that ...
[But that's one way ...]

Is this logical and sufficient? And if so, would there be a more concise way to state this?

You're on the right track, but one overlook ...

 

[blue_leaf77]

I know that for the dot product between two vectors to equal zero, the vectors must be perpendicular. The cross product is the vector that is perpendicular to both vectors b and c, but in this case (3D/Three space/R3), aren't there an infinite number of vectors that can be perpendicular to b and c? In that case how would I be able to explain why a•(b x c) is equal to zero?

In addition to the various hints everyone here has tried to lead the OP with, any three random points which are mutually not coinciding and not all situated along a single line uniquely define a plane.

 

[Stavros Kiri]

In addition to the various hints everyone here has tried to lead the OP with, any three random points which are mutually not coinciding and not all situated along a single line uniquely define a plane.

In general, that would not a good hint, I think, (although true) because in that case you should use the particular values for the vector coordinates and show that b and c truly define a plane (e.g. by showing that they are not linearly dependent). In any case I think that would involve calculations.

But in this case, your hint works, because you can go around avoiding calculations by examining the two available possibilities (about b and c), and doing them separately ... (What are these 2 possibilities?) ...

[Note: also cf. post #5]

 

[ehild]

[a•(b x a)] + [a•(b x 2b)]

Using my knowledge of the dot and cross product, the cross product of b and a will be perpendicular to both vectors. This means that the first term will equal zero, as the dot product of perpendicular vectors is zero. Logically, the second term is also zero as (b x 2b) also results in a vector perpendicular to a.

The vector product of two parallel vectors is zero, so bx2b=0.

 

[tnich]

There is a neat trick for this problem. Are you familiar with the method of calculating a cross product like this?
For the cross product of $\vec b$ and $\vec c$, form a matrix:
$M=\begin{pmatrix} 1 & 1 & 1\\ b_x & b_y & b_z\\ c_x & c_y & c_z \end{pmatrix}$

If you find the determinant of this matrix using expansion by minors, you get:
$det(M) = 1(b_y c_z - b_z c_y) + 1(b_z c_x - b_x c_y) + 1(b_x c_y - b_y c_x)$

You will notice that the terms in the sum are exactly the elements of $\vec b \times \vec c$.

Now try the same thing with matrix
$N = \begin{pmatrix} a_x & a_y & a_z\\ b_x & b_y & b_z\\ c_x & c_y & c_z \end{pmatrix}$

Finding the determinant in the same way, do you get something related to your problem?
How can you use your knowledge of matrices to get your desired result?

What I was hinting at was this:
The problem statement tells us that $\vec c = \vec a + 2 \vec b$, implying that the vectors are linearly dependent. Then 0 = det(N) = $\vec a \cdot \vec b \times \vec c$.

 

[Stavros Kiri]

What I was hinting at was this:
The problem statement tells us that $\vec c = \vec a + 2 \vec b$, implying that the vectors are linearly dependent. Then 0 = det(N) = $\vec a \cdot( \vec b \times \vec c)$.

That's one way, but/providing he has to know that method and equations first. There is a more direct and (perhaps) simpler way, already underlying almost full in previous posts ...
Note: I added an obvious parenthesis in quoting your post.

 

[Mark44]

any three random points which are mutually not coinciding and not all situated along a single line uniquely define a plane.

because in that case you should use the particular values for the vector coordinates and show that b and c truly define a plane (e.g. by showing that they are not linearly dependent).

The problem statement tells us that $\vec c = \vec a + 2 \vec b$, implying that the vectors are linearly dependent.

Since it is given that $\vec c = \vec a + 2 \vec b$, then clearly $\vec c$ is a linear combination of $\vec a$ and $\vec b$, which shows that the three vectors are linearly dependent (and hence lie in the same plane), but these are linear algebra concepts that the OP might not have seen yet.

The solution already posted, using the basic properties of the dot and cross products, is IMO the approach that the author of the problem was looking for.

 

[Sandro Romualdez]

Thanks for all the responses, I'll mark the problem as solved since I will just go with the solution using linear combinations and the properties of dot and cross products to prove the equation, since it seems like the simplest solution in terms of what we have studied recently.