# Solution to vector equation involving vector product

1. Mar 22, 2015

### PcumP_Ravenclaw

I want to find the solution of vector X. I am using text from Alan F. Beardon Algebra and Geometry as attached. I dont know how the solution is derived for the following equation.

$x + (x × a) = b$

The second solution when $a \times b \neq 0$ then X cannot be b. Is it possible to factorize x out from the Left-hand-side of the above equation?

#### Attached Files:

• ###### Vector_Equations.jpg
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2. Mar 22, 2015

### Staff: Mentor

It seems to me that you're missing the point, which is not to solve the equation, but only to determine whether it has one or more solutions.

The assumption is that you have two solutions -- let's call them x1 and x2. How they were obtained is not important. In the page whose image you included, they assume that y is the difference of any two solutions, so y = x1 - x2. Continue from here.

3. Mar 22, 2015

### mathman

When you expand the vector equation as 3 scalars in terms of the components of x, you will have 3 linear equations in 3 unknowns. There is no way to factor out x.

4. Mar 22, 2015

For solving vector equations like the one you got, there is a method. Since a,b and axb are 3 non-coplanar vectors, you can take
$x=ma+nb+p(a*b)$
Just like you take x=ai+bj+ck.
Where m,n,p are scalars.

Try finding the values of m,n,p by substituting the value of x in the given vector equation.

I am sorry @Mark44 for providing the solution.

5. Mar 22, 2015

### WWGD

There is no natural multiplication of vectors in $\mathbb R^3$ that allows you to factor.
EDIT I meant other than the obvious dot and cross products; then some expressions containing
both may be at times factorable, but not otherwise.

6. Mar 23, 2015

### PcumP_Ravenclaw

Hey @AdityaDev, A & B are not necessarily orthogonal so it is not exactly like ai + bj + ck where i, j & k are mutually orthogonal! so??

7. Mar 23, 2015

It doesn't matter. I have solved many questions using this method.
In certain questions, you have to use logic.
Example: Solve $$\vec{r}\times\vec{b}=\vec{a}\times\vec{b}$$
Given r.c=0

8. Mar 24, 2015

### PcumP_Ravenclaw

Aditya's missing explanations in the above post can be found in this attachment!!

Why do we want to find the vector X with basis $a, b, a \times b$ ?

Why not in simple $i, j, k$? Is it because it is simplest and cancels out easily in the equation $x + (x \times a) = b$

Then why is each component $x_1 , x_2 , x_3$ not divided by the scalar triple product $[a, b, a \times b]$ e.g.

$X = [X, b, a \times b]/[a, b, a \times b]*x_1 + [a, X, a \times b]/[a, b, a \times b]*x_2 + [a, b,X]/[a, b, a \times b]*x_3$

#### Attached Files:

• ###### Vector_Equations2.jpg
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Last edited: Mar 24, 2015
9. Mar 24, 2015

It was a question for you to practice. I already knew the solution.

10. Mar 24, 2015

### PcumP_Ravenclaw

I have attached my attempt at finding $x_1 , x_2, x_3$. How do I simplify further??

#### Attached Files:

• ###### Vector_Equations3.jpg
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127
11. Mar 24, 2015

What you have done is correct.

Now, expand (axb)xa.
(axb)xa=(a.a)b-(a.b)a ---> Vector triple product.
Now you will get :

Then expand the last term to get $X_3|a|^2\vec{b}-X_3(a.b)\vec{a}$
Now you will get this expression :

then group axb and bxa terms as you can express axb as -bxa.
then group the terms with vector a.
also group the terms of vector b.
So you have three terms.
Then equate the coefficients of all vectors other than b like a and axb to find x1,x2,x3.

12. Mar 25, 2015

### PcumP_Ravenclaw

Hey @AdityaDev, thanks for your help. I was not that familiar with vector triple product so I could not spot the expansion of $a \times b \times a$ into (a.a)b - (a.b)a.

13. Mar 25, 2015

EDIT: equate the coefficients of all vectors other than b "to zero". Now you can find x1, x2, x3.
Did you get the correct answer?

14. Mar 25, 2015

### Ananya0107

First take the cross product with a on both sides,
This should give you , (b-x) + (a.x)a - (|a|^2)x = b×a.......(equation 1)here I have substituted x × a as b-x,
Substitute value of a.x by using a second equation , in which you take dot product on both sides of the original equation , x.a + (x×a).a= b.a
Which gives x.a + 0 = b.a
Now substitute a.x as b.a

This should allow you to collect terms of x.

Last edited: Mar 25, 2015
15. Mar 25, 2015

You will get (a.x)a=(b-x)×a + kx
also, you cant find a.x from this equation because it is grouped with a vector and if you thought this: $$a.x=\frac{(b-x)\times\vec{a}+kx}{\vec{a}}$$
Then you are wrong.
You can however find the value of vector a.

16. Mar 25, 2015

### Ananya0107

Please see the edited post above .

17. Mar 26, 2015

### PcumP_Ravenclaw

Hi, can you all answer my questions??

Why do we want to find the vector X with basis $a, b, a \times b$ ?

Why not in simple $i,j,k$? Is it because it is simplest and cancels out easily in the equation$x+(x×a)=bx + (x \times a) = b$

Then why is each component $x_1 , x_2 , x_3$ not divided by the scalar triple product $[a, b, a \times b]$ e.g.

$X=[X,b,a \times b]/[a,b,a \times b]∗x_1+[a,X,a \times b]/[a,b,a \times b]∗x_2+[a,b,X]/[a,b,a \times b]∗x_3$

Thanks!!

Last edited: Mar 26, 2015
18. Mar 27, 2015

### Ananya0107

a, b, and a×b are a set of non coplanar vectors and any vector in three dimensional space can be expressed as a linear combination of three non-coplanar vectors,( just like i, j, k)

19. Mar 28, 2015

### PcumP_Ravenclaw

Hey @Ananya0107, Can you explain why was a, b and a x b selected why not any other vector as you say " any vector in three dimensional space can be expressed as a linear combination of three non-coplanar vectors,( just like i, j, k)"?

Thanks!

20. Mar 30, 2015

### Ananya0107

The question contained the vectors a and b so it is easier to express the required vector x in terms of a , b and a×b , it is just easier for computation .