# Vector problem

1. Apr 29, 2015

### Raghav Gupta

1. The problem statement, all variables and given/known data

If the vectors a and b are non-collinear and the vectors c = (x-2)a + b and d = ( 2x + 1)a - b , are collinear then x is equal to
2. Relevant equations
c x d = |c| |d| sin θ n where n is unit vector
3. The attempt at a solution
Since c and d are collinear there cross product magnitude would be zero.
Also cross product of c and d is abx + 3ab
Now | abx + 3ab | = 0
What would be | abx + 3ab | = ?

2. Apr 29, 2015

### HallsofIvy

Staff Emeritus
I would not use the cross product. If two vectors are collinear then one is a multiple of the other: $(x- 2)a+ b= \alpha ((2x+ 1)a- b$ for some number, $\alpha$. Reduce that to Xa= Yb, for X and Y functions of x and $\alpha$. Since a is NOT a multiple of b, we must have X= 0 and Y= 0. That gives two equations to solve for x and $\alpha$.

3. Apr 29, 2015

### SammyS

Staff Emeritus
Don't forget, a × b = - (b × a) .

(In my opinion, the cross product works just fine here.)

4. Apr 29, 2015

### Ray Vickson

We are not told what is the dimensionality of the vectors. Cross-products exist only for 3-dimensional vectors (or 7-dimensional vectors). If the vectors here are in $\mathbb{R}^2$ then the cross-product is meaningless.

5. Apr 29, 2015

### Raghav Gupta

Thanks, I had forgotten that and was doing a mistake.
Getting finally the answer as 1/3
Don't know about dimensionality thing of vectors . Maybe that is taught in university.

6. Apr 29, 2015

### Ray Vickson

The floor in your room is 2-dimensional (a flat plane). Your whole room is 3-dimensional. You can draw line segments on your floor, and they would be 2-dimensional vectors. You can use sticks that point off in different directions in your room, and they would be 3-dimensional vectors. You don't need to go to university for that!

Anyway, in your problem which types of vectors are you talking about? Or, is this another one of those problems you have been given where important pieces of information have been omitted?

7. Apr 30, 2015

### Raghav Gupta

You must see this [URL [Broken] Wikipedia [/URL] ( Why the link is not coming as Wikipedia ? )
What you are saying is true that cross - product has meaning for 3-dimensions and 7- dimensions,
but no matter when the vectors are parallel or have 0° between them or are collinear, the cross product is zero.

Last edited by a moderator: May 7, 2017
8. Apr 30, 2015

### Ray Vickson

This statement is false unless your vectors are in 3-dimensions. In a plane it is not true, because you cannot even talk at all about a cross product in that case. Unless you are looking at vectors in 3 dimensions it is meaningless to say the cross-product is zero --- there is no cross-product!

Lesson for you: just avoid using the cross-product completely in this problem. You do not need it; it does not help, and it just gets in the way. It is too easy to mis-use, as you have done.

Last edited by a moderator: May 7, 2017
9. Apr 30, 2015

### Raghav Gupta

But why in Wikipedia they have written that if two vectors are parallel or make 0° then their cross product is zero, not specifying in which dimensions they should be?
Also, why the answer is coming right by using that?

10. Apr 30, 2015

### Ray Vickson

Getting a right answer by a wrong method is not unusual. However, when you are trying to learn a subject you should try very hard to not do that.

11. Apr 30, 2015

### Raghav Gupta

But I get almost every answer correct for questions involving collinearity of vectors where they are not specifying dimensions. What is the reason for that or the flaw?

12. Apr 30, 2015

### Raghav Gupta

Hey how you can argue with a khan academy video?

At 15: 04 , he is saying that when two vectors are collinear their cross product is zero.
[ Edit ] - sorry, the video person is saying that the cross product is defined for R3.
But I am not getting it.
If vectors are collinear, then definitely their cross product would be a vector perpendicular to them, however it is all looking confusing since it's magnitude would be zero.

Also why in the problem statement we cannot assume that vectors are in R3 ?

Last edited: Apr 30, 2015
13. Apr 30, 2015

### Ray Vickson

It depends on what your textbook does. If this problem occurs in a chapter that is mostly about 3-dimensional vectors, then it is likely OK to assume 3-d here. However, if it occurs in a chapter mostly about 2-dimensional vectors, they probably want you to assume 2 dimensions. Finally, if it is in a chapter dealing with general linear algebra in an unspecified number of dimensions, n, then they likely intend that you use n-dimensional vectors.

Anyway, using the cross-product is doing it the hard way; it is much faster to just recognize that if c = (x-2)a+b and d = (2x+1)a - b are colinear, so are c and -d = b -(2x+1) a. Since these both contain the common term b, it must be the case that the a-coefficients are equal; that is, we must have x-2 = -(2x+1).

Last edited: Apr 30, 2015
14. Apr 30, 2015

### Staff: Mentor

The do say that the vectors are in three dimensions, right at the beginning of the article. From the link you gave (emphasis added).
They couldn't be much plainer than that.

15. May 1, 2015

### Raghav Gupta

Thanks, this is pretty fast.

16. May 1, 2015

### Raghav Gupta

Hmm.. that's true.