# Cross product ,why doesnt this work

In summary, when finding the value of c that makes the vector (A-B) perpendicular to vector B, it is necessary to set the dot product of these two vectors to zero. This can be solved for c, and in this case, c=11. Using the cross product method can also work, but it is more complicated and may lead to errors, as seen when trying to use it to find c=5.

## Homework Statement

A= ci-2j+k
B=i+2j-k
Find c that makes the vector(A-B) perpendicular to the vector B

## Homework Equations

AXB = (AxBy-AyBx)k+(AzBx-AxBz)j+(AyBz-AzBy)i

## The Attempt at a Solution

A-B=(c-1)i-4j+2k
I said that since (A-B) is perpendicular to B then
|A-B| * |B| = |AXB| because sine 90 =1
then I got
5((c-1)^2+20)=5(c+1)^2
which gave me a solution of C=5
but
then A dot B didnt turn out to be zero why?

Isn't is much simpler with the dot product ?

yes it is I already did it with dot product c is equal to 11 I know that but why doesn't it work with cross product?
Btw I meant A in the attempt at solution as the vector A-B

## Homework Statement

A= ci-2j+k
B=i+2j-k
Find c that makes the vector(A-B) perpendicular to the vector B

## Homework Equations

AXB = (AxBy-AyBx)k+(AzBx-AxBz)j+(AyBz-AzBy)i

## The Attempt at a Solution

A-B=(c-1)i-4j+2k
I said that since (A-B) is perpendicular to B then
|A-B| * |B| = |AXB| because sine 90 =1
I don't think this is true at all. On the left you have the product of the magnitudes of A - B and B, and on the right, you have the magnitude of A X B, which is |A||B| sin(theta). You are not given that the angle between A and B is 90 degrees, so sin(theta) isn't going to be 1.
then I got
5((c-1)^2+20)=5(c+1)^2
which gave me a solution of C=5
but
then A dot B didnt turn out to be zero why?

A . B = 0 if A and B are perpendicular, which isn't given.

What you want is for A - B to be perpendicular to B, so set the dot product of these two vectors to zero to solve for c. (I get c = 11, too.)

yes I mentioned that because when I was in a hurry I forgot to write that I mean|A-B|X|B|
but I calculated it but I don't know whether I messed up the the calculation or the method doesn't work.
I am saying (A-B)X(B) = |(A-B| *|B| * (sin90=1)

I got the cross product of B X (A-B)=(c+1)j + 2(c+1)k ( checked more than once)
then sqrt(5(c+1)^2)= sqrt 5 * sqrt(((c-1)^2+20)) then I squared

OK, that will work, and will give you c = 11, but it's a lot more work than using the dot product. It took me half a page using this technique vs about 3 lines using the dot product.

If you got c = 5, you must have made a mistake somewhere.

then I got
5((c-1)^2+20)=5(c+1)^2
What is |B|2 ?

(Hint: It is not 5.)

B=i+2j-k
|b|= sqrt(1^2 + (2)^2 + (1)^2)= sqrt(1+4+1) =sqrt6
|B|^2=6
loll I never thought that I would do a mistake in addition so the only step I didnt check was this and it turned out to be the wrong one... I got 11 now

## 1. Why is the cross product only defined for 3-dimensional vectors?

The cross product is a mathematical operation that calculates a vector perpendicular to two given vectors. It is only defined for 3-dimensional vectors because it is the minimum number of dimensions needed to determine a unique perpendicular direction.

## 2. What happens when two vectors used in the cross product are parallel?

If two vectors are parallel, the cross product will result in a zero vector. This is because there is no unique perpendicular direction that can be determined between two parallel vectors.

## 3. How do you calculate the magnitude of the cross product?

The magnitude of the cross product can be calculated using the formula |a x b| = |a| * |b| * sin(θ), where a and b are the two vectors and θ is the angle between them.

## 4. Can the cross product be used for vectors in higher dimensions?

No, the cross product is only defined for 3-dimensional vectors. In higher dimensions, a similar operation called the wedge product is used.

## 5. Why is the cross product useful in physics and engineering?

The cross product is useful in physics and engineering because it can be used to calculate torque, angular momentum, and the normal vector to a plane. It also has applications in electromagnetism and fluid mechanics, among other fields.

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