Cross product ,why doesnt this work

[madah12]

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?

 

[Quinzio]

Isn't is much simpler with the dot product ?

 

[madah12]

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

 

[Mark44]

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.)

 

[madah12]

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)

 

[madah12]

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

 

[Mark44]

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.

 

[D H]

Your error:

then I got
5((c-1)^2+20)=5(c+1)^2

What is |B|² ?

(Hint: It is not 5.)

 

[madah12]

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