Expert: Ahmed Salami - 10/10/2010

Question
1. The problem statement, all variables and given/known data
A= ci-2j+k
B=i+2j-k
Find c that makes the vector(A-B) perpendicular to the vector B
2. Relevant equations
AXB = (AxBy-AyBx)k+(AzBx-AxBz)j+(AyBz-AzBy)i



3. 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?

Answer
Hi Hamad,
For two vectors A and B, the dot product is defined as
A.B = |A||B|cosθ
Now, this is a scalar as the right hand side is simply a number.
The cross product of two vectors, on the other hand, is another vector whose magnitude is
|A||B|sinθ
and is in a plane perpendicular to that containing vectors A and B.

The easier way to solve the problem here is to use the dot product since this is zero for perpendicular vectors.
A = ci - 2j + k
B = i + 2j - k
A-B = (c-1)i - 4j + 2k
(A-B).B = (c-1)(1) + (-4)(2) + (2)(-1) = (c-1) - 8 - 2 = c - 11
The vectors are then perpendicular when c = 11

Attempting to use the cross product,
(A-B)xB = [(c-1)(2) - (-4)(1)]k + [(2)(1) - (c-1)(-1)]j + [(-4)(-1) - (2)(2)]i
= (2c+2)k + (c+1)j
and its magnitude is
√[(2c+2)² + (c+1)²] = √[(2c+2)² + (c+1)²] = (√5)(c+1)
This should be equal to  |(A-B)||B| since sin90 = 1
|(A-B)||B| = √[(c-1)² + (-4)² + (2)²] . √[(1)² + (2)² + (-1)²]
= √[(c-1)² + 20] . √6
Equating and squaring gives
5(c+1)² = 6[(c-1)² + 20]
5c² + 10c + 5 = 6c² - 12c + 6 + 120
c² - 22c + 121 = 0
(c - 11)² = 0
c = 11
as before.

Clearly the first method is simpler and faster. But you seem to know what you're doing and i think the mistake you made was writing 5 instead of 6 somewhere in your attempted solution.

Regards