Vector Products & Properties

1. Jan 16, 2009

x^2

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

Which of the following are true?
A) A vect. prod. C=C vect. prod. A
B) The x-component of a vector can be +, -, or zero
C) If A=BxC and C=64i, then Ax=0
D) If A perp to B, then B dot A =0
E) The magnitude of a vector is sometimes negative.
F) The scalar (or dot) product of two vectors can be +, -, or zero.

2. Relevant equations

BxC = BC*cos(theta) where theta = angle between vectors

3. The attempt at a solution
Below are my attempts at proving or disproving each statements.

A) False: Using the right hand rule, the direction of the resulting product will be inverted when comparing AxC and CxA
B) True: A vector could cover horizontal distance to the right (positive), to the left (negative), or be a vertical vector with an x component of zero.
C) False: 64i could potentially result in an answer where Ax = 64i
D) True: BxA = BAcos(90) = BA*0 = 0
E) False: A magnitude is the absolute value of a vector and therefore can never be negative.
F) False: A scalar product must always be positive as all scalars are positive--they would be the absolute value of a vector.

I tried answering B & D as true and the rest false but this was not correct. I am not sure which statements I am confused on. I'm fairly confident of my answers to A, B, D, and E while a little less sure on C and F.

Thanks,
x^2

2. Jan 16, 2009

rl.bhat

C)Since A=BXC. A is perpendicular to B and C. Since C = 64i is along x axis, Ax is zero.
F) A.B = ABcos(theta). ANd cos(theta) can be +, - or zero.

3. Jan 16, 2009

NoMoreExams

define 2 vectors A = (-1, 1) and B = (1, -1)

$$A \cdot B = -1 - 1 = -2$$

Certainly negative.

4. Jan 16, 2009

x^2

Thanks for the replies! C definitely makes sense when using the right hand rule. For some reason I was still relating Ci to Ax without thinking that they were perpendicular. As for F I see what they are asking for--I guess I was thinking that scalar had to mean magnitude but in actuality they just meant dot product.

Thanks for the help,

x^2