What is the magnitude of the vector product?

[Equilibrium]

Homework Statement

Find the vector product $$\vec{A} \times \vec{B}$$ (expressed in unit vectors ) of the two vectors given. What is the magnitude of the vector product?
$$\vec{A} = 5.00\hat{i} + 2.00\hat{j}$$
$$\vec{B} = 3.00\hat{i} - 1.00\hat{j}$$

Homework Equations

$$\vec{A}\vec{B}\sin(\theta)$$

The Attempt at a Solution

using pythagorean theorem i found that the resultant of the two is
5.39 and 3.16

plus the total angle w/c is 40.2

using the formula mentioned above i found the answer as 11k and magnitude of 11.0

what i am confused is that it has a different answer from the back of my book w/c is -$$-17\hat{k},17.0$$
who is wrong the book or me?
and pls show me why?

 

[Doc Al]

Your answer seems correct to me. As a double check, find the cross product by multiplying the components.

 

[Equilibrium]

my answer is wrong
the unit vectors should be
using the determinant rule
z =AxBy - AyBy
z = 5(-1)-2(3)

z = -11

-11\hat{k}

and the magnitude is where u use ABsin(theta) =11.0

 

[Doc Al]

Why do you say your answer is wrong? (Note that both methods gave you the same answer.)

 

[Equilibrium]

how?

...
the components shud be positive??
now I am confused

 

[Doc Al]

I'm not sure what you are asking. In your first post, you showed how you arrived at your solution. (Yes, your sign was wrong, but all you needed was the magnitude.) I said it was correct and asked you to double-check using a different method. You did and got the same answer for the magnitude. So why do you think your answer is wrong?

 

[Equilibrium]

i thought this has two questions
Finding vector product AXB expressed in unit vectors and magnitude

 

[D H]

Are you sure that you have both vectors stated properly? For example, flipping the x and y components of B yields the answer in the book: $$\vec B = 1.00\hat i - 3.00\hat j$$. This is an easy mistake to make -- either by you or by the authors.

 

[Doc Al]

Good eye, D H. I'll bet that's it.

 

[Equilibrium]

so the book is wrong then...

 

[Doc Al]

so the book is wrong then...

One of you must be. For the problem as you described it here, your solution is correct. (Make sure you copied it correctly, as D H suggested.)