Cross Vectors A & B: Solving for J

[webz]

Homework Statement

Cross vector A with vector B.
A = <0,3,4> B = <4,-4,7>

Homework Equations

a x b = (a2b3 - a3b2) i + (a3b1 - a1b3) j + (a1b2 - a2b1) k
(I think)

The Attempt at a Solution

AxB= i(21-16)-j(16-0)+k(0-12) = <5,-16,12>

I'm unsure if the component 'j' is correct. First, my teacher negated the value in their formula (i-j+k), and then he cross multiplied backwards (instead of (a1b3-a3b1) he used (a3b1-a1b3)). I have done this in the above problem to mimic what he did, however, I don't know if it is correct. Can someone help to clarify component 'j'?

 

[HallsofIvy]

A good way to remember it is the "determinant"
$$\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{array}\right|$$

Expanded by minors along the first row this is
$$\vec{i}\left|\begin{array}{cc}a_2 & a_3 \\ b_2 & b_3\end{array}\right|- \vec{j}\left|\begin{array}{cc}a_1 & a_3 \\ b_1 & b_3\end{array}\right|+ \vec{k}\right|\begin{array}{cc}a_1 & a_2 \\ b_1 & b_3\end{array}\right|$$
$$= (a_2b_3- a_3b_2)\vec{i}- (a_1b_3- a_3b_1)\vec{j}+ (a_1b_2- a_2b_1)\vec{k}$$
$$= (a_2b_3- a_3b_2)\vec{i}+ (a_3b_1- a_1b_3)\vec{j}+ (a_1b_2- a_2b_1)\vec{k}$$
which is just what you say!

However, in calculating that you have missed a few signs.
$$a_2b_3- a_3b_2= 3(7)-(4)(-4)= 21+ 16= 37$$
$$a_3b_1- a_1b_3= 4(4)- (0)(7)= 16- 0= 16$$
$$a_1b_2- a_2b_1= 0(-4)- (3)(4)= 0- 12= -12$$

 

[tiny-tim]

Welcome to PF!

a x b = (a2b3 - a3b2) i + (a3b1 - a1b3) j + (a1b2 - a2b1) k
(I think)
…
I'm unsure if the component 'j' is correct.

Hi webz! Welcome to PF!

Yes, (a3b1 - a1b3) j is correct.

A good way to check any formula like this is to take the first term, (a2b3 - a3b2) i,

and just make everything "cycle" to the next one …

so a2 goes to a3, b3 goes to b1, and i goes to j.

 

[webz]

Thanks for the quick replies! You guys might be seeing me around here, I have to relearn all the calc and chemistry I've ever taken before next semester starts! :P

Appreciate your help!