Cross Vectors A & B: Solving for J

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The discussion focuses on calculating the cross product of vectors A and B, where A = <0,3,4> and B = <4,-4,7>. The user expresses uncertainty about the correctness of the 'j' component in their solution, which they calculated as -16. Clarification is provided that the formula for the 'j' component, (a3b1 - a1b3), is indeed correct. The conversation also highlights the importance of checking calculations by cycling through the components of the vectors. Overall, the user receives confirmation and guidance on the correct approach to solving the problem.
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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'?
 
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A good way to remember it is the "determinant"
\left|\begin{array}{ccc}\vec{i} &amp; \vec{j} &amp; \vec{k} \\ a_1 &amp; a_2 &amp; a_3 \\ b_1 &amp; b_2 &amp; b_3 \end{array}\right|

Expanded by minors along the first row this is
\vec{i}\left|\begin{array}{cc}a_2 &amp; a_3 \\ b_2 &amp; b_3\end{array}\right|- \vec{j}\left|\begin{array}{cc}a_1 &amp; a_3 \\ b_1 &amp; b_3\end{array}\right|+ \vec{k}\right|\begin{array}{cc}a_1 &amp; a_2 \\ b_1 &amp; 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
 
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Welcome to PF!

webz said:
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! :smile:

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. :wink:
 
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!
 
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