Cross Vectors A & B: Solving for J

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Homework Help Overview

The problem involves calculating the cross product of two vectors, A and B, defined as A = <0, 3, 4> and B = <4, -4, 7>. The original poster expresses uncertainty about the correctness of the 'j' component in their calculation.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to compute the cross product using a formula but questions the accuracy of their 'j' component. Some participants provide alternative methods for remembering the formula and suggest checking for sign errors in calculations.

Discussion Status

Participants are actively engaging in clarifying the formula for the cross product and discussing potential errors in the original poster's calculations. There is no explicit consensus, but guidance has been offered regarding the correct interpretation of the components.

Contextual Notes

There are references to differing methods of applying the formula and potential sign errors, indicating that participants are exploring various interpretations of the cross product calculation.

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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
 
Last edited by a moderator:
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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