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#1
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numerical that is bothering me.
If A = A_1^i + A_2 i + A_3 j and B = B_1 i + B_2 j + B_3 k, then prove that A,B = A_(1 ) B_1 + A_2 B_2 + A_3 A_3
Somebody please solve it and also which type of numerical is this. |
#2
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Its a simple vector numerical.
1. Multiply A with B and i*i=-1 j*j=-1 2. Then take magnitude and the statement will be proved. |
The Following User Says Thank You to lovereflector For This Useful Post: | ||
waqas izhar (Tuesday, May 12, 2015) |
#3
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The Following 2 Users Say Thank You to lovereflector For This Useful Post: | ||
doctorali (Wednesday, June 17, 2015), waqas izhar (Tuesday, May 12, 2015) |
#4
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@ love reflector
u sure one can write i squared and j squared? aren't they vectors which require to be written as i.i and j.j? |
#5
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Solution
Please use the vector notation. Furthermore, iota-squared = -1 is an incorrect notion here in this context. Please do not solve it the way you have done in FSc or, maybe, in A-levels. "i" has nothing to do with iota. It is just a direction vector(unit vector) along the x-axis. Regards. |
The Following User Says Thank You to IslamabadKid For This Useful Post: | ||
lovereflector (Tuesday, May 12, 2015) |
#6
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Sorry my mistake and dear just take the magnitude rest is ok.
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#7
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well i think vectors are cross products and scalers are dot products. mentioned wrongly in above paper
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#8
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Dot/Cross products
Quote:
Sorry buddy! You have taken vectors and scalars wrongly! Please come out of the FSc/A-level's physics; otherwise, you score will be ruined for sure! Regards. Bro, at the end where you have taken the magnitude needs your attention. You cannot cancel the squared-power of each term with the (1/2) of square-root isolately. You are violating the rules of squareroot. |
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