A=3i + 2j +k and Bz = -1i +2j +Bk. |A + B| =6. Find Bk

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The discussion revolves around solving for the vector B in the equation A + B = 6, where A = 3i + 2j + k and Bz = -1i + 2j + Bk. Participants explore the calculation of the magnitude of the resultant vector and the conditions for B to satisfy the equation. One user initially calculates a value for Bz but struggles with the online check, leading to confusion about the absolute value and the correct approach to solving the equation. Ultimately, it is revealed that the correct value for B is 3, although there is a suggestion to start a new thread for further assistance on bisecting a vector. The conversation highlights the importance of clarity in vector calculations and the potential for online systems to accept approximate answers.
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Homework Statement



2 vectors are given by A=3i + 2j +k and Bz = -1i +2j +Bk. A + B =6

Homework Equations



Find the two possible vlaues of Bz

The Attempt at a Solution



6 = sqrt (4 + 20 +Bzsquared)

I get one number to equal -4.87 but when I solve for the other I get it wrong on the online check
 
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Is that the absolute value of A + B = 6?
 
yes it is the absolute value 6, I thought the answer would be to just take the positive 3.87 and subtract 1 from it but that was wrong online
 
Ok well in the equation you've come up with you wouldn't just have B2. When you add the vectors you would end up with a term (1+B)2. What values can B take to make everything under square root add up to 36?
 
Did that and for one answer I got -4.87 which when you put it in the () and square it you get 36 or close enough that the online problem accepted the answer, I assumed that for the other number I just needed to take 3.87 and subtract the 1 which left me with 2.87, however when I plug that in it says it is incorrect, where is my mistake?
 
Can you formulate the problem in a better way? For example you know that:

(B+1)^2 = 12

How would you solve that for B.
 
Never mind but thanks, I figured it out the answer is 3, but I need help bisecting a vector can you help?
 
3? I would not have thought it would be, but the online thing might be accepting answers within certain ranges. It would be best to post a new thread to receive the maximum attention.
 

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