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Two vectors A and B have precisely equal magnitudes.If magnitudes of A and B to be larger than the magnitude of A-B by a factor n,the angle between them is

Two vectors A and B have precisely equal magnitudes.If magnitudes of A and B to be larger than the magnitude of A-B by a factor n,the angle between them is

Grade:10

3 Answers

Mukul
11 Points
6 years ago
 
 
the condition can be represdented by.  (A + B+ 2 ab cos@)1/2= ( …..- 2abcos@ )1/2  * -n 
Then u will be left with n= -1 
Therefore they are anti parallel and @ will be 1800
 
samson kalumpha
24 Points
5 years ago
let ∂ be an angle between the two vectors, then the magnitude, R=A+B form isoscelence triangle with angles 180-∂, ∂/2 and ∂/2
using consine rule, and concept of dot product we have R=2Acos(∂/2)
on the other hand R1=A-B also form isoscelence triangle since A=-B wiht angles 180-∂, ∂/2 and ∂/2
similary R2=2Asin(∂/2)
but for the magnitude of the two vectoers to be exactly equal it requires that R2 must be greater by 100
hence R2=200sin(∂/2)
equating the two vectors gives
2cos(∂/2)=200sin(∂/2)
1=100tan(∂/2)
tan-1(1/100)=∂/2
∂=1.146o
kindalem
13 Points
2 years ago
ual magnitudes. in order for the magnitude of  A - B to be larger than the magnitude of A + B by the factor  n, what must be the angle between them ?

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