Calculating 2A + 3(B+C): Vector Addition

AI Thread Summary
To calculate 2A + 3(B+C) with vectors A, B, and C, first compute 2A as (4, -2, 2). Next, add vectors B and C to get (4, 4, 3), then multiply by 3 to obtain (12, 12, 9). Adding these results gives (16, 10, 11) for 2A + 3(B+C). The method used is correct, as scalar multiplication simply scales the vector's magnitude while maintaining its direction. The discussion confirms that this approach adheres to vector addition and scalar multiplication rules.
klm
Messages
165
Reaction score
0
vector A= (2,-1,1)
vec. B = ( 3, 0, 5)
vec. C = (1,4,-2)
what is 2A + 3(B+C)

this is what i did:
2A = 2(2,-1,1) = (4,-2, 2)

(B+C) = (4,4,3) x 3 = (12,12, 9)

2A + 3(B+C) = (16,10,11)

is this the correct way to think and do this problem?
 
Physics news on Phys.org
Yes, you just add up the components because all vectors are the sum of their components; and of course, a scalar times a vector is just a vector of scaled up magnitude in the direction of the original vector.
 
okay thank you! and is it all right to multiply like that? or is there some other rule for multiplying vectors?
 
It is all right, because that is just a case of a scalar multiplying a vector.
 
ok! thank you very much!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top