vectors

[xc630]

Hi i am having trouble with this problem.

Draw parallelogram ABCS with AB =10, BC = 6, and angle A =60 degrees.

A) [vector AB + vector BC] = ?
the [] signs are supposed to absolute value.

Thanks for any help

 

[arthur ohlsten]

drop a perpindicular from B to AS not necessary but will help "see" the solution

then vector AB can be written as to vectors,one in direction of BC and one perpindicular

from right triangle with AB as hypoteneuse
AB PARALLEL TO BC =5 AB perpindicular to BC= sqrt75

sum of the vectors AB + BC = 6+5 in BC direction sqrt75 perpindicular

absolute value of A+B = sqrt[ 11^2+[sqrt75]^2[
absolute value of AB= sqrt [121+75]
absolute value A+B= sqrt 196
absolute value A+B=14 answer
Arthur
please check math for errors

 

[pka]

Here is a second way to do this problem.
What you are calling absolute value is really vector length.
|vector AB + vector BC| is the length of the vector AC.
Note that angle B is 120°. Now use the law of cosines.
|vector AB + vector BC|<SUP>2</SUP>=(10)<SUP>2</SUP>+(6)<SUP>2</SUP>−2(10)(6)cos(120°).

 

[soroban]

Hello, xc630!

Here's yet another approach . . .

Draw parallelogram ABCD with AB =10, BC = 6, and angle A =60°.

A) |vec{AB} + vec{BC}| = ?

         D       10        C
          *---------------*
         /               /
       6/              6/
       /60°            /60°
      *---------------* - -
     A       10       B

vec{AB} .= .10i
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . _
vec{BC} .= .(6 cos60°)i + (6 sin60°)j .= .3i + 3√3j
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ._
Hence: .vec{AB} + vec{BC} .= .13i + 3√3j
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .__________ . . . . .___
Therefore: . |vec{AB} + vec{BC}| . = . √13² + (3√3)² . = . √196 . = . 14

 

[xc630]

Hi PKA can you explain to me the law of cosines part. I am a little confused. I know cosine is Adjacent/hypotenuse in rt tringles but thats it. Where did you get 2(10) (6) cos (120)? Thanks

 

[pka]

LAW of COSINES:
In any triangle, if a & b are the lengths two sides and Θ is the angle between those two sides
then if the third side is c we have: c<SUP>2</SUP>=a<SUP>2</SUP>+b<SUP>2</SUP>−2(a)(b)cos(Θ).
Note this is for any triangle. If Θ=90° then you just have the Pythagorean Theorem.