If #vec B# is added to #vec C=4.8hati+4.3hatj#, the result is a vector in the positive direction of the y axis, with a magnitude equal to that of #vecC#. What is the magnitude of #vecB?#

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Answer 1 · 2017-09-13T05:41:49

The magnitude of #vecB# is #=5.25#

The magnitude of #vecC# is

#||vecC||=||4.8hati+4.3hatj||=sqrt(4.8^2+4.3^2)=6.44#

The angle is #theta=arctan(4.3/4.8)=41.86^@#

Applying the cosine rule to the isoceles triangle,

#B^2=C^2+C^2-2C^2cos(90-41.86)#

#B^2=2C^2(1-cos(90-41.86))=27.6#

#B=sqrt(27.6)=5.25#

The magnitude of #vecB# is #=5.25#