Question

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?`

Answer 1

The magnitude of `vecB` is `=5.25`

Explanation:

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`