help me plz

Question

|a|=11 |b|=23  |a-b|=30  |a+b|=?

 

Answer 1

     I did this in the xy plane. The same thing would happen if you were studying 3 space. This involves the law of cosines, used twice. For convenience, I'll suppose that your vectors are both in the first quadrant. They don't have to be but just so I can describe a diagram to you, lets pretend. Let's let vector a start at the origin and proceed into the first quadrant close to the x axis ending at point M. Let's let vector b start at the origin and proceed into the first quadrant close to the y axis ending at pt N. The geometric interpretation of a+b is the diagonal of a parallelogram having 3 vertices at pt N, origin and pt M. you need to complete the parallelogram to find the 4th pt,call it P. So, a+b is segment OP. The geometric interpretation of a-b is the other diagonal MN.

We know

OM=11=NP

ON=23=MP

MN=30

Hope you're still with me.

Focus on triangle OMN. You have 3 sides and can therefore use the law of cosines to find angle MON.

30^2 = 11^2 + 23^2 - 2*11*23cosø.

Hopefully you can solve this to get ø approx = 120 degrees. (This is why I knew the diagram is inaccurate, but it doesn't matter)

Using parallelogram MONP, angle ONP is the supplement of 120 or 60..

Using triangle ONP and the law of cosines we get,

let |a+b| = x,

x^2 = 11^2 + 23^2 - 2*11*23cos60. .

Solving you get x = 20.

 

Hope this helps.