∥v∥=3 ∥w∥=4 The angle between v and w is 1.2 radians. How to calculate ? a) v⋅w = ,(b) ∥1v+1w∥= , (c) ∥4v−4w∥=

1 Answer
Trevor Ryan.
Oct 4, 2015

#v*w=4,348#
#||v+w||=4,038#
#||4v-4w||=16,152#

Explanation:

Assuming we are working in a real 3 dimensional inner product space (otherwise the inner product would not be defined by angles), we have that

#v*w=||v||.||w||costheta#, where #theta#is the angle (in degrees) between v and w.
So #v*w=3xx4xxcos68,755^@=4,348#

To find #v+w# you will need to know what the co-ordinates of each vector is as you currently just have their norms.
Then you can add them according to the rules and obtain

#v=(v_1,v_2,v_3) and w=(w_1,w_2,w_3) =>#

#v+w=(v_1+w_1,v_2+w_2,v_3+w_3)#

Then #||v+w||=sqrt((v_1+w_1)^2+(v_2+w_2)^2+(v_3+w_3)^2#

Similarly, #||4v-4w||=4sqrt((v_1-w_1)^2+(v_2-w_2)^2+(v_3-w_3)^2)#

Alternatively, you may use the cosine rule since we are working in #RR^3# and this will yield :

#||v+w||^2=||v||^2+||w||^2-2||v||||w||cos68,755^@#
#therefore||v+w||=sqrt(3^2+4^2-2*3*4cos68,755^@)=4,038#

#||4v-4w||=4||v-w||=4||v+(-w)||=4||v+w||=4xx4,038=16,152#

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