Question

∥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∥=

Answer 1

`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`