Question

Prove mathematically that for any two vectors u and w lying in the xy plane of a given reference frame
mag²(u+w)=
mag²(u)+mag²(w)+2mag(u)mag(w)cos θ
where θ is the angle between directions of p and q?

Answer 1

You can check these two links:

http://physics.stackexchange.com/questions/11132/the-resultant-of-two-forces-acting-at-any-angle

https://www.academia.edu/1621682/The_Resultant_of_Two_Vectors

Answer 2

let the angle \(u\) makes with the x-axis be \(\phi\)\[\vec u=u_x\,\hat x+u_y\hat y=u(\cos\phi\,\hat x+\sin\phi\,\hat y)\]

and the angle \(w\) makes with the x-axis be \(\psi\)\[\vec w=w_x\,\hat x+w_y\,\hat y=w(\cos\psi\,\hat x+\sin\psi\,\hat y)\]

Answer 3

\[\vec u+\vec w=u(\cos\phi\,\hat x+\sin\phi\,\hat y)+w(\cos\psi\,\hat x+\sin\psi\,\hat y)\\
|\vec u+\vec w|=\sqrt{(u(\cos\phi\,\hat x+\sin\phi\,\hat y))^2+(w(\cos\psi\,\hat x+\sin\psi\,\hat y))^2}\\
|\vec u+\vec w|^2=(u(\cos\phi\,\hat x+\sin\phi\,\hat y))^2+(w(\cos\psi\,\hat x+\sin\psi\,\hat y))^2\]

Answer 4

now simplify and use
\[\vec x\cdot\vec x = 1\\\vec y\cdot\vec y = 1\\
\vec x\cdot\vec y = 0\\
\\
\cos^2\phi+\sin^2\phi= 1\\
\cos^2\psi+\sin^2\psi= 1\]

Answer 5

and \(\phi -\psi= \theta\)

Answer 6

do you think that will work?

Answer 7

Can't you just use the distributive law for taking dot products of vectors ?

Answer 8

\[\left| u+w^2 \right|=(u+w).(u+w)=\left| u \right|^2+\left| w \right|^2+2\left| u \right|\left| w \right|\cos(\theta)\]
QED