Vector Squared: Solving for c in (a + b)^2 = c with Vectors

  • Thread starter Thread starter MadMax
  • Start date Start date
  • Tags Tags
    Vector
MadMax
Messages
98
Reaction score
0
If I have (a + b)^2 = c, where a and b are vectors, is c = a^2 + 2(a . b) + b^2, or simply is c=a^2 +b^2 + 2ab?

My motivation behind considering the former is that q^2 = q.q, however my motivation behind considering the latter is that if I have q^2=c then it does not matter whether q ia a vector or not, c is the same...

:/
 
Physics news on Phys.org
Yes the a b should be dot product.

|a|*|b|cos(w) w is angle between a and b.

so: c = (a + b) ^2 = [a|^2 + |b|^2 + 2|a|*|b|cos(w)
 
thanks malawi :)
 
Last edited:
np good luck!

ps. this is not a home work question in "Advanced physics" , but in precalclus maths hehe
 
Oh yeah you are right, sorry about that. I didn't realize there was such a section in the forums :) thanks for the heads up.
 
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
Back
Top