Finding Vector v Given a, b, and c

  • Thread starter Thread starter duoshikunli
  • Start date Start date
  • Tags Tags
    Vector
duoshikunli
Messages
4
Reaction score
0

Homework Statement



The unknown vector v satisfies b . v = a and b x v = c, where a, b, and c are fixed and known. Find v in terms of a, b, and c.

Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
And what have you tried? Personally, I would start with finding the dot and cross products there...
 
I understand dot and cross product. I've played with it a lot, mostly with cross product, but I'm not getting anywhere. I have a feeling that I'm suppose to manipulate it using some other trick. Can someone guide me in the right way?
 
Have you tried

b \cdot v = |b||v|cos \theta
| b \times v | = |b||v|sin \theta
 
(|b||v|)^2 = a^2 + c^2
v . v = |v|^2 = (a^2 + c^2) / |b|^2

And I'm stuck...
 
duoshikunli said:
(|b||v|)^2 = a^2 + c^2
v . v = |v|^2 = (a^2 + c^2) / |b|^2

And I'm stuck...
Check your work c^2 means nothing.
I assume you mean |c|^2.

A vector is uniquely determined by it's magnitude and direction.

You have |v| all you need now is a unit vector in v's direction.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top