Calculate Vector V with Given Magnitude and Direction | Vector Calc Questions

  • Thread starter Thread starter r34racer01
  • Start date Start date
  • Tags Tags
    Vector
r34racer01
Messages
62
Reaction score
0
1.) find the vector v w/ the given magnitude and the same direction as u.
//v// = 2 u = < sq. rt. 3, 3>

2.) find the component form of v given its magnitude and the angle it makes w/ the positive x-axis.
//v// = 2 theta = 150 degrees

3.) find the component form of u + v given the lengths of u and v and the angles that u and v make w/ the positive x-axis
//u// = 2 theta u = 4 //v// = 1 theta v = 2

I already know the answers are <1,sqr rt. 3> < -sqr. rt. 3, 1> and <2 cos 4 + cos 2, 2 sin 4 + sin 2>, I 'd just like an explanation of how to get those answers.
 
Physics news on Phys.org
It's immaterial whether you know the answers or not; you need to show that you have tried something before you'll get any help.
 
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