Calculating Vector Components with Given Magnitude: Solving for x and y

  • Thread starter Thread starter emma3001
  • Start date Start date
  • Tags Tags
    Magnitude Vector
emma3001
Messages
41
Reaction score
0
Find a vector whose magnitude is 4 and whose x component is twice its y component.

I know that to calculate magnitude it is r= sqrt (x1 + x2)2 + (y1 + y2)2.

I also know that k(vector)= 4

I just am confused by the components (so x=2y ...)
 
Physics news on Phys.org
Let a=xi+yj

now |a|=4
so that 4=\sqrt{x^2+y^2}...(*)

You are told x=2y...can you use this fact and (*) to help you find x and y now?
 
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