Vector Problems: Resultant Magnitude and Cross Product Calculations

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Homework Statement



1. Two forces of equal magnitude are acting on a particle along two different directions. Let \theta
be the angle between them. If direction of one vector is reversed , the magnitude of resultant is halved. Find tan\theta.

2.The resultant of the two vectors having magnitudes 5 and 4 is 1. what is the magnitude of their cross product?

Homework Equations





The Attempt at a Solution


for 1st one-
tan\theta = Acos\theta\frac{}{}A + Asin\theta
and tan2\theta = 2Acos\theta\frac{}{}A + Asin\theta
2nd one -
i get cos\theta as -1.
 
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i can't really work out what you have attempted, note that you can put a whole equation in the tex tags.

first if they are parallel, the resultant would go to zero when one is reversed
second, if they are pepindicular, the resultant would not change.

so you know the original angle is some where between 0 & pi/2

i would start by setting up a coordinate axis, where x is parallel to the vector that will be unchanged, then
F_x = F + F cos(\theta)
F_y = F sin(\theta)
 
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...
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