Vrftsjtryk

If line makes an angle of $alpha,beta,gamma$ with positive $x,y$ and $z$

axis respectively. Then Range of

$sin alphacdot sin beta+sin beta cdot sin gamma+sin gamma cdot sin alpha$ is

Try: If line makes an angle of $alpha,beta,gamma$ with positive $x,y$ and

$z$ axis respectively. Then $cos^2 alpha+cos^2 beta+cos^2 gamma = 1$

means $sin^2 alpha+sin^2 beta+sin^2 gamma = 2.$

Using Cauchy Schwarz Inequality

$$(sin^2 alpha+sin^2 beta+sin^2 gamma )cdot (sin^2 beta+sin^2 gamma+sin^2 alpha)geq (sin alphacdot sin beta+sin beta cdot sin gamma+sin gamma cdot sin alpha)^2$$

So $$sin alphacdot sin beta+sin beta cdot sin gamma+sin gamma cdot sin alphaleq 2$$

Now i did not understand how i find minimum value of $$sin alphacdot sin beta+sin beta cdot sin gamma+sin gamma cdot sin alpha$$

could some help me, Thanks

Let $O(0,0,0)$, $A(sqrt{a},0,0),$ $B(0,sqrt{b},0)$ and $C(0,0,sqrt{c})$, where $a$, $b$ and $c$ are non-negatives such that $a^2+b^2+c^2neq0.$

Thus, we can assume that
$$sinalpha=sqrt{frac{b+c}{a+b+c}},$$
$$sinbeta=sqrt{frac{a+c}{a+b+c}}$$ and
$$singamma=sqrt{frac{a+b}{a+b+c}}.$$
Now, for $b=c=0$ we obtain:
$$sum_{cyc}sinalphasingamma=1.$$
We'll prove that it's a minimal value.

Indeed, we need to prove that
$$sum_{cyc}sqrt{frac{(a+b)(a+c)}{(a+b+c)^2}}geq1$$ or
$$sum_{cyc}sqrt{(a+b)(a+c)}geq a+b+c,$$ which is obvious because
$$sqrt{(a+b)(a+c)}geqsqrt{acdot a}=a.$$

For the upper limit of the range, you've proved that $2$ is an upper bound on the range, but not that $2$ actually equals the upper bound. To show that $2$ is the upper limit of the range, you must exhibit an example (or otherwise prove) that $2$ can be attained. In this case, the upper bound can be attained with the line in the direction $langle 1,1,1rangle$. In this case, the angle with all axes is the same, so, by the equality condition for Cauchy-Schwarz, you get the value of $2$ (or, alternately, use the cross product to calculate that the sine of the angle is $frac{sqrt{2}}{sqrt{3}}$ and calculate the desired value directly).

For the lower bound, this occurs when the line is in the direction of one of the axes, e.g., $langle 1,0,0rangle$. In this case, one of the sines is zero and the other two are $1$. This leads to a value of $1$. This can be proved with a bit of multivariate calculus, if you wish, or a little figuring.

Here's the sketch of how to argue that $1$ is the minimum value (without calculus): Observe that
begin{align}
(sinalpha+sinbeta+singamma)^2&=(sin^2alpha+sin^2beta+sin^2gamma)+2(sinalphasinbeta+sinalphasingamma+sinbetasingamma)\
&=2+2(sinalphasinbeta+sinalphasingamma+sinbetasingamma).
end{align}
Therefore, since everything is nonnegative, minimizing the desired quantity is the same as minimizing $sinalpha+sinbeta+singamma$. Since each sine is between $0$ and $1$, we know that $sin^2alphaleqsinalpha$, $sin^2betaleqsinbeta$, and $sin^2gammaleqsingamma$. Therefore,
$$
2=sin^2alpha+sin^2beta+sin^2gammaleq sinalpha+sinbeta+singamma.
$$
The minimum is attained for the example above.