Given $(\sin\alpha)\vec a+(2\sin2\beta)\vec b+(3\sin3\gamma)\vec c-\vec d=0$. Find min. of $\sin^2\alpha+\sin^2\beta+\sin^33\gamma$.

Question

Points $\vec a, \vec b, \vec c$ and $\vec d$ are coplanar and $(\sin\alpha)\vec a+(2\sin2\beta)\vec b+(3\sin3\gamma)\vec c-\vec d=0$. Find the least value of $\sin^2\alpha+\sin^2\beta+\sin^33\gamma$.

If I add the coefficients of the vectors, equate the sum to zero and then apply Cauchy-Schwarz inequality then I am getting the answer.

But I am not sure why exactly I summed the coefficients to zero. I did that because they are coplanar but I am not able to explain further.

I understand if four vectors are coplanar then one vector can be written as a linear combination of other three. So, $$\vec d=l\vec a+m\vec b+n\vec c$$

But why is $l+m+n=1?$

Answer 1

Given four points $A(\vec a),$ $B(\vec b),$ $C(\vec c),$ $D(\vec d)$ are coplanar, WLOG assume $AC$ and $BD$ intersect at $X$. Let $AX:XC=m:n$ and $BX:XD=p:q$. Now the position vector of $X$ is given by $$\vec x=\frac{n\vec a+m\vec c}{m+n}.$$ Also $$\vec x=\frac{q\vec b+p\vec d}{p+q}.$$ Equating both expressions and rearranging, $$\left(\frac n{m+n}\right)\vec a+\left(\frac m{m+n}\right)\vec b-\left(\frac q{p+q}\right)\vec c-\left(\frac p{p+q}\right)\vec d=0.$$ Now check the sum of coefficients and you will get the condition.

Answer 2

Let's draw a vector $\vec x$, from the origin, perpendicular to the plane $P$ formed by $\vec a$, $\vec b$, $\vec c$. Then we can write $$\vec a=\vec x+\vec a_p$$Here $\vec a_p$ is a vector contained in $P$. Similarly to $\vec b$ and $\vec c$. Then a vector contained in the plane $P$ is a linear combination of $\vec a_p$, $\vec b_p$, and $\vec c_p$, so $$\vec d_p=l\vec a_p+m\vec b_p+n\vec c_p$$ Now a linear combination of $\vec a$, $\vec b$, $\vec c$ with the same coefficients can be written as $$\vec d=l(\vec a_p+\vec x)+m(\vec b_p+\vec x)+n(\vec c_p+\vec x)=\vec d_p+(l+m+n)\vec x$$ Now if we want $\vec d$ to be in the same plane, we will write it as a sum of a vector in $P$, in this case $\vec d_p$, and $\vec x$. So the coefficient of $\vec x$ has to be $1$. Therefore $l+m+n=1$.

Answer 3

Since $\vec{d}$ lies in the same plane that passes through $\vec{a}, \vec{b}, \vec{c} $, then it can be written as

$ \vec{d} = \vec{a} + c_1 ( \vec{b} - \vec{a} ) + c_2 (\vec{c}-\vec{a}) $

Re-arranged, this becomes,

$\vec{d} = (1 - c_1 - c_2) \vec{a} + c_1 \vec{b} + c_2 \vec{c} $

Clearly the sum of coefficients of the three vectors is $1$.