Relation between sum of coplanar vectors and coefficients of each vector

Question

I was unable to proceed solving the problem given below

I don't even know where to begin with this, is there a relation between coefficients of coplanar vectors which add up to give a null vector?

Answer 1

Given points $\vec{a}, \vec{b}, \vec{c} $ in $3D$, a point $\vec{d}$ is coplanar with them if and only if, it can be expressed as follows:

$ \vec{d} = \vec{a} + x_1 (\vec{b} - \vec{a}) + x_2 (\vec{c} - \vec{a}) $

where it is assumed that points $\vec{a}, \vec{b}, \vec{c}$ are not collinear.

Then, we have

$ \vec{d} = (1 - x_1 - x_2) \vec{a} + x_1 \vec{b} + x_2 \vec{c} $

And we want to find the minimum of

$f(x_1, x_2) = (1 - x_1 - x_2)^2 + \left(\dfrac{x_1}{2}\right)^2 + \left(\dfrac{x_2}{3}\right)^2$

Expanding, gives us

$ f(x_1, x_2) = x^T Q x + L^T x + c $

where

$ x = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} $

$Q = \begin{bmatrix} \dfrac{5}{4} && 1 \\ 1 && \dfrac{10}{9} \end{bmatrix} $

$L = \begin{bmatrix} -2 \\ -2 \end{bmatrix} $

$c = 1$

Since $Q$ is positive definite, then $f$ is a convex function, and it has a minimum at

$ x = - \frac{1}{2} Q^{-1} L $

And that minimum value is given by

$f_{min} = c - \dfrac{1}{4} L^T Q^{-1} L $

We have

$Q^{-1} = \dfrac{18}{7} \begin{bmatrix} \dfrac{10}{9} && - 1 \\ -1 && \dfrac{5}{4} \end{bmatrix} $

Therefore,

$f_{min} = 1 - \dfrac{18}{7} ( \dfrac{10}{9} + \dfrac{5}{4} - 2 ) = \dfrac{1}{14} $

This minimum value of the function $f$ is achieved at

$x = -\frac{1}{2} Q^{-1} L = \dfrac{18}{7} \begin{bmatrix} \dfrac{1}{9} \\ \dfrac{1}{4} \end{bmatrix} = \begin{bmatrix} \dfrac{2}{7} \\ \dfrac{9}{14} \end{bmatrix} $

So that,

$ \sin(2 \beta) = \dfrac{1}{2} \left( \dfrac{2}{7} \right) = \dfrac{1}{7} $

$ \sin(3 \gamma) = \dfrac{1}{3} \left( \dfrac{9}{14} \right) = \dfrac{3}{14} $

$ \sin(\alpha) = 1 - x_1 - x_2 = 1 - \dfrac{2}{7} - \dfrac{9}{14} = \dfrac{1}{14} $