Linear combinations, linear dependence of vectors. Collinear and coplanar vectors.
Linear combinations, linear dependence of vectors. Collinear and coplanar vectors.
Literature: Collection of problems in mathematics. Part 1. Ed. A. V. Efimov, B. P. Demidovich.
A system of vectors $a_1, a_2, ..., a_n$ is called linearly dependent if there exist numbers $\lambda_1, \lambda_2, ..., \lambda_n$ such that at least one of them is different from zero and $\lambda_1 a_1+\lambda_2 a_2+...+\lambda_n a_n=0.$ Otherwise, the system is called linearly independent.
Two vectors $a_1$ and $a_2$ are called collinear if their directions coincide or are opposite.
Three vectors $a_1, a_2,$ and $a_3$ are called coplanar if they are parallel to some plane.
Geometric criteria of linear dependence:
a) The system ${a_1,, a_2}$ is linearly dependent if and only if the vectors $a_1$ and $a_2$ are collinear.
b) The system ${a_1,, a_2,, a_3}$ is linearly dependent if and only if the vectors $a_1,, a_2,$ and $a_3$ are coplanar.
Examples.
2.19.
Decompose the vector $s=a+b+c$ into three non-coplanar vectors: $p=a+b-2c,$ $q=a-b,$ $r=2b+3c.$
Solution.
Let's find such $\alpha, \beta,$ and $\gamma,$ that $s=\alpha p+\beta q+\gamma r:$
$$s=a+b+c=\alpha(a+b-2c)+\beta(a-b)+\gamma(2b+3c)=$$
$$=a(\alpha+\beta)+b(\alpha-\beta+2\gamma)+c(-2\alpha+3\gamma).$$
From this equality, equating coefficients of $a, b,$ and $c$, we obtain the system of equations:
$$\left\{\begin{array}{lcl}1=\alpha+\beta\\ 1=\alpha-\beta+2\gamma\\ 1=-2\alpha+3\gamma\end{array}\right.$$
Let's solve this system of equations using Cramer's method:
$$\Delta=\begin{vmatrix}1&1&0\\1&-1&2\\-2&0&3\end{vmatrix}=-3-4-3=-10, $$
$$\Delta_1=\begin{vmatrix}1&1&0\\1&-1&2\\1&0&3\end{vmatrix}=-3+2-3=-4, $$
$$\Delta_2=\begin{vmatrix}1&1&0\\1&1&2\\-2&1&3\end{vmatrix}=3-4-2-3=-6, $$
$$\Delta_3=\begin{vmatrix}1&1&1\\1&-1&1\\-2&0&1\end{vmatrix}=-1-2-2-1=-6, $$
$$\alpha=\frac{\Delta_1}{\Delta}=\frac{-4}{-10}=\frac{2}{5};\quad\beta=\frac{\Delta_2}{\Delta}=\frac{-6}{-10}=\frac{3}{5};\quad\gamma=\frac{\Delta_3}{\Delta}=\frac{-6}{-10}=\frac{3}{5}.$$
Thus, $s=\frac{2}{5} p+\frac{3}{5} q+\frac{3}{5}r.$
Answer: $s=\frac{2}{5} p+\frac{3}{5} q+\frac{3}{5}r.$
2.22.
Prove that for any given vectors $a,, b,$ and $c$, the vectors $a+b,,, b+c,,, c-a$ are coplanar.
Proof.
The sets of vectors ${a, b, a+b}; ,,{b, c, b+c},,, {a, c, c-a}$ are coplanar since they are linearly dependent: $a+b-(a+b)=0; ,,$ $b+c-(b+c)=0;,,$ $ -c+a+(c-a)=0.$ Hence, if the vectors $a,, b,$ and $c$ are coplanar, then the vectors $a+b,,, b+c,,, c-a$ are also coplanar.
Let the vectors $a, b,$ and $c$ not be coplanar.
Since vectors $a_1,, a_2,$ and $a_3$ are coplanar if and only if the system ${a_1,, a_2,, a_3}$ is linearly dependent, we need to show that the set of vectors $a+b,,, b+c,,, c-a$ is linearly dependent. To do this, we will show that there exist numbers $\alpha,, \beta,$ and $\gamma$ such that at least one of them is non-zero and $\alpha(a+b)+\beta(b+c)+\gamma(c-a)=0.$
Suppose the opposite: the vectors $a+b,,, b+c,,, c-a$ are non-coplanar. Then the equality $\alpha(a+b)+\beta(b+c)+\gamma(c-a)=0$ holds true only if $\alpha=\beta=\gamma=0.$
Rewriting the last equation as $a(\alpha-\gamma)+b(\alpha+\beta)+c(\beta+\gamma)=0,$ since we are considering the case where vectors $a, b,$ and $c$ are non-coplanar, the following equations must hold:
$$\left\{\begin{array}{lcl}\alpha-\gamma=0\\ \alpha+\beta=0\\ \beta+\gamma=0\end{array}\right.$$
This is a degenerate system:
$$\begin{vmatrix}1&0&-1\\1&1&0\\0&1&1\end{vmatrix}=0,$$ therefore, this system has a non-trivial solution, for example $\alpha=\gamma=1;,,\beta=-1.$ This leads to a contradiction.
Thus, for any given vectors $a,, b,$ and $c$, the vectors $a+b,,, b+c,,, c-a$ are coplanar. This completes the proof.
Homework.
№2.18
On the side $AD$ of parallelogram $ABCD$, the vector $\overline{AK}$ is drawn with length $|\overline{AK}|=1/5|\overline{AD}|$, and on the diagonal $AC$ the vector $\overline{AL}$ is drawn with length $|\overline{AL}|=1/6|\overline{AC}|$. Prove that the vectors $\overline{KL}$ and $\overline{LB}$ are collinear and find $\lambda$ such that $\overline{KL}=\lambda\overline{LB}$.
Answer: $\lambda=5$
№2.20
Find the linear dependency among the given four non-coplanar vectors: $p=a+b,, q=b-c,,r=a-b+c,, s=b+1/2c.$
Answer: $3p-4q-3r-2s=0$
