Linear combinations, linear dependence of vectors. Collinear and coplanar vectors.

Linear combinations, linear dependence of vectors. Collinear and coplanar vectors.

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.

1.

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:

$$\{\begin{array}{l}1=\alpha +\beta \\ 1=\alpha -\beta +2\gamma \\ 1=-2\alpha +3\gamma \end{array}$$

Let's solve this system of equations using Cramer's method:

$$\mathrm{\Delta }=\left|\begin{array}{ccc}1& 1& 0\\ 1& -1& 2\\ -2& 0& 3\end{array}\right|=-3-4-3=-10,$$

$${\mathrm{\Delta }}_1=\left|\begin{array}{ccc}1& 1& 0\\ 1& -1& 2\\ 1& 0& 3\end{array}\right|=-3+2-3=-4,$$

$${\mathrm{\Delta }}_2=\left|\begin{array}{ccc}1& 1& 0\\ 1& 1& 2\\ -2& 1& 3\end{array}\right|=3-4-2-3=-6,$$

$${\mathrm{\Delta }}_3=\left|\begin{array}{ccc}1& 1& 1\\ 1& -1& 1\\ -2& 0& 1\end{array}\right|=-1-2-2-1=-6,$$

$$\alpha =\frac{{\mathrm{\Delta }}_1}{\mathrm{\Delta }}=\frac{-4}{-10}=\frac{2}{5};\beta =\frac{{\mathrm{\Delta }}_2}{\mathrm{\Delta }}=\frac{-6}{-10}=\frac{3}{5};\gamma =\frac{{\mathrm{\Delta }}_3}{\mathrm{\Delta }}=\frac{-6}{-10}=\frac{3}{5}.$$

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.

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:

$$\{\begin{array}{l}\alpha -\gamma =0\\ \alpha +\beta =0\\ \beta +\gamma =0\end{array}$$

This is a degenerate system:

$$\left|\begin{array}{ccc}1& 0& -1\\ 1& 1& 0\\ 0& 1& 1\end{array}\right|=0,$$ 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.

3.

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: $\lambda =5$

4.

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: $3p-4q-3r-2s=0$

Tags: coplanar vectors, linear algebra, vector, linear combinations of vectors, linear dependence of vectors. collinear vectors