Basis of a linear space. Decomposition of a vector by basis.
Literature: Collection of problems in mathematics. Part 1. Edited by A.V. Efimov, B.P. Demidovich.
An ordered triple of non-coplanar vectors $e_1, e_2, e_3$ is called a basis in the space of all geometric vectors. Any geometric vector $a$ can be uniquely represented as $$a=X_1e_1+X_2e_2+X_3e_3.\qquad\qquad\qquad\qquad\qquad (1)$$ The numbers $X_1, X_2, X_3$ are called the coordinates of the vector in the basis $B=\{e_1, e_2, e_3\}.$ The expression (1) is called the decomposition of vector $a$ with respect to basis $B.$
Similarly, an ordered pair of non-collinear vectors $e_1, e_2$ is called a basis $B=(e_1, e_2)$ in the set of geometric vectors coplanar with some plane.
Finally, any non-zero vector $e$ forms a basis $B=(e)$ in the set of geometric vectors collinear with some direction.
If vector $a$ is a linear combination of vectors $a_1, a_2, ..., a_n$ with coefficients $\lambda_1, \lambda_2, ..., \lambda_n$, i.e., $a = \sum_{k=1}^n \lambda_k a_k$, then each coordinate $X_i(a)$ of vector $a$ is equal to the sum of the products of coefficients $\lambda_1, \lambda_2, ..., \lambda_n$ by the corresponding coordinates of vectors $a_1, a_2, ..., a_n$: $X_i(a) = \sum_{k=1}^n \lambda_k X_i(a_k)$, $(i=1, 2, 3)$.
A basis $B=(e_1, e_2, e_3)$ is called rectangular if vectors $e_1, e_2$, and $e_3$ are pairwise perpendicular and have unit length. In this case, the following notation is adopted: $e_1 = i$, $e_2 = -j$, $e_3 = k$.
Examples.
2.26. Given tetrahedron $OABC.$ Find the coordinates in the basis formed by the edges $\overline{OA}, \overline{OB},$ and $\overline{OC}$:
a) Vector $\overline{DE},$ where $D$ and $E$ are the midpoints of edges $OA$ and $BC,$ respectively.
b) Vector $\overline{OF},$ where $F$ is the point of intersection of the medians of the base $ABC.$
Solution.
а)
OABCD1
Expressing the vector $\overline{DE}$ in terms of vectors $\overline{OA}, \overline{OB},$ and $\overline{OC}:$
From triangle $ODE,$ we have $\overline{DE}=\overline{DO}+\overline{OE}.\qquad\qquad\qquad (1)$
Next, $\overline{DO}=-\overline{OD}=-\frac{1}{2}\overline{OA};$
we find vector $\overline{OE}$ from triangle $OBE:$
$\overline{OE}=\overline{OB}+\overline{BE}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad(2),$
where $\overline{BE}=\frac{1}{2}\overline{BC},$ and we find vector $\overline{BC}$ from triangle $OBC:$
$\overline{BC}=\overline{BO}+\overline{OC}=\overline{OC}-\overline{OB}.$
Thus, from (2), we obtain $\overline{OE}=\overline{OB}+\frac{1}{2}(\overline{OC}-\overline{OB}).$
Finally, from (1), we have $$\overline{DE}=\overline{DO}+\overline{OE}=-\frac{1}{2}\overline{OA}+\overline{OB}+\frac{1}{2}(\overline{OC}-\overline{OB})=$$ $$=-\frac{1}{2}\overline{OA}+\frac{1}{2}\overline{OB}+\frac{1}{2}\overline{OC}.$$
Thus, the coordinates of vector $\overline{DE}$ in the basis formed by edges $\overline{OA}, \overline{OB},$ and $\overline{OC}$ are $\left(-\frac{1}{2},\frac{1}{2},\frac{1}{2}\right).$
Answer: $\left(-\frac{1}{2}; \frac{1}{2}; \frac{1}{2}\right).$
б)
OABCD2
Let's express the vector $\overline{OF}$ in terms of vectors $\overline{OA}, \overline{OB},$ and $\overline{OC}:$
From triangle $OFB$, we have $\overline{OF}=\overline{OB}+\overline{BF}.\qquad\qquad\qquad (1)$
Next, $\overline{BF}=\frac{2}{3}\overline{BM};$
the vector $\overline{BM}$ can be found from triangle $BMC:$
$\overline{BM}=\overline{BC}+\overline{CM}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad (2)$
Here, $\overline{CM}=\frac{1}{2}\overline{CA},$ and the vector $\overline{CA}$ can be found from triangle $OCA:$
$\overline{CA}=\overline{CO}+\overline{OA}=-\overline{OC}+\overline{OA};$
$\overline{BC}=\overline{BO}+\overline{OC}=\overline{OC}-\overline{OB}.$
Thus, from (2), we obtain: $$\overline{BM}=\overline{BC}+\overline{CM}=\overline{OC}-\overline{OB}+\frac{1}{2}\overline{CA}=$$ $$=\overline{OC}-\overline{OB}+\frac{1}{2}(-\overline{OC}+\overline{OA}).$$
Finally, from (1), we have: $$\overline{OF}=\overline{OB}+\overline{BF}=\overline{OB}+\frac{2}{3}\overline{BM}=$$ $$=\overline{OB}+\frac{2}{3}\left(\overline{OC}-\overline{OB}+\frac{1}{2}(-\overline{OC}+\overline{OA})\right)=$$ $$=\overline{OB}+\frac{2}{3}\overline{OC}-\frac{2}{3}\overline{OB}+\frac{1}{3}(-\overline{OC}+\overline{OA})=\frac{1}{3}\overline{OA}+\frac{1}{3}\overline{OB}+\frac{1}{3}\overline{OC}.$$
Thus, the coordinates of the vector $\overline{OF}$ in the basis formed by the edges $\overline{OA}, \overline{OB}, \overline{OC}$ are $\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)$.
Answer: $\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)$.
2.27. In the tetrahedron $OABC$, the median $AL$ of the face $ABC$ is divided by the point $M$ in the ratio $|\overline{AM}|:|\overline{ML}|=3:7$. Find the coordinates of the vector $\overline{OM}$ in the basis formed by the edges $\overline{OA}, \overline{OB}, \overline{OC}$.
Solution.
We find the vector $\overline{OM}$ from the triangle $AOM$:$$\overline{OM}=\overline{OA}+\overline{AM}.\qquad\qquad\qquad (1)$$
From the condition $|\overline{AM}|:|\overline{ML}|=3:7$, we have $\overline{AM}=\frac{3}{10}\overline{AL}$. From the triangle $ABL$, we find $\overline{AL}=\overline{AB}+\overline{BL}=\overline{AB}+\frac{1}{2}\overline{BC}$.
Next, from the triangles $AOB$ and $BOC$, we obtain
$\overline{AB}=\overline{AO}+\overline{OB}=-\overline{OA}+\overline{OB}.$
$\overline{BC}=\overline{BO}+\overline{OC}=\overline{OC}-\overline{OB}.$
Therefore,
$$\overline{AM}=\frac{3}{10}\overline{AL}=\frac{3}{10}\left(\overline{AB}+\frac{1}{2}\overline{BC}\right)=\frac{3}{10}\left(-\overline{OA}+\overline{OB}+\frac{1}{2}(\overline{OC}-\overline{OB})\right)=$$ $=-\frac{3}{10}\overline{OA}+\frac{3}{20}\overline{OB}+\frac{3}{20}\overline{OC}.$
Hence, and from (1), we get$$\overline{OM}=\overline{OA}+\overline{AM}=\overline{OA}+\frac{3}{10}\overline{OA}+\frac{3}{20}\overline{OB}+\frac{3}{20}\overline{OC}=$$ $$=\frac{7}{10}\overline{OA}-\frac{3}{20}\overline{OB}+\frac{3}{20}\overline{OC}.$$
Answer: $\left(\frac{1}{1+\lambda}, \frac{\lambda}{1+\lambda}, 0\right)$.
2.29. In trapezoid $ABCD$, the ratio of the lengths of the bases $|\overline{AB}|/|\overline{CD}| = \lambda$. Find the coordinates of vector $\overline{CB}$ in the basis of vectors $\overline{AB}$ and $\overline{AD}$.
Solution.
Vector $\overline{CB}$ can be found from triangle $ABC$: $\overline{CB} = \overline{CA} + \overline{AB}$.
$\overline{CA}$ находим из треугольника $ACD:$ $\overline{CA}=\overline{CD}+\overline{DA}=\overline{CD}-\overline{AD}.$
From the condition $|\overline{AB}|/|\overline{CD}| = \lambda$, we find the vector $\overline{CD}$: $\overline{CD} = -\overline{AB}/\lambda$.
Thus, $\overline{CA} = -\overline{AB}/\lambda - \overline{AD}$.
$\overline{CB}=-\overline{AB}/\lambda-\overline{AD}+\overline{AB}=\left(1-\frac{1}{\lambda}\right)\overline{AB}-\overline{AD}.$
Answer: $\left(1-\frac{1}{\lambda}; -1\right).$
2.36. Заданы векторы $e(-1, 1, 1/2)$ и $a(2, -2, -1).$ Убедиться, что они коллинеарны и найти разложение вектора $a$ по базису $B(e). $
Решение.
The vectors are collinear if their directions coincide or are opposite, i.e., if and only if their coordinates are proportional. Let's check: $$\frac{-1}{2}=\frac{1}{-2}=\frac{1/2}{-1}=-\frac{1}{2},$$ which means that the vectors $e$ and $a$ are collinear.
To find the decomposition of vector $a$ with respect to the basis $B(e)$, i.e., to find the number $\lambda$ such that $a = \lambda e$, we solve the following system of equations:
$$\left\{\begin{array}{lcl}2=-\lambda\\ -2=\lambda\\-1=\frac{1}{2}\lambda\end{array}\right.\Rightarrow \lambda=-2,$$
Hence, $a = -2e$.
Answer: $a = -2e$.
Homework:
2.28. Outside the plane of parallelogram $ABCD$, point $O$ is taken. In the basis of vectors $\overline{OA}, \overline{OB}$, and $\overline{OC}$, find the coordinates of:
a) vector $\overline{OM}$, where $M$ is the point of intersection of the diagonals of the parallelogram;
b) vector $\overline{OK}$, where $K$ is the midpoint of side $AD$.
Answer: a) $(1/2; 0; 1/2);$ b) $(1, -1/2, 1/2).$
2.31. In triangle $ABC$, $\overline{AK}=\alpha\overline{AB}; \overline{BM}=\beta\overline{BC};$ $\overline{CF}=\gamma\overline{CA}.$ Let $P, Q$, and $R$ be the points of intersection of the lines $BF$ and $CK;$ $CK$ and $AM;$ $AM$ and $BF$, respectively. In the basis of vectors $\overline{AB}$ and $\overline{AC}$, find the coordinates of vectors $\overline{AP},$ $\overline{BQ}$, and $\overline{CR}.$
Answer: $\overline{AP}\left(\frac{\alpha(1-\gamma)}{1-\alpha\gamma}; \frac{\gamma(1-\alpha)}{1-\alpha\gamma}\right);$ $\overline{BQ}\left(\frac{2\alpha\beta-\alpha-\beta}{1-\alpha\beta}; \frac{\beta(1-\alpha)}{1-\alpha\beta}\right);$ $\overline{CQ}\left(\frac{\beta(1-\gamma)}{1-\beta\gamma}; \frac{2\beta\gamma-\beta-\gamma}{1-\beta\gamma}\right).$
2.37. On the plane, the vectors are given as $e_1(-1,2),$ $e_2(2,1),$ and $a(0,-2).$ Verify that the basis $B=e_1, e_2$ in the set of all vectors on the plane. Plot the given vectors and find the decomposition of vector $a$ with respect to basis $B.$
Answer: $a=-\frac{4}{5}e_1-\frac{2}{5}e_2.$
2.38. Show that the triplet of vectors $e_1(1,0,0),$ $e_2(1,1,0),$ and $e_3(1,1,1)$ form a basis in the set of all vectors in space. Calculate the coordinates of the vector $a=-2i-k$ in the basis $B(e_1, e_2, e_3)$ and write down the corresponding decomposition of the vector with respect to the basis.
Answer: $a=-2e_1+e_2-e_3.$
