Basis of a linear space. Decomposition of a vector by basis.

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 $$a=X_1{e}_1+X_2{e}_2+X_3{e}_3.(1)$$ 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.

1. 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 $\overline{DE}=\overline{DO}+\overline{OE}.(1)$

Next, $\overline{DO}=-\overline{OD}=-\frac{1}{2}\overline{OA};$

we find vector $\overline{OE}$ from triangle $OBE:$

$\overline{OE}=\overline{OB}+\overline{BE}(2),$

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 $(-\frac{1}{2},\frac{1}{2},\frac{1}{2}).$

Answer: $(-\frac{1}{2};\frac{1}{2};\frac{1}{2}).$

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 $\overline{OF}=\overline{OB}+\overline{BF}.(1)$

Next, $\overline{BF}=\frac{2}{3}\overline{BM};$

the vector $\overline{BM}$ can be found from triangle $BMC:$

$\overline{BM}=\overline{BC}+\overline{CM}(2)$

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: $$\overline{OF}=\overline{OB}+\overline{BF}=\overline{OB}+\frac{2}{3}\overline{BM}=$$ $$=\overline{OB}+\frac{2}{3}(\overline{OC}-\overline{OB}+\frac{1}{2}(-\overline{OC}+\overline{OA}))=$$ $$=\overline{OB}+\frac{2}{3}\overline{OC}-\frac{2}{3}\overline{OB}+\frac{1}{3}(-\overline{OC}+\overline{OA})=\frac{1}{3}\overline{OA}+\frac{1}{3}\overline{OB}+\frac{1}{3}\overline{OC}.$$

Thus, the coordinates of the vector $\overline{OF}$ in the basis formed by the edges $\overline{OA},\overline{OB},\overline{OC}$ are $(\frac{1}{3},\frac{1}{3},\frac{1}{3})$.

Answer: $(\frac{1}{3},\frac{1}{3},\frac{1}{3})$.

3. 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 $AOM$:$$\overline{OM}=\overline{OA}+\overline{AM}.(1)$$

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,

$$\overline{AM}=\frac{3}{10}\overline{AL}=\frac{3}{10}(\overline{AB}+\frac{1}{2}\overline{BC})=\frac{3}{10}(-\overline{OA}+\overline{OB}+\frac{1}{2}(\overline{OC}-\overline{OB}))=$$ $=-\frac{3}{10}\overline{OA}+\frac{3}{20}\overline{OB}+\frac{3}{20}\overline{OC}.$

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: $(\frac{1}{1+\lambda },\frac{\lambda }{1+\lambda },0)$.

4. 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 $ABC$: $\overline{CB}=\overline{CA}+\overline{AB}$.

$\overline{CA}$ is found from triangle $ACD:$ $\overline{CA}=\overline{CD}+\overline{DA}=\overline{CD}-\overline{AD}.$

From the condition $|\overline{AB}|/|\overline{CD}|=\lambda$, we find the vector $\overline{CD}$: $\overline{CD}=-\overline{AB}/\lambda$.

Thus, $\overline{CA}=-\overline{AB}/\lambda -\overline{AD}$.

$\overline{CB}=-\overline{AB}/\lambda -\overline{AD}+\overline{AB}=(1-\frac{1}{\lambda })\overline{AB}-\overline{AD}.$

Answer: $(1-\frac{1}{\lambda };-1).$

5. Given the vectors $e(-1,1,1/2)$ and $a(2,-2,-1).$ verify that they are collinear and find the decomposition of vector $a$in the basis $B(e).$

Solution.

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:

$$\{\begin{array}{l}2=-\lambda \\ -2=\lambda \\ -1=\frac{1}{2}\lambda \end{array}\Rightarrow \lambda =-2,$$

Hence, $a=-2e$.

Answer: $a=-2e$.

Homework:

1. 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. 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: $\overline{AP}(\frac{\alpha (1-\gamma )}{1-\alpha \gamma };\frac{\gamma (1-\alpha )}{1-\alpha \gamma });$ $\overline{BQ}(\frac{2\alpha \beta -\alpha -\beta }{1-\alpha \beta };\frac{\beta (1-\alpha )}{1-\alpha \beta });$ $\overline{CQ}(\frac{\beta (1-\gamma )}{1-\beta \gamma };\frac{2\beta \gamma -\beta -\gamma }{1-\beta \gamma }).$

3. 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: $a=-\frac{4}{5}{e}_1-\frac{2}{5}{e}_2.$

4. 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: $a=-2e_1+e_2-e_3.$

Tags: Basis, linear space, vector