Example. Find the decomposition of the vector $\vec{x}=\{3,-1,2\}$ in terms of the vectors $\vec{p}=\{2,0,1\}, \vec{q}=\{1,-1,1\}$ and $\vec{r}=\{1,-1,-2\}$.
## Solution.

1. The desired decomposition of vector $\vec{x}$ has the form

$$
\vec{x}=\alpha \vec{p}+\beta \vec{q}+\gamma \vec{r}
$$

2. This vector equation with respect to $\alpha, \beta$, and $\gamma$ is equivalent to a system of three linear equations with three unknowns

$$
\left\{\begin{aligned}
2 \alpha+\beta+\gamma & =3 \\
-\beta-\gamma & =-1 \\
\alpha+\beta-2 \gamma & =2
\end{aligned}\right.
$$

3. The system has a unique solution $\alpha=1, \beta=1, \gamma=0$.

Answer. $\vec{x}=\vec{p}+\vec{q}$.

Conditions of the Problem. Write the decomposition of vector $\vec{x}$ in terms of vectors $\vec{p}, \vec{q}$, and $\vec{r}$.

1. $\vec{x}=\{-2,0,9\}, \quad \vec{p}=\{0,-1,2\}, \vec{q}=\{1,0,-1\}, \quad \vec{r}=\{-1,2,4\}$.
2. $\vec{x}=\{5,-12,-1\}, \vec{p}=\{1,-3,0\}, \vec{q}=\{1,-1,1\}, \quad \vec{r}=\{0,-1,2\}$.
3. $\vec{x}=\{0,2,4\}, \quad \vec{p}=\{3,1,-1\}, \vec{q}=\{0,-3,1\}, \quad \vec{r}=\{1,1,1\}$.
4. $\vec{x}=\{-1,5,5\}, \quad \vec{p}=\{2,1,1\}, \quad \vec{q}=\{-2,0,-3\}, \vec{r}=\{-1,2,1\}$.
5. $\vec{x}=\{-1,-2,3\}, \quad \vec{p}=\{2,0,1\}, \quad \vec{q}=\{1,2,-1\}, \quad \vec{r}=\{0,4,-1\}$.
6. $\vec{x}=\{-5,2,-1\}, \vec{p}=\{-1,1,0\}, \vec{q}=\{2,-1,3\}, \vec{r}=\{1,0,1\}$.
7. $\vec{x}=\{1,-5,7\}, \quad \vec{p}=\{0,-1,1\}, \vec{q}=\{2,0,1\}, \quad \vec{r}=\{3,-1,0\}$.
8. $\vec{x}=\{5,1,4\}, \quad \vec{p}=\{2,0,2\}, \quad \vec{q}=\{0,-1,1\}, \quad \vec{r}=\{3,-1,4\}$.
9. $\vec{x}=\{1,1,-1\}, \quad \vec{p}=\{1,1,0\}, \quad \vec{q}=\{-1,0,1\}, \quad \vec{r}=\{-1,0,2\}$.
10. $\vec{x}=\{-3,7,4\}, \quad \vec{p}=\{-2,2,1\}, \quad \vec{q}=\{2,0,1\}, \quad \vec{r}=\{1,1,1\}$.

Answers. 1. $\{2,-1,1\}$. 2. $\{4,1,-1\}$. 3. $\{-1,0,3\}$. 4. $\{-1,-1,3\}$. 5. $\{1,-3,1\}$. 6. $\{3,1,-4\}$. 7. $\{2,5,-3\}$. 8. $\{1,-2,1\}$. 9. $\{1,1,-1\}$. 10. $\{2,-1,3\}$.

## 1.2. Collinearity of Vectors

Problem Statement. Are the vectors $\vec{p}=\lambda_{1} \vec{a}+\lambda_{2} \vec{b}$ and $\vec{q}=\mu_{1} \vec{a}+\mu_{2} \vec{b}$, where $\vec{a}=\left\{a_{1}, a_{2}, a_{3}\right\}$ and $\vec{b}=\left\{b_{1}, b_{2}, b_{3}\right\}$, collinear?

Plan of Solution. Vectors are collinear if and only if there exists a number $\alpha$ such that $\vec{p}=\alpha \vec{q}$. In other words, vectors are collinear if and only if their coordinates are proportional,

1. We find the coordinates of vectors $\vec{p}$ and $\vec{q}$, using the fact that when vectors are added, their coordinates are added, and when multiplied by a number, the coordinates are multiplied by that number.
2. If the coordinates of vectors $\vec{p}=\left\{p_{1}, p_{2}, p_{3}\right\}$ and $\vec{q}=\left\{q_{1}, q_{2}, q_{3}\right\}$ are proportional, i.e.,

$$
\frac{p_{1}}{q_{1}}=\frac{p_{2}}{q_{2}}=\frac{p_{3}}{q_{3}}
$$

then the vectors $\vec{p}$ and $\vec{q}$ are collinear. If the equalities

$$
\frac{p_{1}}{q_{1}}=\frac{p_{2}}{q_{2}}=\frac{p_{3}}{q_{3}}
$$

do not hold, then the vectors $\vec{p}$ and $\vec{q}$ are not collinear.