Example. Calculate the area of the parallelogram constructed on the vectors $\vec{a}=3 \vec{p}+2 \vec{q}$ and $\vec{b}=2 \vec{p}-\vec{q}$, given that $|\vec{p}|=4,|\vec{q}|=3$ and the angle between the vectors $\vec{p}$ and $\vec{q}$ is $3 \pi / 4$.
## Solution.

1. We calculate $[\vec{a}, \vec{b}]$ using the properties of the vector product

$$
[\vec{a}, \vec{b}]=[3 \vec{p}+2 \vec{q}, 2 \vec{p}-\vec{q}]=6[\vec{p}, \vec{p}]-3[\vec{p}, \vec{q}]+4[\vec{q}, \vec{p}]-2[\vec{q}, \vec{q}]=-7[\vec{p}, \vec{q}]
$$

2. We calculate the magnitude of the vector product

$$
|[\vec{a}, \vec{b}]|=|-7[\vec{p}, \vec{q}]|=7|\vec{p}||\vec{q}| \sin \frac{3 \pi}{4}=42 \sqrt{2}
$$

3. We find the area of the parallelogram using formula (1)

$$
S=|[\vec{a}, \vec{b}]|=42 \sqrt{2}
$$

Answer. The area of the parallelogram is $42 \sqrt{2}$ (units of length) ${ }^{2}$.

Conditions of the Problem. Calculate the area of the parallelogram constructed on vectors $\vec{a}$ and $\vec{b}(\widehat{\vec{p} \vec{q}}-$ the angle between vectors $\vec{p} u \vec{q})$.

1. $\quad \vec{a}=\vec{p}+3 \vec{q}, \quad \vec{b}=2 \vec{p}-\vec{q}, \quad|\vec{p}|=2, \quad|\vec{q}|=1, \quad \widehat{\vec{p} \vec{q}}=\pi / 6$.
2. $\vec{a}=2 \vec{p}+\vec{q}, \quad \vec{b}=\vec{p}-3 \vec{q}, \quad|\vec{p}|=2, \quad|\vec{q}|=2, \quad \widehat{\vec{p} \vec{q}}=\pi / 4$.
3. $\quad \vec{a}=\vec{p}-2 \vec{q}, \quad \vec{b}=\vec{p}+3 \vec{q}, \quad|\vec{p}|=1, \quad|\vec{q}|=2, \quad \widehat{\vec{p} \vec{q}}=\pi / 2$.
4. $\vec{a}=3 \vec{p}-5 \vec{q}, \quad \vec{b}=\vec{p}+2 \vec{q}, \quad|\vec{p}|=2, \quad|\vec{q}|=1, \quad \widehat{\vec{p} \vec{q}}=5 \pi / 6$
5. $\quad \vec{a}=\vec{p}-\vec{q}, \quad \vec{b}=2 \vec{p}+2 \vec{q}, \quad|\vec{p}|=1, \quad|\vec{q}|=6, \quad \widehat{\vec{p} \vec{q}}=3 \pi / 4$.
6. $\quad \vec{a}=\vec{p}+2 \vec{q}, \quad \vec{b}=3 \vec{p}-2 \vec{q}, \quad|\vec{p}|=3, \quad|\vec{q}|=2, \quad \widehat{\vec{p} \vec{q}}=\pi / 3$.
7. $\quad \vec{a}=2 \vec{p}-2 \vec{q}, \quad \vec{b}=\vec{p}+\vec{q}, \quad|\vec{p}|=2, \quad|\vec{q}|=3, \quad \widehat{\vec{p} \vec{q}}=\pi / 2$.
8. $\quad \vec{a}=\vec{p}+\vec{q}, \quad \vec{b}=\vec{p}-4 \vec{q}, \quad|\vec{p}|=7, \quad|\vec{q}|=4, \quad \widehat{\vec{p} \vec{q}}=\pi / 4$.
9. $\quad \vec{a}=4 \vec{p}-4 \vec{q}, \quad \vec{b}=\vec{p}+3 \vec{q}, \quad|\vec{p}|=2, \quad|\vec{q}|=1, \quad \widehat{\vec{p} \vec{q}}=\pi / 6$.
10. $\vec{a}=\vec{p}+\vec{q}, \quad \vec{b}=2 \vec{p}-\vec{q}, \quad|\vec{p}|=2, \quad|\vec{q}|=3, \quad \widehat{\vec{p} \vec{q}}=\pi / 3$.

Answers. 1. $S=7$. 2. $S=14 \sqrt{2}$. 3. $S=10$. 4. $S=11$. $5 . S=15 \sqrt{2}$. 6. $S=24 \sqrt{3}$. 7. $S=24$. 8. $S=70 \sqrt{2}$. 9. $S=16$. 10. $S=9 \sqrt{3}$.

## 1.5. Coplanarity of Vectors

Problem Statement. Are the vectors $\vec{a}=\left\{a_{1}, a_{2}, a_{3}\right\}$, $\vec{b}=\left\{b_{1}, b_{2}, b_{3}\right\}$, and $\vec{c}=\left\{c_{1}, c_{2}, c_{3}\right\}$ coplanar?

Plan of Solution. For three vectors to be coplanar (lie in the same plane or in parallel planes), it is necessary and sufficient that their scalar triple product $(\vec{a}, \vec{b}, \vec{c})$ be equal to zero.

1. The scalar triple product of vectors is expressed through their coordinates by the formula

$$
(\vec{a}, \vec{b}, \vec{c})=\left|\begin{array}{ccc}
a_{1} & a_{2} & a_{3} \\
b_{1} & b_{2} & b_{3} \\
c_{1} & c_{2} & c_{3}
\end{array}\right|
$$

2. If the determinant on the right side of this equation is zero, then the vectors are coplanar; if the determinant is not zero, then the vectors are non-coplanar.