Operations with geometric vectors.

Literature: Collection of problems in mathematics. Part 1. Edited by A.V. Efimov, B.P. Demidovich.

The set of all directed line segments with the same length and direction is called a vector, or geometric vector, denoted by $\overline{a}$. Any segment $\overline{AB}$ from this set is said to represent vector $\overline{a}$ (obtained by applying vector $\overline{a}$ to point $A$). The length of segment $\overline{AB}$ is called the length (or magnitude) of the vector, denoted by $|\overline{a}|=|\overline{AB}|$. A vector of zero length is called the zero vector and denoted by $\overline{0}$.

Vectors $\overline{a}$ and $\overline{b}$ are said to be equal ($\overline{a}=\overline{b}$) if the sets representing them coincide.

Let $\overline{AB}$ be a directed line segment representing vector $\overline{a}$. Applying vector $\overline{b}$ to point $B$, we get another directed line segment $\overline{BC}$. The vector represented by directed line segment $\overline{AC}$ is called the sum of vectors $\overline{a}$ and $\overline{b}$, denoted by $\overline{a+b}$.

The product of vector $\overline{a}$ by a real number $\lambda$ is a vector denoted by $\overline{\lambda a}$, such that:

$|\overline{\lambda a}|=|\lambda|\cdot|\overline{a}|;$

vectors $\overline{a}$ and $\overline{\lambda a}$ are collinear for $\lambda>0$ and antiparallel for $\lambda<0$.

Examples:

2.4.

Given vectors $a_1$ and $a_2$. Construct:

a) $3a_1;$

b) $1/2 a_2;$

c) $a_1+2a_2;$

d) $1/2 a_1-a_2.$

Solution.

a) $|\overline{3a_1}|=3|\overline{a_1}|;$ the directions of vectors $\overline{3a_1}$ and $\overline{a_1}$ coincide.

b) $|\overline{0.5a_2}|=0.5|\overline{a_2}|;$ the directions of vectors $\overline{0.5a_2}$ and $\overline{a_2}$ coincide.

c) First, let's construct vector $\overline{2 a_2}$: vector $\overline{2a_2}$ is directed the same as $\overline{a_2}$ and $|\overline{2a_2}|=2|\overline{a_2}|.$

The vector $\overline{a_1+2a_2};$ can be constructed as the diagonal of the parallelogram constructed on vectors $\overline{a_1}$ and $\overline{2a_2}.$

d) First, let's construct vector $\overline{0.5 a_1}:$ $|\overline{0.5a_1}|=0.5|\overline{a_1}|;$ the directions of vectors $\overline{0.5a_1}$ and $\overline{a_1}$ coincide.

The vector $\overline{0.5 a_1-a_2}$ is such a vector that when added to $\overline{a_2} $ gives $\overline{0.5 a_1}.$

2.8.

$\overline{AK}$ and $\overline{BM}$ are medians of triangle $ABC$. Express vectors $\overline{AB}$, $\overline{BC}$, and $\overline{CA}$ in terms of $p=\overline{AK}$ and $q=\overline{BM}.$

Solution.

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It is known that medians intersect at a point dividing each other in a ratio of 2:1 counting from the vertex. Therefore, $|\overline{AO}|=\frac{2}{3}|\overline{AK}|.$ Since the directions of vectors $\overline{AO}$ and $\overline{AK}$ coincide, $\overline{AO}=\frac{2}{3}\overline{AK}.$

Similarly, $\overline{BO}=\frac{2}{3}\overline{BM}.$

From triangle $AOB$ we have $$\overline{AB}=\overline{AO}+\overline{OB}=\overline{AO}-\overline{BO}= \frac{2}{3}(\overline{AK}-\overline{BM})=\frac{2}{3}(p-q).$$

Next, from triangle $ABK$, we find $BK:$

$$\overline{BK}=\overline{BA}+\overline{AK}=-\overline{AB}+\overline{AK}=-\frac{2}{3}(p-q)+p=\frac{1}{3}p+\frac{2}{3}q.$$

$$\overline{BC}=2\overline{BK}=\frac{2}{3}p+\frac{4}{3}q.$$

From triangle $ABC$ we find $$\overline{AC}=\overline{AB}+\overline{BC}=\frac{2}{3}(p-q)+\frac{2}{3}p+\frac{4}{3}q=\frac{4}{3}p+\frac{2}{3}q\Rightarrow\overline{CA}=-\overline{AC}=-\frac{4}{3}p-\frac{2}{3}q.$$

Answer: $\overline{AB}=\frac{2}{3}(p-q); $ $\overline{BC}=\frac{2}{3}p+\frac{4}{3}q;$ $\overline{CA}=-\frac{4}{3}p-\frac{2}{3}q.$

2.10.

In triangle ABC, $\overline{AM}=\alpha\overline{AB}$ and $\overline{CN}=\beta\overline{CM}.$ Assuming $\overline{AB}=a$ and $\overline{AC}=b$, express $\overline{AN}$ and $\overline{BN}$ in terms of vectors $a$ and $b$.

Solution.

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Since $\overline{AM}=\alpha\overline{AB}$, and $\overline{AB}=a$, then $\overline{AM}=\alpha a$.

From triangle $AMC$, we have

$$\overline{CM}=\overline{CA}+\overline{AM}=-\overline{AC}+\overline{AM}=-b+\alpha a.$$

According to the condition $\overline{CN}=\beta\overline{CM}$, hence $\overline{CN}=\beta(-b+\alpha a).$

From triangle $ANC$, we get $$\overline{AN}=\overline{AC}+\overline{CN}=b+\beta(-b+\alpha a)=b(1-\beta)+\alpha\beta a.$$

From triangle $ABN$, we have

$$\overline{BN}=\overline{BA}+\overline{AN}=-\overline{AB}+\overline{AN}=-a+b(1-\beta)+\alpha\beta a=b(1-\beta)+a(\alpha\beta-1).$$

Answer: $\overline{AN}=\alpha\beta a+b(1-\beta); $ $\overline{BN}=a(\alpha\beta-1)+b(1-\beta).$