Proofs with vectors

Vectors, grade 12 (5 units) Lesson 10 of 23Learning

Proofs with vectors

How to prove with vectors

Theorems about parallel lines, about ratios between segments and about meeting points are proved like this:

choose two independent vectors leaving the same point (in space – three);
express the vectors in the claim with them;
parallel: one vector is a multiple of the other, $\displaystyle {\underline{v}}={t\underline{u}}$ – and the number $\displaystyle t$ is also the ratio of the lengths;
meeting point and ratios: express the same vector in two ways and compare coefficients;
the same point: two points have the same vector from a fixed point.

Proof: the midsegment of a triangle

In the triangle $\displaystyle ABC$ , $\displaystyle M$ is the midpoint of $\displaystyle AB$ and $\displaystyle N$ the midpoint of $\displaystyle AC$ . We prove: $\displaystyle MN\parallel BC$ and $\displaystyle {MN}={\frac{1}{2}BC}$ .

The proof below expresses $\displaystyle \overrightarrow{MN}$ and $\displaystyle \overrightarrow{BC}$ with $\displaystyle \overrightarrow{AB}$ and $\displaystyle \overrightarrow{AC}$ .

$\displaystyle {\overrightarrow{AM}}={\frac{1}{2}\overrightarrow{AB}}$ and $\displaystyle {\overrightarrow{AN}}={\frac{1}{2}\overrightarrow{AC}}$ .
$\displaystyle {\overrightarrow{MN}}={\overrightarrow{AN}-\overrightarrow{AM}}={\frac{1}{2}\left(\overrightarrow{AC}-\overrightarrow{AB}\right)}={\frac{1}{2}\overrightarrow{BC}}$ .
$\displaystyle \overrightarrow{MN}$ is a multiple of $\displaystyle \overrightarrow{BC}$ , so $\displaystyle MN\parallel BC$ and $\displaystyle {MN}={\frac{1}{2}BC}$ .

Proof: diagonals that bisect each other

The diagonals of the quadrilateral $\displaystyle ABCD$ bisect each other at the point $\displaystyle O$ . We prove: $\displaystyle ABCD$ is a parallelogram.

The proof below shows that $\displaystyle {\overrightarrow{AB}}={\overrightarrow{DC}}$ .

$\displaystyle O$ is the midpoint of $\displaystyle AC$ and of $\displaystyle BD$ , so $\displaystyle {\overrightarrow{OC}}={-\overrightarrow{OA}}$ and $\displaystyle {\overrightarrow{OD}}={-\overrightarrow{OB}}$ .
$\displaystyle {\overrightarrow{AB}}={\overrightarrow{OB}-\overrightarrow{OA}}$ .
$\displaystyle {\overrightarrow{DC}}={\overrightarrow{OC}-\overrightarrow{OD}}={-\overrightarrow{OA}+\overrightarrow{OB}}$ .
$\displaystyle {\overrightarrow{AB}}={\overrightarrow{DC}}$ : the sides $\displaystyle AB$ and $\displaystyle DC$ are parallel and equal, so $\displaystyle ABCD$ is a parallelogram.

Exercise 1

In the triangle $\displaystyle ABC$ : $\displaystyle {\overrightarrow{AB}}={\underline{u}}$ , $\displaystyle {\overrightarrow{AC}}={\underline{v}}$ .
$\displaystyle D$ is on $\displaystyle AB$ and $\displaystyle {AD:DB}={1:2}$ ; $\displaystyle E$ is on $\displaystyle AC$ and $\displaystyle {AE:EC}={1:2}$ .

\displaystyle \overrightarrow{DE}

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Prove that $\displaystyle DE\parallel BC$ and $\displaystyle {DE}={\frac{1}{3}BC}$ .

Exercise 2

In the quadrilateral $\displaystyle ABCD$ : $\displaystyle {\overrightarrow{AB}}={\underline{u}}$ , $\displaystyle {\overrightarrow{AD}}={\underline{v}}$ , and $\displaystyle {\overrightarrow{DC}}={\underline{u}}$ .

\displaystyle \overrightarrow{AK}

, where

\displaystyle K

is the midpoint of

\displaystyle AC

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

\displaystyle \overrightarrow{AL}

, where

\displaystyle L

is the midpoint of

\displaystyle BD

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Prove: in a parallelogram the diagonals bisect each other.

Exercise 3

In the trapezoid $\displaystyle ABCD$ ( $\displaystyle AB\parallel DC$ ): $\displaystyle {\overrightarrow{AB}}={\underline{u}}$ , $\displaystyle {\overrightarrow{AD}}={\underline{v}}$ , and $\displaystyle {\overrightarrow{DC}}={\frac{1}{2}\underline{u}}$ .
$\displaystyle M$ is the midpoint of the diagonal $\displaystyle AC$ , and $\displaystyle N$ the midpoint of the diagonal $\displaystyle BD$ .

\displaystyle \overrightarrow{AM}

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

\displaystyle \overrightarrow{AN}

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

\displaystyle \overrightarrow{MN}

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

What can you conclude about $\displaystyle MN$ ? Compare its length to the difference of the bases $\displaystyle AB-DC$ .

Exercise 4

$\displaystyle ABCD$ is a quadrilateral – perhaps not in one plane. $\displaystyle E$ is the midpoint of $\displaystyle AD$ , and $\displaystyle F$ the midpoint of $\displaystyle BC$ .
$\displaystyle {\overrightarrow{AB}}={\underline{u}}$ , $\displaystyle {\overrightarrow{DC}}={\underline{v}}$ .

Prove: $\displaystyle {\overrightarrow{EF}}={\frac{1}{2}(\underline{u}+\underline{v})}$ . (Hint: express $\displaystyle \overrightarrow{EF}$ along a path through $\displaystyle A$ and $\displaystyle B$ , and along a path through $\displaystyle D$ and $\displaystyle C$ , and add.)
What do you get when $\displaystyle ABCD$ is a trapezoid and $\displaystyle AB\parallel DC$ ?

Formulas: vectors, grade 12 (5 units)

Open as a page

Solids in space

The volume of a prism and a cylinder – $\displaystyle S$ the base’s area, $\displaystyle h$ the solid’s height
$\displaystyle V$ $\displaystyle =$ $\displaystyle S\cdot h$
The volume of a pyramid and a cone – $\displaystyle S$ the base’s area, $\displaystyle h$ the solid’s height
$\displaystyle V$ $\displaystyle =$ $\displaystyle \frac{S\cdot h}{3}$
The lateral area of a right cone – $\displaystyle R$ the base’s radius, $\displaystyle \ell$ the slant height
$\displaystyle M$ $\displaystyle =$ $\displaystyle \pi R\ell$
The volume of a sphere – $\displaystyle R$ its radius
$\displaystyle V$ $\displaystyle =$ $\displaystyle \frac{4}{3}\pi R^3$
The surface area of a sphere
$\displaystyle M$ $\displaystyle =$ $\displaystyle 4\pi R^2$
The diagonal of a box of dimensions $\displaystyle a$ , $\displaystyle b$ , $\displaystyle c$
$\displaystyle d$ $\displaystyle =$ $\displaystyle \sqrt{a^2+b^2+c^2}$

Trigonometry in the triangle

The law of sines – $\displaystyle R$ the radius of the circumscribed circle
$\displaystyle \frac{a}{\sin\alpha}=\frac{b}{\sin\beta}=\frac{c}{\sin\gamma}=2R$
The law of cosines
$\displaystyle c^2$ $\displaystyle =$ $\displaystyle a^2+b^2-2ab\cos\gamma$
The area of a triangle – two sides and the angle between them
$\displaystyle S$ $\displaystyle =$ $\displaystyle \frac{1}{2}ab\sin\gamma$

Vectors in the geometric approach

The triangle rule
$\displaystyle \overrightarrow{AB}+\overrightarrow{BC}$ $\displaystyle =$ $\displaystyle \overrightarrow{AC}$
A vector between two points – the end minus the start
$\displaystyle \overrightarrow{AB}$ $\displaystyle =$ $\displaystyle \overrightarrow{OB}-\overrightarrow{OA}$
Collinear (parallel) vectors
$\displaystyle \underline{v}=t\cdot\underline{u}$
The point $\displaystyle M$ divides the segment $\displaystyle AB$ in the ratio $\displaystyle m:n$
$\displaystyle \overrightarrow{OM}=\frac{n}{m+n}\overrightarrow{OA}+\frac{m}{m+n}\overrightarrow{OB}$ $\displaystyle AM:MB=m:n$
$\displaystyle M$ the midpoint of the segment $\displaystyle AB$
$\displaystyle \overrightarrow{OM}=\frac{1}{2}\left(\overrightarrow{OA}+\overrightarrow{OB}\right)$

The scalar product

The length of a vector $\displaystyle {\underline{u}}={(u_1,u_2,u_3)}$
$\displaystyle |\underline{u}|=\sqrt{\underline{u}\cdot\underline{u}}=\sqrt{u_1^2+u_2^2+u_3^2}$
The scalar product in coordinates
$\displaystyle \underline{u}\cdot\underline{v}$ $\displaystyle =$ $\displaystyle u_1v_1+u_2v_2+u_3v_3$
The scalar product – $\displaystyle \alpha$ the angle between the vectors
$\displaystyle \underline{u}\cdot\underline{v}$ $\displaystyle =$ $\displaystyle |\underline{u}|\cdot|\underline{v}|\cdot\cos\alpha$
Perpendicular vectors
$\displaystyle \underline{u}\perp\underline{v}\iff\underline{u}\cdot\underline{v}=0$
A short scalar product
$\displaystyle (\underline{u}+\underline{v})^2$ $\displaystyle =$ $\displaystyle \underline{u}^2+2\underline{u}\cdot\underline{v}+\underline{v}^2$

Points and vectors in axes

A vector between two points
$\displaystyle \overrightarrow{AB}$ $\displaystyle =$ $\displaystyle (x_B-x_A,\ y_B-y_A,\ z_B-z_A)$
The distance between two points
$\displaystyle AB$ $\displaystyle =$ $\displaystyle \sqrt{(x_B-x_A)^2+(y_B-y_A)^2+(z_B-z_A)^2}$
The midpoint of the segment $\displaystyle AB$
$\displaystyle M=\left(\frac{x_A+x_B}{2},\frac{y_A+y_B}{2},\frac{z_A+z_B}{2}\right)$

The line and the plane

A line – $\displaystyle \underline{v}$ a point on the line, $\displaystyle \underline{u}$ the direction vector
$\displaystyle \underline{x}=\underline{v}+t\cdot\underline{u}$
A plane in parametric form – $\displaystyle \underline{u}$ , $\displaystyle \underline{v}$ not parallel
$\displaystyle \underline{x}=\underline{p}+t\cdot\underline{u}+s\cdot\underline{v}$
A plane’s equation – $\displaystyle {\underline{n}}={(a,b,c)}$ the normal vector
$\displaystyle ax+by+cz+d=0$
The plane through $\displaystyle P(x_1,y_1,z_1)$ whose normal vector is $\displaystyle (a,b,c)$
$\displaystyle a(x-x_1)+b(y-y_1)+c(z-z_1)=0$

Distances

The distance of a point $\displaystyle P(x_1,y_1,z_1)$ from the plane $\displaystyle {ax+by+cz+d}={0}$
$\displaystyle d$ $\displaystyle =$ $\displaystyle \frac{|a\cdot x_1+b\cdot y_1+c\cdot z_1+d|}{\sqrt{a^2+b^2+c^2}}$
The distance between the parallel planes $\displaystyle {ax+by+cz+d_1}={0}$ and $\displaystyle {ax+by+cz+d_2}={0}$
$\displaystyle d$ $\displaystyle =$ $\displaystyle \frac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}}$

Angles

The angle between two lines – $\displaystyle \underline{u}$ , $\displaystyle \underline{v}$ the direction vectors
$\displaystyle \cos\alpha$ $\displaystyle =$ $\displaystyle \frac{|\underline{u}\cdot\underline{v}|}{|\underline{u}|\cdot|\underline{v}|}$
The angle between the line $\displaystyle {\underline{x}}={\underline{v}+t\cdot\underline{u}}$ and the plane whose normal is $\displaystyle \underline{n}$
$\displaystyle \sin\beta$ $\displaystyle =$ $\displaystyle \frac{|\underline{n}\cdot\underline{u}|}{|\underline{n}|\cdot|\underline{u}|}$
The angle between two planes – $\displaystyle \underline{n_1}$ , $\displaystyle \underline{n_2}$ the normals
$\displaystyle \cos\alpha$ $\displaystyle =$ $\displaystyle \frac{|\underline{n_1}\cdot\underline{n_2}|}{|\underline{n_1}|\cdot|\underline{n_2}|}$