Linear dependence and the uniqueness of a representation

Vectors, grade 12 (5 units) Lesson 9 of 23Learning

Linear dependence and the uniqueness of a representation

Collinear vectors

Two vectors $\displaystyle \underline{u}\neq\underline{0}$ and $\displaystyle \underline{v}$ are parallel (collinear) if and only if $\displaystyle {\underline{v}}={t\cdot\underline{u}}$ for some number $\displaystyle t$ .

So three points $\displaystyle A$ , $\displaystyle B$ , $\displaystyle C$ lie on one line if and only if $\displaystyle {\overrightarrow{AC}}={t\cdot\overrightarrow{AB}}$ : the two vectors are parallel, and they share the point $\displaystyle A$ .

Linear dependence

A vector is linearly dependent on a set of vectors if it can be expressed as a linear combination of them.
A set of vectors is linearly independent if no vector in it is a linear combination of the others.

What this means in a drawing:

two dependent vectors are collinear.
three dependent vectors lie in one plane (or in parallel planes).
any three vectors in a plane are dependent, and any four vectors in space are dependent.

The uniqueness of a representation

In the plane: if $\displaystyle \underline{u}$ and $\displaystyle \underline{v}$ are independent, every vector in the plane is a linear combination $\displaystyle s\underline{u}+t\underline{v}$ in one way only.
In space: if $\displaystyle \underline{u}$ , $\displaystyle \underline{v}$ , $\displaystyle \underline{w}$ are independent, every vector is $\displaystyle s\underline{u}+t\underline{v}+r\underline{w}$ in one way only.

The reason, in the plane: if $\displaystyle {s\underline{u}+t\underline{v}}={s'\underline{u}+t'\underline{v}}$ and $\displaystyle s\neq s'$ , then $\displaystyle {\underline{u}}={\frac{t'-t}{s-s'}\underline{v}}$ – that is, $\displaystyle \underline{u}$ and $\displaystyle \underline{v}$ are dependent.

Comparing coefficients: express the same vector in two ways with independent vectors, and set the coefficients of each vector equal. This is how ratios between segments are found.

Example: a meeting point

$\displaystyle ABCD$ is a parallelogram, $\displaystyle {\overrightarrow{AB}}={\underline{u}}$ , $\displaystyle {\overrightarrow{AD}}={\underline{v}}$ .
$\displaystyle E$ is the midpoint of $\displaystyle BC$ . The segments $\displaystyle AE$ and $\displaystyle BD$ meet at the point $\displaystyle K$ .
In what ratio does $\displaystyle K$ divide $\displaystyle AE$ , and $\displaystyle BD$ ?

The solution below expresses $\displaystyle \overrightarrow{AK}$ in two ways, and compares coefficients.

$\displaystyle K$ is on $\displaystyle AE$ : $\displaystyle {\overrightarrow{AK}}={t\cdot\overrightarrow{AE}}={t\left(\underline{u}+\frac{1}{2}\underline{v}\right)}={t\underline{u}+\frac{t}{2}\underline{v}}$ .
$\displaystyle K$ is on $\displaystyle BD$ : $\displaystyle {\overrightarrow{AK}}={\overrightarrow{AB}+s\cdot\overrightarrow{BD}}={\underline{u}+s(\underline{v}-\underline{u})}={(1-s)\underline{u}+s\underline{v}}$ .
$\displaystyle \underline{u}$ and $\displaystyle \underline{v}$ are independent, so the coefficients are equal: $\displaystyle {t}={1-s}$ and $\displaystyle {\frac{t}{2}}={s}$ .
Solving the system: $\displaystyle {t}={\frac{2}{3}}$ , $\displaystyle {s}={\frac{1}{3}}$ . So $\displaystyle {AK:KE}={2:1}$ and $\displaystyle {BK:KD}={1:2}$ .

Example: three points on a line

$\displaystyle ABCD$ is a parallelogram, $\displaystyle {\overrightarrow{AB}}={\underline{u}}$ , $\displaystyle {\overrightarrow{AD}}={\underline{v}}$ .
$\displaystyle E$ is on $\displaystyle AB$ and $\displaystyle {AE}={\frac{1}{3}AB}$ ; $\displaystyle F$ is on $\displaystyle AC$ and $\displaystyle {AF}={\frac{1}{4}AC}$ .
Prove that $\displaystyle D$ , $\displaystyle F$ , $\displaystyle E$ lie on one line.

$\displaystyle {\overrightarrow{DE}}={\overrightarrow{AE}-\overrightarrow{AD}}={\frac{1}{3}\underline{u}-\underline{v}}$ .
$\displaystyle {\overrightarrow{DF}}={\overrightarrow{AF}-\overrightarrow{AD}}={\frac{1}{4}(\underline{u}+\underline{v})-\underline{v}}={\frac{1}{4}\underline{u}-\frac{3}{4}\underline{v}}$ .
$\displaystyle {\overrightarrow{DF}}={\frac{3}{4}\left(\frac{1}{3}\underline{u}-\underline{v}\right)}={\frac{3}{4}\overrightarrow{DE}}$ : the vectors are collinear and both leave $\displaystyle D$ , so $\displaystyle D$ , $\displaystyle F$ , $\displaystyle E$ lie on one line, and $\displaystyle {DF:FE}={3:1}$ .

Exercise 1

In the triangle $\displaystyle ABC$ : $\displaystyle {\overrightarrow{AB}}={\underline{u}}$ , $\displaystyle {\overrightarrow{AC}}={\underline{v}}$ .
$\displaystyle M$ is the midpoint of $\displaystyle BC$ and $\displaystyle N$ the midpoint of $\displaystyle AC$ . The medians $\displaystyle AM$ and $\displaystyle BN$ meet at the point $\displaystyle G$ .

\displaystyle {\overrightarrow{AG}}={t\cdot\overrightarrow{AM}}

– with

\displaystyle t

\displaystyle \underline{u}

\displaystyle \underline{v}

/ – 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

\displaystyle {\overrightarrow{AG}}={\overrightarrow{AB}+s\cdot\overrightarrow{BN}}

– with

\displaystyle s

\displaystyle \underline{u}

\displaystyle \underline{v}

/ – 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

Compare coefficients, and solve the system.

In what ratio does $\displaystyle G$ divide the median $\displaystyle AM$ ?

Exercise 2

$\displaystyle {\overrightarrow{OA}}={\underline{u}}$ , $\displaystyle {\overrightarrow{OB}}={\underline{v}}$ , and $\displaystyle {\overrightarrow{OC}}={3\underline{u}-2\underline{v}}$ , where $\displaystyle \underline{u}$ and $\displaystyle \underline{v}$ are independent.

\displaystyle \overrightarrow{AB}

/ – 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{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

Do $\displaystyle A$ , $\displaystyle B$ , $\displaystyle C$ lie on one line? Explain.

Exercise 3

$\displaystyle \underline{u}$ and $\displaystyle \underline{v}$ are independent, and $\displaystyle {(s+1)\underline{u}+(2t-3)\underline{v}}={4\underline{u}-\underline{v}}$ .
Find $\displaystyle s$ and $\displaystyle t$ .

Exercise 4

In the tetrahedron $\displaystyle ABCD$ : $\displaystyle {\overrightarrow{AB}}={\underline{b}}$ , $\displaystyle {\overrightarrow{AC}}={\underline{c}}$ , $\displaystyle {\overrightarrow{AD}}={\underline{d}}$ .
$\displaystyle K$ , $\displaystyle L$ , $\displaystyle P$ , $\displaystyle Q$ , $\displaystyle R$ , $\displaystyle T$ are the midpoints of the edges $\displaystyle AB$ , $\displaystyle CD$ , $\displaystyle AC$ , $\displaystyle BD$ , $\displaystyle AD$ , $\displaystyle BC$ respectively.
Each of the segments $\displaystyle KL$ , $\displaystyle PQ$ , $\displaystyle RT$ joins the midpoints of two opposite edges.

\displaystyle \overrightarrow{AX}

, where

\displaystyle X

is the midpoint of

\displaystyle KL

/ – 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{AY}

, where

\displaystyle Y

is the midpoint of

\displaystyle PQ

/ – 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{AZ}

, where

\displaystyle Z

is the midpoint of

\displaystyle RT

/ – 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 the three segments meet at one point, which bisects each of them.

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
$\displaystyle \cos\alpha$ $\displaystyle =$ $\displaystyle \frac{|\underline{n_1}\cdot\underline{n_2}|}{|\underline{n_1}|\cdot|\underline{n_2}|}$