The parametric form of a line

Vectors, grade 12 (5 units) Lesson 16 of 23Learning

The parametric form of a line

A line: a point plus multiples of a direction

Every point $\displaystyle X$ on the line through $\displaystyle A$ in the direction $\displaystyle \underline{u}$ satisfies $\displaystyle {\overrightarrow{AX}}={t\underline{u}}$ for some number $\displaystyle t$ . So:

\displaystyle \underline{x}

\displaystyle =

\displaystyle \underline{v}+t\cdot\underline{u}

$\displaystyle \underline{v}$ – the position vector of a point on the line; $\displaystyle \underline{u}\neq\underline{0}$ – the direction vector; $\displaystyle t$ – the parameter.
Each number $\displaystyle t$ gives one point: $\displaystyle {t}={0}$ gives $\displaystyle A$ , $\displaystyle {t}={1}$ gives $\displaystyle A+\underline{u}$ , $\displaystyle {t}={-1}$ gives $\displaystyle A-\underline{u}$ .
A line through two points $\displaystyle A$ , $\displaystyle B$ : $\displaystyle {\underline{x}}={A+t(B-A)}$ .
The same line has many forms: any point on it, and any multiple of $\displaystyle \underline{u}$ (other than 0) as a direction.

For example, $\displaystyle {\underline{x}}={(1,2,3)+t(2,0,-1)}$ and $\displaystyle {\underline{x}}={(3,2,2)+s(-4,0,2)}$ are the same line: $\displaystyle (3,2,2)$ comes from $\displaystyle {t}={1}$ , and $\displaystyle {(-4,0,2)}={-2(2,0,-1)}$ . In the plane it is the same, with two coordinates: $\displaystyle {\underline{x}}={(0,1)+t(1,2)}$ is the line $\displaystyle {y}={2x+1}$ .

A point on a line

The point $\displaystyle P$ is on the line if one $\displaystyle t$ gives all three of its coordinates.
Is $\displaystyle P(5,2,1)$ on the line $\displaystyle {\underline{x}}={(1,2,3)+t(2,0,-1)}$ ?

$\displaystyle {(1+2t,\ 2,\ 3-t)}={(5,2,1)}$ : the first coordinate gives $\displaystyle {t}={2}$ , and the third gives $\displaystyle {t}={2}$ ; the second holds for every $\displaystyle t$ .
So $\displaystyle P$ is on the line, for $\displaystyle {t}={2}$ . For the point $\displaystyle (5,2,2)$ the third coordinate gives $\displaystyle {t}={1}$ , and it is not on the line.

The relative position of two lines

For the lines $\displaystyle {\underline{x}}={A+t\underline{u}}$ and $\displaystyle {\underline{x}}={B+s\underline{v}}$ :

the direction vectors are parallel ( $\displaystyle {\underline{v}}={k\underline{u}}$ ): the lines coincide if $\displaystyle B$ is on the first line, and otherwise they are parallel.
the direction vectors are not parallel: solve $\displaystyle {A+t\underline{u}}={B+s\underline{v}}$ $A + t \underline{u} = B + s \underline{v}$ – three equations in two unknowns.
- there is a solution – the lines intersect, and the meeting point comes from $\displaystyle t$ ;
- there is no solution – the lines are skew.

Example: intersecting lines

$\displaystyle {\ell_1:\ \underline{x}}={(1,2,3)+t(1,0,1)}$ , $\displaystyle {\ell_2:\ \underline{x}}={(2,1,0)+s(0,1,4)}$ .
What is their relative position?

$\displaystyle (1,0,1)$ and $\displaystyle (0,1,4)$ are not parallel.
$\displaystyle {(1+t,\ 2,\ 3+t)}={(2,\ 1+s,\ 4s)}$ : $\displaystyle {1+t}={2}$ , $\displaystyle {2}={1+s}$ , $\displaystyle {3+t}={4s}$ .
The first two equations give $\displaystyle {t}={1}$ , $\displaystyle {s}={1}$ , and the third holds: $\displaystyle {4}={4}$ .
The lines intersect at the point $\displaystyle (2,2,4)$ .

Exercise 1

The line through $\displaystyle A(1,-2,3)$ and $\displaystyle B(3,0,2)$ is $\displaystyle {\underline{x}}={A+t(B-A)}$ .
Write it with a worked-out direction vector, and then in another form: through $\displaystyle B$ , with the opposite direction vector.

Exercise 2

Is the point on the line $\displaystyle {\underline{x}}={(1,-2,3)+t(2,2,-1)}$ ? Solve.

\displaystyle P(5,2,1)

\displaystyle Q(3,0,3)

Exercise 3

What is the relative position of the lines? If they intersect, find the meeting point.

$\displaystyle {\underline{x}}={(2,0,1)+t(1,2,-1)}$ , $\displaystyle {\underline{x}}={(1,0,2)+s(1,1,-1)}$

$\displaystyle {\underline{x}}={(1,0,2)+t(1,1,1)}$ , $\displaystyle {\underline{x}}={(0,1,0)+s(1,0,1)}$

Write the conclusion for each part, and in part 1 – the meeting point.

Exercise 4

$\displaystyle {\ell_1:\ \underline{x}}={(1,1,0)+t(2,-4,6)}$ , $\displaystyle {\ell_2:\ \underline{x}}={(3,-3,6)+s(-1,2,-3)}$ , $\displaystyle {\ell_3:\ \underline{x}}={s(1,-2,3)}$ .

\displaystyle (3,-3,6)

\displaystyle \ell_1

\displaystyle (0,0,0)

\displaystyle \ell_1

What is the relative position of $\displaystyle \ell_1$ and $\displaystyle \ell_2$ , and of $\displaystyle \ell_1$ and $\displaystyle \ell_3$ ? Explain.

Formulas: vectors, grade 12 (5 units)

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