Lines and planes in space

Vectors, grade 12 (5 units) Lesson 1 of 23Learning

Lines and planes in space

Determining a plane

A plane is determined by any one of these:

three points not on one line;
a line and a point outside it;
two intersecting lines;
two parallel lines.

So a three-legged stool never wobbles: the ends of its legs are three points, and only one plane passes through them – the floor.

Two lines

Two lines in space have four relative positions:

intersecting – they have one common point, and they lie in one plane;
parallel – they lie in one plane, and have no common point;
coinciding – they are the same line;
skew – no plane holds both of them: they do not meet, and they are not parallel either.

In the box in the drawing: $\displaystyle AB$ and $\displaystyle DC$ are parallel, $\displaystyle AB$ and $\displaystyle BB'$ intersect, and $\displaystyle AB$ and $\displaystyle CC'$ are skew.

A line and a plane

A line and a plane have three positions:

the line lies in the plane – all its points are in the plane;
the line cuts the plane – they have one common point;
the line is parallel to the plane – they have no common point.

A line parallel to some line in the plane, and not lying in it, is parallel to the plane. In the box: $\displaystyle A'B'$ is parallel to the plane $\displaystyle ABCD$ , since $\displaystyle A'B'\parallel AB$ .

Two planes

Two different planes are parallel (they have no common point) or they intersect along a line – the line of intersection.

In the box: the planes $\displaystyle ABCD$ and $\displaystyle A'B'C'D'$ are parallel, and the planes $\displaystyle ABCD$ and $\displaystyle BCC'B'$ intersect along $\displaystyle BC$ .

A line perpendicular to a plane

A line is perpendicular to a plane when it is perpendicular to every line in the plane through the point where they meet, the foot of the line.

Checking every line in the plane is impossible, so we use a theorem:

a line is perpendicular to a plane if and only if it is perpendicular to two intersecting lines in the plane through its foot.

We prove it with vectors in the lesson on using the scalar product. In the box: $\displaystyle AA'\perp AB$ and $\displaystyle AA'\perp AD$ , so $\displaystyle AA'$ is perpendicular to the plane $\displaystyle ABCD$ – and in particular $\displaystyle AA'\perp AC$ .

Example: the diagonal of a box

In the box $\displaystyle ABCDA'B'C'D'$ : $\displaystyle {AB}={3}$ , $\displaystyle {AD}={4}$ , $\displaystyle {AA'}={12}$ .
Find the length of the diagonal $\displaystyle AC'$ .

The solution below first finds $\displaystyle AC$ in the triangle $\displaystyle ABC$ , and then $\displaystyle AC'$ in the triangle $\displaystyle ACC'$ .

In the right-angled triangle $\displaystyle ABC$ : $\displaystyle {AC^2}={3^2+4^2}={25}$ , $\displaystyle {AC}={5}$ .
$\displaystyle CC'$ is perpendicular to the base’s plane, so $\displaystyle CC'\perp AC$ : the triangle $\displaystyle ACC'$ is right-angled.
$\displaystyle {AC'^2}={AC^2+CC'^2}={25+144}={169}$ , so $\displaystyle {AC'}={13}$ .

In any box with dimensions $\displaystyle a$ , $\displaystyle b$ , $\displaystyle c$ :

\displaystyle d

\displaystyle =

\displaystyle \sqrt{a^2+b^2+c^2}

Exercise 1

What is the relative position of each pair of lines in the box $\displaystyle ABCDA'B'C'D'$ ? Explain.

$\displaystyle AB$ and $\displaystyle C'D'$
$\displaystyle AD$ and $\displaystyle CC'$
$\displaystyle A'C$ and $\displaystyle BD'$
$\displaystyle AC$ and $\displaystyle B'D'$

Exercise 2

What is the relative position of the line and the plane in the same box? Explain.

$\displaystyle B'D'$ and the plane $\displaystyle ABCD$
$\displaystyle AC'$ and the plane $\displaystyle ABCD$
$\displaystyle C'D'$ and the plane $\displaystyle ABC'D'$

Exercise 3

The edge of a cube is 6 long.
Find the length of its diagonal, exactly.

/ – 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

Exercise 4

In the box $\displaystyle ABCDA'B'C'D'$ : $\displaystyle {AB}={2}$ , $\displaystyle {AD}={3}$ , and the diagonal $\displaystyle {AC'}={7}$ .
Find the height $\displaystyle AA'$ .

/ – 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

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