Points, Straight Lines and Shapes in 3D (Senior 5 Mathematics): free lessons | Ganzaa

Points, Straight Lines and Shapes in 3D

Lessons

Senior 5 · Mathematics

Points, Straight Lines and Shapes in 3D

5 lessons

Lesson 1 of 5

Points in space: distance, dividing points and centroids

35 min

Corners of a roof

An architect's plan of a house shows the floor, but the builder also needs heights: where the ridge of the roof is, how long each sloping rafter must be, where a beam should meet a wall. Every corner of the frame is a point with three coordinates, and every timber is a segment between two of them. In unit 4 you found distances, midpoints and lines in the plane. This unit carries all of it into space, using the vectors of unit 7.

Important

Distance between two points. For $A (x_{1}, y_{1}, z_{1})$ and $B (x_{2}, y_{2}, z_{2})$ ,

AB = |\overrightarrow{AB}| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} .

Reason. In the figure, $C$ is directly below $B$ and level with $A$ . In the horizontal right triangle with legs $\Delta x$ and $\Delta y$ , $AC^2 = \Delta x^2 + \Delta y^2$ . The triangle $A C B$ has a right angle at $C$ , so $AB^2 = AC^2 + \Delta z^2 = \Delta x^2 + \Delta y^2 + \Delta z^2$ : Pythagoras' theorem twice.

Definition

A point $P$ divides the segment $A B$ internally in the ratio $m : n$ ( $m, n > 0$ ) if $P$ lies on $A B$ and $A P : P B = m : n$ . It divides $A B$ externally in the ratio $m : n$ ( $m \ne n$ ) if $P$ lies on the line $A B$ , outside the segment, and $A P : P B = m : n$ .

Important

Ratio (section) formula. With position vectors $a, b$ :

\text{internal: } p = \frac{n\,a + m\,b}{m + n}, \qquad \text{external: } p = \frac{-n\,a + m\,b}{m - n} .

Taking $m = n = 1$ internally gives the midpoint $M\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right)$ .

Reason (internal). $P$ is $\frac{m}{m + n}$ of the way from $A$ to $B$ : $p = a + \frac{m}{m + n}(b - a) = \frac{(m + n)a - ma + mb}{m + n} = \frac{na + mb}{m + n}$ .

Centroids. The centroid (meeting point of the medians) of triangle $A B C$ is $g = \frac{a + b + c}{3}$ ; the centroid of the tetrahedron $A B C D$ is $g = \frac{a + b + c + d}{4}$ .

Worked example 1: distance and midpoint

For $A (1, - 2, 3)$ and $B (5, 2, 5)$ , find $A B$ and the midpoint of $A B$ .

Solution. $AB = \sqrt{4^2 + 4^2 + 2^2} = \sqrt{36} = 6$ . The midpoint is $\left(\frac{1 + 5}{2}, \frac{-2 + 2}{2}, \frac{3 + 5}{2}\right) = (3, 0, 4)$ .

Worked example 2: a right-angled triangle in space

Show that $A (2, - 1, 1)$ , $B (1, - 3, - 5)$ , $C (3, - 4, - 4)$ form a right-angled triangle, and say where the right angle is.

Solution.

AB^2 = 1 + 4 + 36 = 41, \qquad BC^2 = 4 + 1 + 1 = 6, \qquad AC^2 = 1 + 9 + 25 = 35 .

Since $B C^{2} + A C^{2} = 6 + 35 = 41 = A B^{2}$ , the converse of Pythagoras' theorem gives a right angle at $C$ , opposite the longest side $A B$ .

Worked example 3: internal and external division

Find the points that divide $A (2, - 1, 4)$ and $B (8, 5, - 2)$ internally and externally in the ratio $1 : 2$ .

Solution. Internally, with $m = 1$ , $n = 2$ : $p = \frac{2a + b}{3} = \frac{(4 + 8,\ -2 + 5,\ 8 - 2)}{3} = (4, 1, 2)$ .

Externally: $q = \frac{-2a + b}{1 - 2} = 2a - b = (4 - 8,\ -2 - 5,\ 8 + 2) = (-4, -7, 10)$ . Check: $\overrightarrow{AQ} = (-6, -6, 6)$ and $\overrightarrow{QB} = (12, 12, -12)$ , so $A Q : Q B = 1 : 2$ with $Q$ beyond $A$ . ✓

Worked example 4: where a coordinate plane cuts a segment

In what ratio does the $x y$ -plane divide the segment joining $A (1, 2, - 3)$ and $B (4, - 1, 6)$ ? Find the point of division.

Solution. Let the ratio be $k : 1$ . The point is $\frac{a + kb}{k + 1}$ , and it lies in the $x y$ -plane when its $z$ -coordinate is 0: $\frac{-3 + 6k}{k + 1} = 0$ , so $k = \frac{1}{2}$ . The ratio is $\frac{1}{2} : 1 = 1 : 2$ , internal because $k > 0$ , and the point is $\frac{2a + b}{3} = \left(\frac{6}{3}, \frac{3}{3}, 0\right) = (2, 1, 0)$ .

Warning

In $\frac{na + mb}{m + n}$ the weights cross over: $a$ is multiplied by $n$ , the part of the ratio next to $B$ . A point with $A P : P B = 1 : 2$ is nearer $A$ , so $A$ gets the bigger weight.
Square every difference, including negative ones: $(- 3 - 1)^{2} = 16$ , not $- 16$ .
For the centroid of a triangle divide by 3, for a tetrahedron by 4: one for each vertex.

Tip

A negative ratio $k$ in worked example 4 would mean external division: the plane meets the line $A B$ outside the segment.

Mario's Math Tutoring: Distance and Midpoint Formula in 3 Dimensions (about 2 min). A quick worked example of both formulas.

✦

Practice

Questions 1 to 3 are routine; 4 to 6 need more care; 7 asks for a proof.

Find the distance between $(3, - 1, 2)$ and $(- 1, 2, 14)$ .
Find $k$ if the distance from $(1, k, 2)$ to $(3, 1, 0)$ is 3.
Show that $O (0, 0, 0)$ , $A (1, 2, 2)$ , $B (2, 1, - 2)$ form a right-angled isosceles triangle.
Find the points that divide $A (- 1, 3, 2)$ and $B (5, 0, 8)$ internally and externally in the ratio $2 : 1$ .
In what ratio does the $y z$ -plane divide the segment from $A (- 2, 4, 7)$ to $B (3, - 5, 8)$ ? Find the point of division.
The centroid of triangle $A B C$ is $G (1, 2, 3)$ , with $A (0, 1, 2)$ and $B (2, 4, 1)$ . Find $C$ . Then find the centroid of the tetrahedron $A (1, 2, 3)$ , $B (3, 0, 1)$ , $C (2, 4, - 1)$ , $D (2, 2, 5)$ .
Prove that the midpoints of the four sides of any quadrilateral $A B C D$ in space are the vertices of a parallelogram.

Show answersHide answers

$\sqrt{16 + 9 + 144} = 13$ .
$4 + (k - 1)^{2} + 4 = 9$ , so $(k - 1)^{2} = 1$ : $k = 0$ or $k = 2$ .
$O A = O B = 3$ and $AB = \sqrt{1 + 1 + 16} = \sqrt{18}$ ; $O A^{2} + O B^{2} = 18 = A B^{2}$ , so the right angle is at $O$ .
Internal: $\frac{a + 2b}{3} = (3, 1, 6)$ . External: $\frac{-a + 2b}{2 - 1} = (11, -3, 14)$ .
$x = 0$ gives $\frac{-2 + 3k}{k + 1} = 0$ , $k = \frac{2}{3}$ : ratio $2 : 3$ internally, at $\left(0, \frac{2}{5}, \frac{37}{5}\right)$ .
$c = 3 g - a - b = (1, 1, 6)$ . The centroid of the tetrahedron is $(2, 2, 2)$ .
The midpoints of $A B$ and $B C$ give $\overrightarrow{PQ} = \frac{b + c}{2} - \frac{a + b}{2} = \frac{c - a}{2}$ ; the midpoints of $D C$ and $A D$ give $\overrightarrow{SR} = \frac{c + d}{2} - \frac{a + d}{2} = \frac{c - a}{2}$ . Equal vectors on opposite sides make a parallelogram.

Sources and credits

Written by Ganzaa: the explanations, the diagram (drawn from computed coordinates), the worked examples, the practice items and the quiz questions. Every distance and dividing point was checked by computer.
Video: Distance and Midpoint Formula in 3 Dimensions, Mario's Math Tutoring, embedded from YouTube.

1Corners of a roof

Corners of a roof