Coordinates in Space

Everything you have done in coordinate geometry so far has lived on a flat sheet. Real objects — a kite over Tundikhel, a drone above Pokhara, the corner of a room — need a third number to fix them.

This unit adds a z-axis perpendicular to the x and y axes, and rebuilds the three basic tools in three dimensions: the distance formula, the section formula, and the direction cosines that describe which way a line points.

13.1 Points in space

Take three mutually perpendicular lines through a point O — the origin — and call them the x-, y- and z-axes. Taken in pairs they determine three coordinate planes:

Plane	Contains the axes	Equation
xy-plane	x and y	z = 0
yz-plane	y and z	x = 0
zx-plane	z and x	y = 0

These three planes cut space into eight octants. The first octant is the one in which all three coordinates are positive; the others are labelled by the sign pattern of (x,y,z).

A point P in space is fixed by an ordered triple (x, y, z): drop a perpendicular from P to the xy-plane meeting it at M, then x and y are the usual plane coordinates of M and z = PM, taken positive above the plane.

13.2 Distance between two points

The section formula

13.3 Direction cosines and direction ratios

A line in space is aimed by the three angles it makes with the positive directions of the axes.

Direction ratios

Any three numbers a, b, c proportional to l, m, n are called direction ratios of the line. Direction ratios are far more convenient, because you may scale them freely. To recover the direction cosines, divide by the length:

The \pm appears because a line has two opposite directions.

(a) A 3 x 3 x 3 cube: positions are described by three coordinates

(b) Face and space diagonals of a cube of side 3

Chapter summary

• Three mutually perpendicular axes give three coordinate planes (x=0, y=0, z=0) and eight octants; a point is an ordered triple (x,y,z).
• Distance of P(x,y,z) from the xy-plane is |z|; from the x-axis is \sqrt{y^{2}+z^{2}}; from the origin is \sqrt{x^{2}+y^{2}+z^{2}}.
• AB = \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}+(z_2-z_1)^{2}}.
• Section formula (internal, ratio m:n): \left(\dfrac{mx_2+nx_1}{m+n},\ \dfrac{my_2+ny_1}{m+n},\ \dfrac{mz_2+nz_1}{m+n}\right); for external division replace n by -n. Midpoint is the case m=n; centroid is the average of the three vertices.
• Direction cosines l = \cos\alpha, m = \cos\beta, n = \cos\gamma satisfy l^{2}+m^{2}+n^{2} = 1, and l = \dfrac{x}{r}, m = \dfrac{y}{r}, n = \dfrac{z}{r} with r = OP.
• Direction ratios (a,b,c) are any numbers proportional to (l,m,n); divide by \sqrt{a^{2}+b^{2}+c^{2}} to get the direction cosines.
• The join of A and B has direction ratios (x_2-x_1, y_2-y_1, z_2-z_1).
• \cos\theta = l_1l_2+m_1m_2+n_1n_2; lines are perpendicular when a_1a_2+b_1b_2+c_1c_2 = 0 and parallel when their direction ratios are proportional.

Practice questions

Group A — Multiple choice (1 mark each)

1. The distance of the point (2,-3,6) from the origin is [1]a5b7c11d\sqrt{41}
2. The number of octants into which the coordinate planes divide space is [1]a4b6c8d12
3. If l, m, n are direction cosines then l^{2}+m^{2}+n^{2} equals [1]a0b1c2d3
4. The direction ratios of the line joining (1,0,2) and (3,4,5) are [1]a(2,4,3)b(4,4,7)c(3,0,10)d(-2,-4,-7)
5. The distance of the point (1,-2,3) from the y-axis is [1]a1b2c\sqrt{10}d\sqrt{13}
6. Two lines with direction ratios (1,2,-1) and (3,-1,1) are [1]aparallelbperpendicularccoincidentdinclined at 60^{\circ}

Answers to Group A

1. (b) — \sqrt{4+9+36} = \sqrt{49} = 7.

2. (c) — three planes, each giving two sides: 2^{3} = 8.

3. (b) — proved in §13.3.

4. (a) — (3-1,\ 4-0,\ 5-2) = (2,4,3).

5. (c) — ignore y: \sqrt{1^{2}+3^{2}} = \sqrt{10}.

6. (b) — 1(3)+2(-1)+(-1)(1) = 3-2-1 = 0.

Group B — Short answer (5 marks each)

1. Derive the formula for the distance between the points A(x_1,y_1,z_1) and B(x_2,y_2,z_2). [5]
2. Show that the points (-2,3,5), (1,2,3) and (7,0,-1) are collinear. [5]
3. Find the ratio in which the zx-plane divides the join of A(-2,4,7) and B(3,-5,8). [5]
4. A line makes angles 90^{\circ} and 135^{\circ} with the positive x- and y-axes. Find the angle it makes with the positive z-axis and write down its direction cosines. [5]
5. Find the angle between the lines whose direction ratios are (1,1,2) and (\sqrt{3}-1,\ -\sqrt{3}-1,\ 4). [5]

Answers to Group B

1. Complete the rectangular box on AB as a space diagonal, with M(x_2,y_2,z_1). A and M lie in the plane z = z_1, so AM^{2} = (x_2-x_1)^{2}+(y_2-y_1)^{2}. The edge MB is parallel to the z-axis, so MB = |z_2-z_1| and MB \perp AM.

Pythagoras in \triangle AMB gives AB^{2} = AM^{2}+MB^{2}, hence AB = \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}+(z_2-z_1)^{2}}.

2. Let A(-2,3,5), B(1,2,3), C(7,0,-1). Direction ratios of AB are (3,-1,-2) and of AC are (9,-3,-6) = 3\times(3,-1,-2). The ratios are proportional and the lines share A, so the three points are collinear.

(Alternatively AB = \sqrt{9+1+4} = \sqrt{14}, BC = \sqrt{36+4+16} = \sqrt{56} = 2\sqrt{14}, AC = \sqrt{81+9+36} = \sqrt{126} = 3\sqrt{14}, and AB+BC = AC.)

3. Let the zx-plane cut AB in the ratio k:1. The dividing point is \left(\frac{3k-2}{k+1},\ \frac{-5k+4}{k+1},\ \frac{8k+7}{k+1}\right). On the zx-plane y = 0, so -5k+4 = 0 and k = \frac45. The ratio is 4:5 internally.

4. l = \cos 90^{\circ} = 0 and m = \cos 135^{\circ} = -\frac{1}{\sqrt{2}}. Then n^{2} = 1-0-\frac12 = \frac12, so n = \pm\frac{1}{\sqrt{2}} and \gamma = 45^{\circ} or 135^{\circ}.

The direction cosines are \left(0,\ -\frac{1}{\sqrt{2}},\ \frac{1}{\sqrt{2}}\right) or \left(0,\ -\frac{1}{\sqrt{2}},\ -\frac{1}{\sqrt{2}}\right).

5. a_1a_2+b_1b_2+c_1c_2 = 1(\sqrt{3}-1)+1(-\sqrt{3}-1)+2(4) = -2+8 = 6. The lengths are \sqrt{1+1+4} = \sqrt{6} and \sqrt{(\sqrt{3}-1)^{2}+(\sqrt{3}+1)^{2}+16} = \sqrt{(4-2\sqrt{3})+(4+2\sqrt{3})+16} = \sqrt{24} = 2\sqrt{6}.

So \cos\theta = \dfrac{6}{\sqrt{6}\cdot 2\sqrt{6}} = \dfrac{6}{12} = \dfrac12, giving \theta = 60^{\circ}.

Group C — Long answer (8 marks each)

1. (a) Define the direction cosines of a line and prove that l^{2}+m^{2}+n^{2} = 1. [5](b) A line makes angles 60^{\circ}, 45^{\circ} with the x- and y-axes. Find the angle it makes with the z-axis. [3]
2. The vertices of a triangle are A(3,-1,2), B(1,-3,-4) and C(-1,1,2).(a) Find the lengths of the three sides and classify the triangle. [4](b) Find the centroid, and the length of the median from A. [4]

Answers to Group C

1. (a) If a directed line through the origin makes angles \alpha, \beta, \gamma with the positive axes, its direction cosines are l = \cos\alpha, m = \cos\beta, n = \cos\gamma. Take P(x,y,z) on the line with OP = r.

Dropping a perpendicular from P to the x-axis at A gives a right angle at A with OA = x, so \cos\alpha = \frac{x}{r}; similarly \cos\beta = \frac{y}{r}, \cos\gamma = \frac{z}{r}. Hence

l^{2}+m^{2}+n^{2} = \frac{x^{2}+y^{2}+z^{2}}{r^{2}} = \frac{r^{2}}{r^{2}} = 1

(b) l = \cos 60^{\circ} = \frac12, m = \cos 45^{\circ} = \frac{1}{\sqrt{2}}, so n^{2} = 1-\frac14-\frac12 = \frac14 and n = \pm\frac12. Therefore \gamma = 60^{\circ} or 120^{\circ}.

2. (a) AB^{2} = (1-3)^{2}+(-3+1)^{2}+(-4-2)^{2} = 4+4+36 = 44, so AB = 2\sqrt{11}. BC^{2} = (-1-1)^{2}+(1+3)^{2}+(2+4)^{2} = 4+16+36 = 56, so BC = 2\sqrt{14}. CA^{2} = (3+1)^{2}+(-1-1)^{2}+(2-2)^{2} = 16+4+0 = 20, so CA = 2\sqrt{5}.

All three sides are different and AB^{2}+CA^{2} = 44+20 = 64 \ne 56, so the triangle is scalene and not right-angled.

(b) Centroid G = \left(\frac{3+1-1}{3},\ \frac{-1-3+1}{3},\ \frac{2-4+2}{3}\right) = \left(1,\ -1,\ 0\right). The midpoint of BC is D = \left(\frac{1-1}{2},\ \frac{-3+1}{2},\ \frac{-4+2}{2}\right) = (0,-1,-1), so the median from A has length

AD = \sqrt{(0-3)^{2}+(-1+1)^{2}+(-1-2)^{2}} = \sqrt{9+0+9} = \sqrt{18} = 3\sqrt{2}\ \text{units}

Group D — Model answers for long questions (10 marks each, for deeper revision)

1. Describe the coordinate system in space, derive the distance formula between two points and use it for collinearity and shape problems. [10]
2. Define direction cosines and direction ratios, prove l^{2}+m^{2}+n^{2}=1, find them for a line joining two points and derive the angle between two lines. [10]

Model answer D1: points in space and the distance formula (10 marks)

Introduction. A point in space is located by three mutually perpendicular axes OX, OY, OZ meeting at the origin O. The ordered triple (x,y,z) gives the distance of the point from the YZ-plane, the ZX-plane and the XY-plane respectively.

Drone navigation and the location of a mobile tower use such coordinates (latitude, longitude, height).

Octants. The three coordinate planes divide space into eight octants.

Octant	Signs of (x,y,z)
I	(+,+,+)
II	(-,+,+)
III	(-,-,+)
IV	(+,-,+)
V	(+,+,-)
VI	(-,+,-)
VII	(-,-,-)
VIII	(+,-,-)

Special points. A point on the X-axis has the form (x,0,0); on the XY-plane, (x,y,0). For P(3,-4,5) the distance from the XY-plane is 5, and from the X-axis is \sqrt{(-4)^{2}+5^{2}}=\sqrt{41}.

Theorem: distance formula. The distance between P(x_1,y_1,z_1) and Q(x_2,y_2,z_2) is

Proof. Through P and Q draw planes parallel to the coordinate planes to form a rectangular box with PQ as a diagonal. The edges of the box are |x_2-x_1|, |y_2-y_1| and |z_2-z_1|.

The diagonal of the base is d with d^{2}=(x_2-x_1)^{2}+(y_2-y_1)^{2}, and PQ is the hypotenuse of a right-angled triangle with sides d and |z_2-z_1|. Pythagoras' theorem gives the formula.

The distance from the origin is \sqrt{x^{2}+y^{2}+z^{2}}. The mid-point of PQ is \left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2},\dfrac{z_1+z_2}{2}\right).

Worked examples.

1. A(1,2,3) and B(4,6,15): AB=\sqrt{9+16+144}=\sqrt{169}=13.
2. Collinearity. A(3,2,-4), B(5,4,-6), C(9,8,-10): AB=\sqrt{4+4+4}=2\sqrt3, BC=\sqrt{16+16+16}=4\sqrt3 and AC=\sqrt{36+36+36}=6\sqrt3. Since AB+BC=AC, the points are collinear.
3. Right-angled triangle. P(0,7,10), Q(-1,6,6), R(-4,9,6): PQ^{2}=1+1+16=18, QR^{2}=9+9+0=18 and PR^{2}=16+4+16=36. As PQ^{2}+QR^{2}=PR^{2}, the triangle is right-angled at Q, and isosceles because PQ=QR.
4. Locus. The point (x,y,z) equidistant from (1,0,0) and (0,1,0) satisfies (x-1)^{2}+y^{2}+z^{2}=x^{2}+(y-1)^{2}+z^{2}, which gives x=y.

Test	Condition
Collinear	the largest distance equals the sum of the other two
Right angle	a^{2}+b^{2}=c^{2}
Equilateral	all three sides equal

Conclusion. The distance formula is Pythagoras' theorem applied twice, and it settles questions about collinearity, shapes and loci in space.

Model answer D2: direction cosines, direction ratios and the angle between lines (10 marks)

Introduction. A line through the origin makes angles \alpha,\beta,\gamma with the positive X, Y, Z axes. The cosines l=\cos\alpha, m=\cos\beta, n=\cos\gamma are the direction cosines (d.c.s) of the line.

Any three numbers a,b,c proportional to l,m,n are direction ratios (d.r.s).

Theorem: l^{2}+m^{2}+n^{2}=1.

Proof. Let P(x,y,z) be a point on the line with OP=r. The projections of OP on the axes give x=lr, y=mr, z=nr. Since OP^{2}=x^{2}+y^{2}+z^{2}, r^{2}=r^{2}(l^{2}+m^{2}+n^{2}), so l^{2}+m^{2}+n^{2}=1.

Relation between d.r.s and d.c.s. If a,b,c are d.r.s, then

The d.c.s are unique up to the sign for the two opposite senses of the line; d.r.s are not unique.

Line joining two points. For P(x_1,y_1,z_1) and Q(x_2,y_2,z_2) the d.r.s of PQ are x_2-x_1,\ y_2-y_1,\ z_2-z_1, and the d.c.s are these divided by PQ.

Example. For P(1,2,3) and Q(3,-1,9): d.r.s 2,-3,6, PQ=\sqrt{4+9+36}=7, so d.c.s are \dfrac27,-\dfrac37,\dfrac67. Check: \dfrac{4+9+36}{49}=1.

Axes and special lines. The d.c.s of the X-axis are 1,0,0; of the Y-axis 0,1,0. A line equally inclined to the axes has l=m=n=\dfrac1{\sqrt3}, so each angle is \cos^{-1}\dfrac1{\sqrt3}\approx54.74^{\circ} (the body diagonal of a cube).

Theorem: angle between two lines. If the lines have d.c.s l_1,m_1,n_1 and l_2,m_2,n_2 then

Proof. Draw lines through the origin parallel to the given lines and take P(l_1,m_1,n_1) and Q(l_2,m_2,n_2) on them at unit distance from O.

Then PQ^{2}=(l_1-l_2)^{2}+(m_1-m_2)^{2}+(n_1-n_2)^{2}=2-2(l_1l_2+m_1m_2+n_1n_2) using l^{2}+m^{2}+n^{2}=1 for each line. The cosine rule in \triangle OPQ gives PQ^{2}=1+1-2\cos\theta. Comparing, we get the result.

With d.r.s a_1,b_1,c_1 and a_2,b_2,c_2,

Conditions. The lines are perpendicular if a_1a_2+b_1b_2+c_1c_2=0 and parallel if \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}.

Worked examples.

1. D.r.s 2,3,6 and 1,2,2: dot product =2+6+12=20; lengths 7 and 3, so \cos\theta=\dfrac{20}{21} and \theta\approx17.75^{\circ}.
2. D.r.s 1,2,2 and 2,-2,1: dot product =2-4+2=0, so the lines are perpendicular.

Conclusion. The d.c.s of a line satisfy l^{2}+m^{2}+n^{2}=1, are found from the d.r.s by dividing by the length, and determine the angle between two lines.