Unit 13 · Analytic Geometry · 6 teaching hours · 16 min read
Coordinates in Space
What the syllabus asks for 3
- Points in space
- Distance between two points
- Direction cosines and direction ratios
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.
Expect one short question on the distance or section formula and one longer question on direction cosines. The two results you must be able to derive are AB = \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}+(z_2-z_1)^{2}} and l^{2}+m^{2}+n^{2} = 1.
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.
For P(x,y,z):
- distance from the yz-, zx- and xy-planes: |x|, |y|, |z|
- distance from the x-, y- and z-axes: \sqrt{y^{2}+z^{2}}, \sqrt{z^{2}+x^{2}}, \sqrt{x^{2}+y^{2}}
- distance from the origin: OP = \sqrt{x^{2}+y^{2}+z^{2}}
Problem. For the point P(3,-4,5) find (i) its distance from the xy-plane, (ii) its distance from the x-axis, (iii) its distance from the origin, and (iv) the octant it lies in.
Solution.
(i) The distance from the xy-plane is |z| = 5 units.
(ii) The distance from the x-axis ignores the x-coordinate:
(iii) OP = \sqrt{3^{2}+(-4)^{2}+5^{2}} = \sqrt{9+16+25} = \sqrt{50} = 5\sqrt{2} \approx 7.07 units.
(iv) The signs are (+,-,+), so P lies in the fourth octant.
Distance from a plane uses the one coordinate perpendicular to it; distance from an axis uses the other two. Students routinely swap these. A quick memory check: a point can be far from the x-axis while having a huge x-coordinate, so x cannot appear in that distance.
13.2 Distance between two points
Let A(x_1,y_1,z_1) and B(x_2,y_2,z_2) be two points. Complete the rectangular box whose edges are parallel to the axes and which has AB as a space diagonal. Let M be the corner (x_2,y_2,z_1) — vertically below B and in the same horizontal plane as A.
Step 1 — the horizontal diagonal. A and M have the same z, so by the two-dimensional distance formula
Step 2 — the vertical edge. MB is parallel to the z-axis, so MB = |z_2-z_1| and MB is perpendicular to the plane containing AM, hence perpendicular to AM.
Step 3 — Pythagoras in \triangle AMB, right-angled at M.
Putting A at the origin gives OB = \sqrt{x_2^{2}+y_2^{2}+z_2^{2}}.
Problem. Show that the points A(0,7,10), B(-1,6,6) and C(-4,9,6) are the vertices of a right-angled isosceles triangle.
Solution. Work with the squares of the lengths — no surds are needed.
Since AB^{2} = BC^{2}, the triangle is isosceles. Since AB^{2}+BC^{2} = 18+18 = 36 = CA^{2}, the converse of Pythagoras' theorem gives a right angle at B. So \triangle ABC is right-angled and isosceles, with AB = BC = 3\sqrt{2} and CA = 6.
Problem. Prove that A(1,2,3), B(2,3,1) and C(3,1,2) form an equilateral triangle.
Solution.
All three sides equal \sqrt{6} units, so the triangle is equilateral.
Problem. Find the point on the y-axis that is equidistant from A(3,1,2) and B(5,5,2).
Solution. Any point on the y-axis has the form P(0,y,0). The condition PA^{2} = PB^{2} gives
The point is (0,5,0).
Check. PA = \sqrt{9+16+4} = \sqrt{29} and PB = \sqrt{25+0+4} = \sqrt{29} ✓.
The section formula
Let P(x,y,z) divide the join of A(x_1,y_1,z_1) and B(x_2,y_2,z_2) internally in the ratio m:n, so AP:PB = m:n.
Drop perpendiculars AL, PN, BM from A, P, B to the xy-plane. These three perpendiculars are parallel, so they lie in one plane, and the line AB cuts them proportionally.
Draw through A a line parallel to LM, meeting PN at R and BM at S. Then RP = z-z_1 and SB = z_2-z_1, and \triangle APR and \triangle ABS are similar (equal angles, since RP and SB are parallel), so
Cross-multiplying, (m+n)(z-z_1) = m(z_2-z_1), so (m+n)z = mz_2+nz_1, that is
The same argument applied to the other two pairs of coordinate planes gives x and y, so
For external division replace n by -n: P = \left(\dfrac{mx_2-nx_1}{m-n},\ \dfrac{my_2-ny_1}{m-n},\ \dfrac{mz_2-nz_1}{m-n}\right).
- Midpoint (m=n=1): \left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2},\ \dfrac{z_1+z_2}{2}\right)
- Centroid of the triangle with vertices A, B, C: \left(\dfrac{x_1+x_2+x_3}{3},\ \dfrac{y_1+y_2+y_3}{3},\ \dfrac{z_1+z_2+z_3}{3}\right)
Problem. Find the coordinates of the points dividing the join of A(2,-1,3) and B(4,3,1) in the ratio 3:2 (i) internally and (ii) externally.
Solution. Here m = 3, n = 2, so m+n = 5 and m-n = 1.
(i) Internally:
The point is \left(\dfrac{16}{5},\ \dfrac{7}{5},\ \dfrac{9}{5}\right).
(ii) Externally:
The point is (8,\ 11,\ -3).
Problem. In what ratio does the yz-plane divide the join of A(2,4,5) and B(3,5,-4)?
Solution. Let the plane cut AB in the ratio k:1 at the point
Every point of the yz-plane has x = 0, so
A negative ratio means external division: the yz-plane divides AB externally in the ratio 2:3.
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.
If a directed line through the origin makes angles \alpha, \beta, \gamma with the positive x-, y- and z-axes, then \alpha, \beta, \gamma are its direction angles and
are its direction cosines, written (l, m, n).
Let P(x,y,z) be a point on the directed line through the origin, with OP = r = \sqrt{x^{2}+y^{2}+z^{2}}.
Drop a perpendicular PA from P to the x-axis. Then OA = x and \triangle OAP is right-angled at A, so
Similarly \cos\beta = \dfrac{y}{r} and \cos\gamma = \dfrac{z}{r}, so
Squaring and adding,
l^{2}+m^{2}+n^{2} = 1 means the three direction cosines are not independent — you cannot pick all three. In particular no set like (\frac12,\frac12,\frac12) can be direction cosines, since \frac14+\frac14+\frac14 \ne 1. Note also \sin^{2}\alpha+\sin^{2}\beta+\sin^{2}\gamma = 3-1 = 2.
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.
- The line through A(x_1,y_1,z_1) and B(x_2,y_2,z_2) has direction ratios
- If two lines have direction cosines (l_1,m_1,n_1) and (l_2,m_2,n_2), the angle \theta between them satisfies
- In terms of direction ratios,
- Perpendicular: a_1a_2+b_1b_2+c_1c_2 = 0. Parallel: \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}.
Problem. A line makes angles 60^{\circ} and 45^{\circ} with the positive x- and y-axes. Find the angle it makes with the positive z-axis, and its direction cosines.
Solution. l = \cos 60^{\circ} = \frac12 and m = \cos 45^{\circ} = \frac{1}{\sqrt{2}}. Using l^{2}+m^{2}+n^{2} = 1,
So \gamma = 60^{\circ} or \gamma = 120^{\circ}, and the direction cosines are
Problem. Find the direction ratios and direction cosines of the line joining A(1,2,3) and B(2,4,5).
Solution. Direction ratios are the differences:
Their length is \sqrt{1^{2}+2^{2}+2^{2}} = \sqrt{9} = 3, so the direction cosines are
Check. \frac19+\frac49+\frac49 = \frac99 = 1 ✓.
Problem. Find the acute angle between two lines whose direction ratios are (2,2,1) and (4,1,8).
Solution.
Problem. Show that the points A(1,2,3), B(4,0,4) and C(-2,4,2) are collinear.
Solution. Direction ratios of AB: (4-1,\ 0-2,\ 4-3) = (3,-2,1). Direction ratios of AC: (-2-1,\ 4-2,\ 2-3) = (-3,2,-1).
Since (-3,2,-1) = -1\times(3,-2,1), the two sets are proportional, so AB and AC are parallel lines. They share the point A, so they are the same line and A, B, C are collinear.
(The negative factor tells us A lies between B and C.)
In three dimensions almost every standard question can be settled without ever taking a square root: compare AB^{2} with BC^{2} for triangles, and compare ratios a_1:b_1:c_1 with a_2:b_2:c_2 for parallel lines. You lose marks to arithmetic slips, not to missing ideas.
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)
- The distance of the point (2,-3,6) from the origin is [1]a5b7c11d\sqrt{41}
- The number of octants into which the coordinate planes divide space is [1]a4b6c8d12
- If l, m, n are direction cosines then l^{2}+m^{2}+n^{2} equals [1]a0b1c2d3
- 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)
- The distance of the point (1,-2,3) from the y-axis is [1]a1b2c\sqrt{10}d\sqrt{13}
- 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)
- 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]
- Show that the points (-2,3,5), (1,2,3) and (7,0,-1) are collinear. [5]
- Find the ratio in which the zx-plane divides the join of A(-2,4,7) and B(3,-5,8). [5]
- 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]
- 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)
- (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]
- 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
(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
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