3.1 Distance, midpoint, volume and surface area

A point in a plane needs two coordinates. A point in space needs three, measured along three axes that meet at the origin $O$ and are each perpendicular to the other two. The $x$- and $y$-axes form a horizontal plane, and the $z$-axis points upwards. The point $(x, y, z)$ is reached from the origin by moving $x$ units parallel to the $x$-axis, $y$ units parallel to the $y$-axis and $z$ units parallel to the $z$-axis.

The axes are drawn in perspective: the $x$-axis comes out of the page towards the reader, so it is drawn sloping down to the left. Three planes contain two of the axes each: the $xy$-plane ($z = 0$), the $xz$-plane ($y = 0$) and the $yz$-plane ($x = 0$). A point such as $(3, 4, 0)$ lies in the $xy$-plane, directly below $P(3, 4, 5)$.

Distance between two points

Distance in two dimensions

For two points $(x_1, y_1)$ and $(x_2, y_2)$ in a plane, the horizontal and vertical differences $x_2 - x_1$ and $y_2 - y_1$ are the two shorter sides of a right-angled triangle whose hypotenuse joins the points. Pythagoras's theorem gives the distance:

The order of subtraction does not matter, because each difference is squared: $(x_1 - x_2)^2 = (x_2 - x_1)^2$.

Distance in three dimensions

In space, the two points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ are opposite corners of a cuboid whose edges are parallel to the axes. The edges have lengths equal to the differences in the three coordinates, and the distance $AB$ is the length of the cuboid's space diagonal.

Pythagoras's theorem is used twice. In the base, $AD^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$. The edge $DB$ is vertical, so it is perpendicular to every line in the base, including $AD$. Triangle $ADB$ therefore has a right angle at $D$, and

Writing the three differences on their own line before squaring keeps the signs under control. Each squared term is positive or zero, so a negative number under the square root always means an arithmetic slip.

Using distances to classify triangles

Three side lengths decide the shape of a triangle. Equal lengths show that a triangle is isosceles or equilateral, and if the squares of the two shorter sides add up to the square of the longest side, the converse of Pythagoras's theorem shows that the triangle is right-angled. Working with squared distances avoids square roots until the end.

Finding an unknown coordinate

When a distance is given and one coordinate is unknown, the distance formula becomes an equation. Squaring both sides removes the square root, and the result is usually a quadratic equation with two solutions. Both are valid unless the context rules one out: geometrically, there are two points on a line at a given distance from a fixed point.

The midpoint of a line segment

The midpoint of a line segment is halfway between its endpoints in every direction at once. Halfway in the $x$-direction is the mean of the two $x$-coordinates, and the same is true for $y$ and $z$.

Finding an endpoint from the midpoint

If the midpoint $M$ and one endpoint $A$ are known, the other endpoint $B$ is as far beyond $M$ as $A$ is before it. Each coordinate satisfies $\frac{x_A + x_B}{2} = x_M$, so $x_B = 2x_M - x_A$, and similarly for $y$ and $z$.

Prisms, cylinders and cuboids

The volume and surface area of a solid measure two different things. Volume is the amount of space inside, in cubic units ($\text{cm}^3$, $\text{m}^3$). Surface area is the total area of the outside, in square units ($\text{cm}^2$, $\text{m}^2$), and equals the area of the solid's net, the flat shape that folds up to make it.

A cuboid is a prism with a rectangular cross-section, and a cylinder behaves like a prism with a circular cross-section. Their standard results are prior knowledge.

The curved surface of a cylinder unrolls into a rectangle whose length is the circumference of the base, $2\pi r$, and whose width is the height $h$. A closed cylinder adds two circular ends, but a pipe or an open tube has none, and a can without a lid has only one, so always decide which faces a real object has.

Right pyramids

A pyramid has a polygon for a base and triangular faces that meet at a single point, the apex. The base may be a triangle, a square, a rectangle or any other polygon.

In a right pyramid on a square or rectangular base, the triangular faces come in congruent pairs, and opposite faces are mirror images of each other. The height of a triangular face, measured from the apex to the midpoint of a base edge, is the slant height $l$ of that face. It is longer than the perpendicular height $h$, because $h$, half the base side and $l$ form a right-angled triangle with $l$ as the hypotenuse.

Volume of a pyramid

The volume of any pyramid, whatever the shape of its base, is one third of the volume of the prism with the same base and height.

where $A$ is the area of the base and $h$ is the perpendicular height. One way to see the factor $\frac{1}{3}$ is to split a cube of side $2a$ into six identical square-based pyramids, each with its apex at the centre of the cube and one face of the cube as its base. Each pyramid has base area $4a^2$ and height $a$, and six of them make up the cube's volume $8a^3$, so each has volume

Surface area of a pyramid

There is no single formula for the surface area of a pyramid. Add the area of the base to the areas of the triangular faces, using the slant height of each face as the height of its triangle. For a square base of side $s$ and slant height $l$, the total is $s^2 + 4 \times \frac{1}{2}sl = s^2 + 2sl$.

Right cones

A cone is the limit of a pyramid whose base has more and more sides: the polygonal base becomes a circle, and the triangular faces merge into one curved surface. The same factor of $\frac{1}{3}$ therefore appears in its volume.

The volume of a right cone is one third of the volume of the cylinder with the same base and height, and its curved surface area depends on the slant height.

Where the curved surface area comes from

Cutting the curved surface of a cone along a slant line and flattening it gives a sector of a circle. The radius of the sector is the slant height $l$, and the length of its arc is the circumference of the cone's base, $2\pi r$.

A full circle of radius $l$ has circumference $2\pi l$ and area $\pi l^2$. The sector's arc is the fraction $\frac{2\pi r}{2\pi l} = \frac{r}{l}$ of the full circumference, so its area is the same fraction of the full area:

A closed (solid) cone also has a circular base, so its total surface area is $\pi r l + \pi r^2$.

Working backwards from a volume or area

A question may give the volume or the surface area and ask for a length. Substitute what is known, then rearrange; the unknown is often under a square or cube root.

Spheres and hemispheres

A sphere has a single curved surface and no edges. A solid hemisphere has two surfaces: a curved surface, which is half of the sphere's, and a flat circular face of radius $r$.

The formulas for the sphere follow from calculus and are given; the hemisphere's are built from them.

Coordinates and solids combine naturally. The centre of a sphere is the midpoint of any diameter, and the radius is half the length of that diameter.

Composite solids

Many real objects are combinations of the basic solids: a grain silo is a cylinder with a hemispherical roof, a pencil is a hexagonal prism with a cone at one end, and a capsule is a cylinder with a hemisphere at each end. Volumes add directly. Surface areas need more care, because a face where two solids are joined is inside the object and is not part of its surface.

A reliable method for any composite solid:

1. Split the solid into basic shapes and write down the dimensions of each.

2. For volume, add the volumes of the parts (or subtract, for a hole or a cut-out).

3. For surface area, list every exposed surface, then add their areas. Leave out any face that is joined to another part or that rests on the ground if the question says so.

The height of a composite solid is often given as a total, and the dimensions of each part must be worked out before any formula is used.

Angles inside solids

An angle in three dimensions is found the same way as in two: identify a triangle that contains the angle, find enough of its sides, and use trigonometry. Inside cuboids, prisms, pyramids and cones the triangle is usually right-angled, because vertical edges and heights are perpendicular to horizontal faces. In a right-angled triangle with an angle $\theta$, the trigonometric ratios are

These are developed fully in 3.2; inside a solid they are used exactly as in a flat triangle.

The angle between two intersecting lines

Two lines that meet at a point always lie in one plane, so the angle between them is an ordinary angle in a triangle. Draw the triangle separately, flat on the page, label the lengths that are known, and look for a right angle. If there is none, the cosine rule from 3.2 finds the angle from three sides.

The angle between a line and a plane

A line that meets a plane at a point makes different angles with different lines in the plane, so the angle with the plane needs a precise definition. The natural choice is the smallest of these angles, the one made with the line's projection: the shadow it would cast on the plane under light shining perpendicular to the plane.

The angle between a line and a plane always sits in a right-angled triangle made of three parts: the line itself (the hypotenuse), the perpendicular from a point on the line to the plane, and the projection.

The method has four steps:

1. Find the point where the line meets the plane (here $A$).

2. Choose another point on the line (here $G$) and find the foot of the perpendicular from it to the plane (here $C$, directly below $G$).

3. Draw the right-angled triangle formed (here $ACG$, with the right angle at $C$) and find two of its sides.

4. Use a trigonometric ratio to find the angle at the point where the line meets the plane.

Angles in a pyramid

In a right pyramid, two different angles with the base are common: the angle between a sloping edge (such as $VA$) and the base, and the angle between a sloping face and the base, measured along the slant height $VN$. Both use the perpendicular height $VM$, because $M$ is the foot of the perpendicular from the apex $V$.

Exam-style questions on three-dimensional geometry

The next question is set without a calculator, so answers stay exact in terms of $\pi$, and the "show that" part must reach the given result with every step visible.

Self-check

• To do:

  I can find the distance between two points in two or three dimensions, and explain why the formula follows from Pythagoras's theorem.

• To do:

  I can find the midpoint of a line segment, and find an endpoint given the midpoint and the other endpoint.

• To do:

  I can find an unknown coordinate from a given distance, giving every possible value.

• To do:

  I can calculate the volume and surface area of right pyramids, right cones, spheres and hemispheres.

• To do:

  I can find the slant height of a cone or a pyramid from the perpendicular height, and use each height in the right formula.

• To do:

  I can find the volume and surface area of a composite solid, leaving out the faces that are joined or hidden.

• To do:

  I can find a length or radius from a given volume or surface area.

• To do:

  I can find the angle between two intersecting lines, and between a line and a plane, inside a solid.

1. (a)

   Find the coordinates of $\mathrm{B}$.

   [2]2 marks

   Markscheme (a)

   $\frac{5 + x}{2} = 2$, $\frac{-1 + y}{2} = 1$, $\frac{2 + z}{2} = 4$ (or $\overrightarrow{\mathrm{MB}} = \overrightarrow{\mathrm{AM}}$) (M1)

   $\mathrm{B}(-1, 3, 6)$ A1

   Note: Award (M1)A0 if one coordinate is incorrect.

2. (b)

   Find the exact value of $\mathrm{AB}$.

   [2]2 marks

   Markscheme (b)

   $\mathrm{AB} = \sqrt{(5 - (-1))^2 + (-1 - 3)^2 + (2 - 6)^2}$

3. (c)

   Find the exact surface area of the sphere.

   [2]2 marks

   Markscheme (c)

   radius $= \sqrt{17}$ (seen anywhere) (A1)

   surface area $= 4\pi\left(\sqrt{17}\right)^2 = 68\pi$

And 8 more in the formula booklet.

Since $AB^2 + AC^2 = 8 + 9 = 17 = BC^2$, the converse of Pythagoras's theorem shows that the triangle has a right angle at $A$, the vertex opposite the longest side $BC$.

The two sides that meet at the right angle are a base and a perpendicular height, so

Take square roots, remembering both signs.

Check $k = 7$: $\sqrt{16 + 16 + 4} = \sqrt{36} = 6$. Both values are correct.

Check: the midpoint of $(1, 2, 5)$ and $(7, -4, 1)$ is $\left(\frac{8}{2}, \frac{-2}{2}, \frac{6}{2}\right) = (4, -1, 3)$, as given.

$G$ is directly above $C$, so triangle $ACG$ has a right angle at $C$.

The three-dimensional distance formula gives the same result in one step: $\sqrt{8^2 + 6^2 + 5^2} = \sqrt{125}$.

(b) The projection of $AG$ onto the base is $AC$. In triangle $ACG$, the side opposite $\theta$ is $CG = 5$ and the adjacent side is $AC = 10$.

In triangle $VMA$ (right angle at $M$):

(b) The face $VAB$ meets the base along $AB$, and $VN$ and $MN$ are both perpendicular to $AB$, so the angle between the face and the base is $V\hat{N}M$. Here $MN = 5$ cm.

(c) The slant height is $VN = \sqrt{12^2 + 5^2} = 13$ cm. Triangle $VAB$ is isosceles, and $VN$ splits it into two right-angled triangles, so angle $AVN$ is half of angle $AVB$. In triangle $VNA$, $AN = 5$ is opposite and $VN = 13$ is adjacent.

(b) Use the distance formula.

(c) The base has area $6^2 = 36$ and the height is $VM = 8$.

(d) The projection of $VA$ onto the base is $AM = \sqrt{3^2 + 3^2} = 3\sqrt{2}$, and $VM = 8$ is perpendicular to the base.

(e) Let $N$ be the midpoint of $AB$, so $N = (3, 0, 0)$. The slant height of each face is

Each triangular face has area $\frac{1}{2} \times 6 \times \sqrt{73} = 3\sqrt{73}$. Add the four faces and the base.

Marks are typically for finding the slant height, the area of one face, multiplying by four, and including the base.

so the volume of the sphere is $\frac{2}{3}$ of the volume of the cylinder.

(c) Use the cylinder's volume to find $r$.

$= \sqrt{68} \; \left(= 2\sqrt{17}\right)$ A1

Note: Accept $\mathrm{AB} = 2\mathrm{AM} = 2\sqrt{9 + 4 + 4}$.

Note: Award (A0)A1FT for $4\pi r^2$ with their radius from part (b). Award A0 for $272\pi$ (diameter used as radius).