IB Mathematics Analysis and Approaches SL 3.1—3d Space, Volume, Angles, Distance, Midpoints Notes

Why 3D Geometry Exists: Measuring What You Cannot Reach

1. You can measure a short line with a ruler, but most real distances cannot be reached directly: the height of a tower, the diagonal across a room, or the angle a ramp makes with the ground.
2. 3D geometry lets you calculate those lengths and angles from a few known measurements and the coordinates of points in space.
3. Almost every SL 3.1 problem comes down to one move: find a right-angled triangle hidden inside the 3D shape, then use Pythagoras or basic trigonometry on that flat triangle.
4. Everything here is built from the 2D tools you already know, stretched into a third direction, so nothing in this topic is truly new; it is the same ideas seen one dimension up.

The 3D Coordinate System: Adding A Third Axis

1. In 2D a point is fixed by two numbers $(x, y)$: how far along, and how far up.
2. In 3D we add a third axis, the $z$-axis, placed at right angles to both the $x$-axis and the $y$-axis.
3. Every point in space is now written as an ordered triple $(x, y, z)$.
4. The three axes are mutually perpendicular, so they meet like the corner of a room where two walls meet the floor.

Distance In 3D: Pythagoras Applied Twice

1. Take two points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$, and imagine the box (cuboid) that has $A$ and $B$ at opposite corners.
2. The horizontal gap between them has two parts: $\Delta x = x_2 - x_1$ across, and $\Delta y = y_2 - y_1$ back.
   1. Step 1: on the base of the box these two form a right-angled triangle, so the base diagonal is $\sqrt{\Delta x^2 + \Delta y^2}$ by ordinary 2D Pythagoras.
   2. Step 2: that base diagonal and the vertical rise $\Delta z = z_2 - z_1$ form a second right-angled triangle, and this time the hypotenuse is the line $AB$ itself.
   3. Step 3: apply Pythagoras again, so $AB^2 = \left(\Delta x^2 + \Delta y^2\right) + \Delta z^2$.
3. Putting it together gives the 3D distance formula: $$ d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2} $$

The Midpoint In 3D: Average Each Coordinate

1. The midpoint of $AB$ is the point exactly halfway along the segment.
2. Halfway means the average, so you average the two $x$-values, then the two $y$-values, then the two $z$-values.

$$ M = \left( \dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2},\ \dfrac{z_1+z_2}{2} \right) $$

Volume And Surface Area Of 3D Solids

1. SL 3.1 expects you to find the volume and surface area of five standard solids, and of shapes made by combining them.
2. Every formula in this section is given in the formula booklet, so the skill is choosing the right one and substituting without slips.
3. Keep the units straight: volume is measured in cubic units (such as $\mathrm{cm^3}$) and surface area in square units (such as $\mathrm{cm^2}$).

Right Pyramid

1. A right pyramid has a flat polygon base and triangular faces that meet at an apex directly above the centre of the base.
2. Volume: $$V = \dfrac{1}{3} B h$$ where $B$ is the area of the base and $h$ is the perpendicular height.
3. The $\frac{1}{3}$ comes from three identical pyramids of the same base and height that fit exactly inside the matching prism $Bh$, so one pyramid is one third of that prism.

Right Cone

1. A right cone has a circular base of radius $r$ and an apex directly above the centre, at perpendicular height $h$.
2. Volume: $$V = \dfrac{1}{3}\pi r^2 h$ which is one third of the cylinder $\pi r^2 h$, exactly the same one-third relationship as the pyramid.
3. The slant height $l$ is the straight-line distance from the apex down to the edge of the base.
4. Deriving the slant height: $r$, $h$ and $l$ form a right-angled triangle, so by Pythagoras $l = \sqrt{r^2 + h^2}$.
5. Curved (lateral) surface area: $$A = \pi r l$$
6. Total surface area of a solid cone adds the circular base: $$\pi r l + \pi r^2$$

Sphere

1. A sphere is the set of all points a fixed distance $r$ from a centre.
2. Volume: $$V = \dfrac{4}{3}\pi r^3$$
3. Surface area: $$A = 4\pi r^2$$

Hemisphere

1. A hemisphere is exactly half of a sphere, and its formulas are not in the booklet, so you derive them from the sphere each time.
2. Volume: half the sphere, so $V = \dfrac{2}{3}\pi r^3$.
3. Curved surface: half the sphere's surface, so $2\pi r^2$.
4. A solid hemisphere also has a flat circular face of area $\pi r^2$, so its total surface area is $2\pi r^2 + \pi r^2 = 3\pi r^2$.

Combinations Of Solids

1. Many exam solids are combinations: an ice-cream shape is a cone plus a hemisphere, a capsule is a cylinder plus two hemispheres, and a rocket is a cone on a cylinder.
2. Volume of a combination: simply add the volumes of the separate parts.
3. Surface area of a combination: add only the surfaces that are actually on the outside, and never count a face that is hidden where two parts join.

Angles In 3D: Between Two Lines And Between A Line And A Plane

1. Two new angle types appear in 3D: the angle between two intersecting lines, and the angle between a line and a plane.
2. Both are found the same way: locate the right-angled triangle that contains the angle, redraw it flat, then use trigonometry.

Angle Between Two Intersecting Lines

1. Two lines that meet form an angle at their point of intersection.
2. If you can find a triangle whose two sides lie along those lines, the angle between the lines is an angle of that triangle.
3. In SL the triangle is always set up so that Pythagoras and right-angled trigonometry (SOHCAHTOA) are enough to find it.

Angle Between A Line And A Plane

1. The angle between a line and a plane is not measured straight down to the flat surface; it is measured to the line's projection onto the plane.
2. Picture the line casting a shadow straight down onto the plane; that shadow is the projection.
3. The line, its projection, and the vertical drop from the line to the plane together form a right-angled triangle.
4. The angle you want is the angle in that triangle between the line (the hypotenuse) and its projection (the adjacent side).

Applying The Strategy: The Diagonal Of A Box

1. The classic version asks for the angle between a cuboid's space diagonal and its base.
2. The projection of the space diagonal onto the base is simply the base diagonal, which you already know how to find with 2D Pythagoras.