IB Mathematics AA Distance , volume and Surface Area Study Notes
IB Mathematics AA Distance , volume and Surface Area Study Notes
IB Mathematics AA Distance , volume and Surface Area Study Notes Offer a clear explanation of Distance , volume and Surface Area , including various formula, rules, exam style questions as example to explain the topics. Worked Out examples and common problem types provided here will be sufficient to cover for topic Distance , volume and Surface Area
Distance & Midpoint Between Two Points in 3D Space
Distance Between Two Points in 3D Space
If two points are \( A(x_1, y_1, z_1) \) and \( B(x_2, y_2, z_2) \), then the distance between them is:
\( d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2} \)
Midpoint Between Two Points in 3D Space
The midpoint \( M \) between the points \( A \) and \( B \) is:
\( M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) \)
Example :
Find the distance between points \( A(1, 2, 3) \) and \( B(4, 6, 8) \).
▶️Answer/Explanation
$ d = \sqrt{(4 – 1)^2 + (6 – 2)^2 + (8 – 3)^2} = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5 \sqrt{2} $
Example :
Find the midpoint between points \( A(1, 2, 3) \) and \( B(4, 6, 8) \).
▶️ Answer/Explanation
$ M = \left( \frac{1 + 4}{2}, \frac{2 + 6}{2}, \frac{3 + 8}{2} \right) = \left( \frac{5}{2}, 4, \frac{11}{2} \right) $
Volume and Surface area of Three-Dimensional Solids
3D Shapes Terminology
Face: A flat or curved surface.
Edge: A line where two faces meet.
Vertex: A point where three or more edges meet.
Cuboid (Rectangular Prism)
Volume: $V = \ell \times w \times h$
Surface Area: $SA = 2\ell h + 2w h + 2\ell w$
where $\ell$ = length, $w$ = width, $h$ = height.
Cube
Volume: $V = s^3$
Surface Area: $SA = 6s^2$
where $s$ is the length of one side.
Pyramid (General)
Volume: $V = \frac{1}{3} A h$
where $A$ = base area, $h$ = height.
Regular Tetrahedron
Volume: $V = \frac{b^3}{6\sqrt{2}}$
Surface Area: $SA = \sqrt{3} b^2$
where $b$ = side of the base.
Square Pyramid
Volume: $V = \frac{1}{3} s^2 h$
Surface Area: $SA = s^2 + 2sh$
where $s$ = base side length, $h$ = vertical height.
Prism (General)
Volume: $V = A h$
Surface Area (Closed Prism):
$SA = 2A + (h \times p)$
where $A$ = base area, $h$ = height, $p$ = base perimeter.
Triangular Prism
Volume: $V = A \ell \quad \text{or} \quad \frac{1}{2} b h \ell$
Surface Area: $SA = bh + 2s + \ell b$
where $A$ = base area, $\ell$ = length, $b$ = base, $h$ = height, $s$ = slant length.
Sphere
Volume: $V = \frac{4}{3} \pi r^3$
Surface Area: $SA = 4\pi r^2$
Hemisphere
Volume: $V = \frac{2}{3} \pi r^3$
Surface Area (including base): $SA = 3\pi r^2$
Surface Area (without base): $SA = 2\pi r^2$
Right Cylinder
Volume: $V = \pi r^2 h$
Surface Area: $SA = 2\pi r (r + h)$
Right Circular Cone
Volume: $V = \frac{1}{3} \pi r^2 h$
Surface Area: $SA = \pi r (r + s)$
where $s$ = slant height.
Example:
A solid consists of a hemisphere of radius \( r = 3 \) cm mounted on top of a right circular cone of height \( h = 4 \) cm and base radius \( 3 \) cm. Find the total volume and the total surface area (excluding the base of the cone).
▶️ Answer/Explanation
Volume
\( V = V_{\text{cone}} + V_{\text{hemisphere}} \)
\( V_{\text{cone}} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (3^2)(4) = 12 \pi \text{ cm}^3 \)
\( V_{\text{hemisphere}} = \frac{1}{2} \left( \frac{4}{3} \pi r^3 \right) = \frac{2}{3} \pi (27) = 18 \pi \text{ cm}^3 \)
\( V_{\text{total}} = 12 \pi + 18 \pi = 30 \pi \approx 94.2 \text{ cm}^3 \)
Surface Area (excluding base of cone)
\( S = S_{\text{cone side}} + S_{\text{hemisphere curved}} \)
First, slant height of cone: \( l = \sqrt{r^2 + h^2} = \sqrt{9 + 16} = 5 \text{ cm} \)
\( S_{\text{cone}} = \pi r l = \pi \times 3 \times 5 = 15 \pi \)
\( S_{\text{hemisphere}} = 2 \pi r^2 = 2 \pi (9) = 18 \pi \)
\( S_{\text{total}} = 15 \pi + 18 \pi = 33 \pi \approx 103.7 \text{ cm}^2 \)
Example :
A solid consists of a sphere of radius \( 5 \) cm placed on top of a right circular cylinder of radius \( 5 \) cm and height \( 10 \) cm. Find the total volume and surface area (excluding cylinder base).
▶️ Answer/Explanation
Volume
\( V = V_{\text{cylinder}} + V_{\text{sphere}} \)
\( V_{\text{cylinder}} = \pi r^2 h = \pi (25)(10) = 250 \pi \text{ cm}^3 \)
\( V_{\text{sphere}} = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (125) = \frac{500}{3} \pi \approx 523.6 \text{ cm}^3 \)
\( V_{\text{total}} = 250 \pi + \frac{500}{3} \pi \approx 785.4 + 523.6 = 1309 \text{ cm}^3 \)
Surface Area
\( S = S_{\text{cylinder side}} + S_{\text{sphere}} \)
\( S_{\text{cylinder}} = 2 \pi r h = 2 \pi (5)(10) = 100 \pi \)
\( S_{\text{sphere}} = 4 \pi r^2 = 4 \pi (25) = 100 \pi \)
\( S_{\text{total}} = 100 \pi + 100 \pi = 200 \pi \approx 628.3 \text{ cm}^2 \)
Example :
A cone of base radius \( 3 \) cm and height \( 5 \) cm is placed on top of a square pyramid of base \( 6 \) cm and height \( 8 \) cm. Find the total volume.
▶️ Answer/Explanation
Volume
\( V_{\text{cone}} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (9)(5) = 15 \pi \)
\( V_{\text{pyramid}} = \frac{1}{3} \times (6^2) \times 8 = \frac{1}{3} \times 36 \times 8 = 96 \)
\( V_{\text{total}} = 15 \pi + 96 \approx 47.1 + 96 = 143.1 \text{ cm}^3 \)
Angle Between Two Lines or Planes
Angle Between Two Intersecting Lines
Two lines have direction vectors 𝐚 and 𝐛.
The angle \( \theta \) between them is given by:
\( \cos \theta = \frac{ \mathbf{a} \cdot \mathbf{b} }{ |\mathbf{a}| |\mathbf{b}| } \)
where:
- \( \mathbf{a} \cdot \mathbf{b} \) is the scalar (dot) product
- \( |\mathbf{a}| \) and \( |\mathbf{b}| \) are the magnitudes of the vectors
The angle \( \theta \) lies between \( 0^\circ \) and \( 90^\circ \).
Angle Between a Line and a Plane
If a line has direction vector 𝐝 and the plane has normal vector 𝐧, the angle \( \alpha \) between the line and the plane is given by:
\( \sin \alpha = \frac{ | \mathbf{d} \cdot \mathbf{n} | }{ |\mathbf{d}| |\mathbf{n}| } \)
Alternatively, compute the angle \( \beta \) between the line and the normal, and then \( \alpha = 90^\circ – \beta \).
The angle between a line and a plane is always between \( 0^\circ \) and \( 90^\circ \).
Example :
Find the angle between the lines with direction vectors:
- \( \mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} \)
- \( \mathbf{b} = \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix} \)
▶️ Answer/Explanation
$ \mathbf{a} \cdot \mathbf{b} = (1)(2) + (2)(1) + (2)(2) = 2 + 2 + 4 = 8 $
$ |\mathbf{a}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 $ $ |\mathbf{b}| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 $
$ \cos \theta = \frac{8}{3 \times 3} = \frac{8}{9} $
The angle is:
$ \theta = \cos^{-1}\left(\frac{8}{9}\right) \approx 27.27^\circ $
Example :
Find the angle between the line with direction vector:
\( \mathbf{d} = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} \)
and the plane with normal vector:
\( \mathbf{n} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \)
▶️ Answer/Explanation
dot product:
$ \mathbf{d} \cdot \mathbf{n} = (2)(1) + (3)(-1) + (1)(2) = 2 – 3 + 2 = 1 $
$ |\mathbf{d}| = \sqrt{2^2 + 3^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14} $ $ |\mathbf{n}| = \sqrt{1^2 + (-1)^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6} $
Then,
$ \sin \alpha = \frac{|1|}{\sqrt{14} \sqrt{6}} = \frac{1}{\sqrt{84}} \approx 0.109 $ $ \alpha = \sin^{-1}(0.109) \approx 6.26^\circ $