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IB Mathematics AI SL The distance between two points MAI Study Notes- New Syllabus

IB Mathematics AI SL The distance between two points MAI Study Notes

LEARNING OBJECTIVE

  • The distance between two points in three dimensional space, and their midpoint.

Key Concepts: 

  • Coordinate Geometry

MAI HL and SL Notes – All topics

THREE-DIMENSIONAL GEOMETRY

◆ 3D COORDINATE GEOMETRY

We know that a point in the Cartesian plane has the form \( P(x, y) \). In 3D space we add one more coordinate, thus a point has the form \( P(x, y, z) \).

The distance between two points \( A(x_1, y_1, z_1) \) and \( B(x_2, y_2, z_2) \) is given by:


$ d_{AB} = \sqrt{(x_1 – x_2)^2 + (y_1 – y_2)^2 + (z_1 – z_2)^2} $

while the midpoint of the line segment \( AB \) is given by:

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

Example

Let \( A(1, 0, 5) \) and \( B(2, 3, 1) \). 

Find: 

(a) the distance between \( A \) and \( B \)
(b) the distance between \( O \) and \( B \)
(c) the coordinates of the midpoint \( M \) of the line segment \( [AB] \)
(d) the coordinates of point \( C \) given that \( B \) is the midpoint of \( [AC] \).

▶️Answer/Explanation

Solution:

(a)
$ d_{AB} = \sqrt{(1-2)^2 + (0-3)^2 + (5-1)^2} = \sqrt{1+9+16} = \sqrt{26} $

(b)
$ d_{OB} = \sqrt{2^2 + 3^2 + 1^2} = \sqrt{14} $

(c)
$ M\left(\frac{1+2}{2}, \frac{0+3}{2}, \frac{5+1}{2}\right) \quad \text{i.e. } M\left(\frac{3}{2}, \frac{3}{2}, 3\right) $

(d)
$ C(3, 6, -3) $

Notice: the coordinates of \( A, B, C \) (\( B \) midpoint) form arithmetic sequences:
\( x: 1, 2, 3 \)
\( y: 0, 3, 6 \)
\( z: 5, 1, -3 \)

VOLUMES AND SURFACE AREAS OF KNOWN 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

The volume and the surface area for the following solids:

▶️Answer/Explanation

Solution:

Cube:
\( V = x \cdot x \cdot x = x^3 \)
\( S = 6x^2 \)

Cuboid of square base:
\( V = x^2 y \)
\( S = 2x^2 + 4xy \)

Example

Given that the volume of a cylinder is 25,

(a) express \( h \) in terms of \( r \)

(b) hence express the surface area in terms of \( r \).

▶️Answer/Explanation

Solution:

(a)
$ V = \pi r^2 h \Rightarrow \pi r^2 h = 25 \Rightarrow h = \frac{25}{\pi r^2} $

(b)
$ S = 2\pi rh + 2\pi r^2 = 2\pi r \cdot \frac{25}{\pi r^2} + 2\pi r^2 = \frac{50}{r} + 2\pi r^2 $

Example

Given that the surface area of a cylinder is \( 100\pi \),

(a) express \( h \) in terms of \( r \)

(b) hence express the volume in terms of \( r \).

▶️Answer/Explanation

Solution:

(a)
$ S = 2\pi rh + 2\pi r^2 \Rightarrow 2\pi rh + 2\pi r^2 = 100\pi \Rightarrow h = \frac{50 – r^2}{r} $

(b)
$ V = \pi r^2 h = \pi r^2 \cdot \frac{50 – r^2}{r} = \pi r(50 – r^2) = 50\pi r – \pi r^3 $

Example

Find the volume and the surface area of a right pyramid of square base of side 6 and vertical height 4 .

▶️Answer/Explanation

Solution:

The vertical height is $h=4$.
For the slant height AM we use the Pythagoras theorem on ANM.

$A M^2=A N^2+N M^2 \Leftrightarrow A M^2=4^2+3^2 \Leftrightarrow A M=5$

The area of the triangle $A E D$ (and any side triangle) is

$A=\frac{1}{2} \times E D \times A M=\frac{1}{2} \times 6 \times 5=15$

The volume is $\quad V=\frac{1}{3}$ (area of base) $\times($ height $)=\frac{1}{3} \times 6^2 \times 4=48$
The surface area is $S=($ area of square base $)+4 A=6^2+4 \times(15)=96$

Angles between lines and planes

Angle Between Two Lines in 3D Space

The angle between two lines in 3-dimensional Euclidean space is the angle they make between each other. Even if the lines are skew (not intersecting and not parallel), they still define an angle between them.

Line 1:

$
\frac{x – x_1}{a_1} = \frac{y – y_1}{b_1} = \frac{z – z_1}{c_1}
$

Direction ratios:

$a_1, b_1, c_1$

Line 2:

$
\frac{x – x_2}{a_2} = \frac{y – y_2}{b_2} = \frac{z – z_2}{c_2}
$

Direction ratios:

$a_2, b_2, c_2$

Direction Vectors

The direction vectors parallel to these lines are:

$\mathbf{m}_1 = a_1 \hat{i} + b_1 \hat{j} + c_1 \hat{k}$
$\mathbf{m}_2 = a_2 \hat{i} + b_2 \hat{j} + c_2 \hat{k}$

Angle Between the Two Vectors

The angle $\theta$ between the two direction vectors is given by the dot product formula:

$
\theta = \cos^{-1} \left( \frac{\mathbf{m}_1 \cdot \mathbf{m}_2}{\|\mathbf{m}_1\| \cdot \|\mathbf{m}_2\|} \right)
$

Where:

$\mathbf{m}_1 \cdot \mathbf{m}_2 = a_1a_2 + b_1b_2 + c_1c_2$
$\|\mathbf{m}_1\| = \sqrt{a_1^2 + b_1^2 + c_1^2}$
$\|\mathbf{m}_2\| = \sqrt{a_2^2 + b_2^2 + c_2^2}$

Example

Find the angle between the lines:

Line 1:

$
\frac{x – 1}{2} = \frac{y + 3}{-1} = \frac{z – 2}{4}
$

Line 2:

$
\frac{x + 2}{-1} = \frac{y – 1}{2} = \frac{z}{-2}
$

▶️Answer/Explanation

Solution:

For Line 1: $\mathbf{m}_1 = 2\hat{i} – 1\hat{j} + 4\hat{k}$
For Line 2: $\mathbf{m}_2 = -1\hat{i} + 2\hat{j} – 2\hat{k}$

$
\mathbf{m}_1 \cdot \mathbf{m}_2 = (2)(-1) + (-1)(2) + (4)(-2) = -2 -2 -8 = -12
$

$
\|\mathbf{m}_1\| = \sqrt{2^2 + (-1)^2 + 4^2} = \sqrt{4 + 1 + 16} = \sqrt{21}
$
$
\|\mathbf{m}_2\| = \sqrt{(-1)^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3
$
$
\cos \theta = \frac{-12}{\sqrt{21} \cdot 3} = \frac{-12}{3\sqrt{21}} = \frac{-4}{\sqrt{21}}
$
$
\theta = \cos^{-1} \left( \frac{-4}{\sqrt{21}} \right) \approx \cos^{-1}(-0.8739)
$
$
\theta \approx 150.5^\circ
$

Angle Between Lines and plane in 3D Space

Angle between line $A M$ and plane $B C D E=$ angle $A \hat{M} N$
Angle between line $A D$ and plane $B C D E=$ angle $A \hat{D} N$
Angle between the planes $A D E$ and $B C D E=$ angle $A \hat{M} N$
Angle between the planes $A C B$ and $A D E=$ angle $M \hat{A} M^{\prime}=2 \times M \hat{A} N$

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