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IB Mathematics AI AHL Vector equation of a line in two and three dimensions MAI Study Notes - New Syllabus

IB Mathematics AI AHL Vector equation of a line in two and three dimensions MAI Study Notes

LEARNING OBJECTIVE

  • Vector equation of a line in two and three dimensions:

Key Concepts: 

  • Vector Equations of Lines
  • Shortest Distances with Lines

MAI HL and SL Notes – All topics

VECTOR EQUATION OF A LINE IN 2D

◆ VECTOR EQUATION
Let:

\( A(a_1, a_2) \) be a point with position vector \( \vec{a} = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} \).
 \( \vec{b} = \begin{pmatrix} b_1 \\ b_2 \end{pmatrix} \) be a direction vector.

The vector equation of the line passing through \( A \) and parallel to \( \vec{b} \) is:

$
\vec{r} = \vec{a} + \lambda \vec{b}, \quad \text{or} \quad \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} + \lambda \begin{pmatrix} b_1 \\ b_2 \end{pmatrix},
$
where \( \lambda \) is a parameter.

◆ PARAMETRIC EQUATIONS

From the vector equation:

$
x = a_1 + \lambda b_1, \quad y = a_2 + \lambda b_2.
$

◆ CARTESIAN EQUATION

Solve both parametric equations for \( \lambda \):

$
\frac{x – a_1}{b_1} = \frac{y – a_2}{b_2}.
$

Example

Given point \( A(1, 2) \) and direction vector \( \vec{b} = \begin{pmatrix} 3 \\ 4 \end{pmatrix} \):

Vector equation:
Parametric equations:
Cartesian equation:

▶️Answer/Explanation

Solution:

Vector equation:
$
\vec{r} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ 4 \end{pmatrix}.
$
Parametric equations:
$
x = 1 + 3\lambda, \quad y = 2 + 4\lambda.
$
 Cartesian equation:
$
\frac{x – 1}{3} = \frac{y – 2}{4} \quad \text{or} \quad 4x – 3y + 2 = 0.
$

◆ GIVEN TWO POINTS

For points \( A(a_1, a_2) \) and \( B(b_1, b_2) \), the line equation is:

$
\vec{r} = \vec{a} + \lambda (\vec{b} – \vec{a}).
$

Example

Find the line through \( A(1, 2) \) and \( B(4, 7) \):

Direction vector: \( \overrightarrow{AB} = \begin{pmatrix} 3 \\ 5 \end{pmatrix} \).
Vector equation:

▶️Answer/Explanation

Solution:

$
\vec{r} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ 5 \end{pmatrix}.
$

◆ INTERSECTION OF TWO LINES

Given lines:

$
\vec{r}_1 = \vec{a}_1 + \lambda \vec{b}_1, \quad \vec{r}_2 = \vec{a}_2 + \mu \vec{b}_2,
$

set \( \vec{r}_1 = \vec{r}_2 \) and solve for \( \lambda \) and \( \mu \).

Example

Find the intersection of: 

$
\vec{r}_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ 4 \end{pmatrix}, \quad \vec{r}_2 = \begin{pmatrix} 2 \\ -2 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 4 \end{pmatrix}.
$

▶️Answer/Explanation

Solution:

$
\begin{cases}
1 + 3\lambda = 2 + \mu, \\
2 + 4\lambda = -2 + 4\mu.
\end{cases} \quad \Rightarrow \quad \lambda = 1, \mu = 2.
$
Intersection point: \( (4, 6) \).

 VECTOR EQUATION OF A LINE IN 3D

◆ VECTOR EQUATION
For a line through \( A(1, 2, 3) \) parallel to \( \vec{b} = \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix} \):

$
\vec{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix}.
$

◆ PARAMETRIC EQUATIONS

$
x = 1 + 4\lambda, \quad y = 2 + 5\lambda, \quad z = 3 + 6\lambda.
$

◆ CARTESIAN EQUATIONS

Solve for \( \lambda \):

$
\frac{x – 1}{4} = \frac{y – 2}{5} = \frac{z – 3}{6}.
$

◆

Example

Find the line through \( A(1, 2, 3) \) and \( B(5, 2, -1) \):

Direction vector: \( \overrightarrow{AB} = \begin{pmatrix} 4 \\ 0 \\ -4 \end{pmatrix} \).
Vector equation:

▶️Answer/Explanation

Solution:

$
\vec{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ 0 \\ -4 \end{pmatrix}.
$

◆ INTERSECTION OF TWO LINES

In 3D, lines can be:

1. Parallel: \( \vec{b}_1 = k \vec{b}_2 \).
2. Intersecting: Solve \( \vec{r}_1 = \vec{r}_2 \).
3. Skew: Neither parallel nor intersecting.

Example

Check if the lines intersect:

$
\vec{r}_1 = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ 4 \\ 5 \end{pmatrix}, \quad \vec{r}_2 = \begin{pmatrix} 1 \\ 4 \\ 4 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ 2 \\ 3 \end{pmatrix}.
$

▶️Answer/Explanation

Solution:

$
\begin{cases}
1 + 3\lambda = 1 + 2\mu, \\
2 + 4\lambda = 4 + 2\mu, \\
3 + 5\lambda = 4 + 3\mu.
\end{cases} \quad \Rightarrow \quad \lambda = 2, \mu = 3.
$
Intersection point: \( (7, 10, 13) \).

VECTOR EQUATION OF A LINE IN 3D (Distance Formulas)

1. Distance Between Two Points
Given points \( A(x_1,y_1,z_1) \) and \( B(x_2,y_2,z_2) \):

$
d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}
$

Example

For A and B

\( A(1,2,3) \) and \( B(2,7,9) \):

Find distance between them.

▶️Answer/Explanation

Solution:

$
d = \sqrt{(2-1)^2 + (7-2)^2 + (9-3)^2} = \sqrt{1+25+36} = \sqrt{62}
$

Example

Find the distance of A from line l.

Point \( A(1,2,3) \)

Line \( L: \vec{r} = \begin{pmatrix}3\\7\\9\end{pmatrix} + \lambda\begin{pmatrix}3\\2\\1\end{pmatrix} \)

▶️Answer/Explanation

Solution:

1. Let \( P(3+3\lambda, 7+2\lambda, 9+\lambda) \) be a general point on L
2. Vector \( \overrightarrow{AP} = \begin{pmatrix}2+3\lambda\\5+2\lambda\\6+\lambda\end{pmatrix} \)
3. \( \overrightarrow{AP} \) must be perpendicular to direction vector \( \vec{b} = \begin{pmatrix}3\\2\\1\end{pmatrix} \):
$
\overrightarrow{AP} \cdot \vec{b} = 0 \Rightarrow 3(2+3\lambda) + 2(5+2\lambda) + 1(6+\lambda) = 0
$
$
6+9\lambda + 10+4\lambda + 6+\lambda = 0 \Rightarrow 14\lambda = -22 \Rightarrow \lambda = -11/7
$
4. Find foot \( P \) and calculate \( |AP| \):
$
P = \left(3-\frac{33}{7}, 7-\frac{22}{7}, 9-\frac{11}{7}\right) = \left(-\frac{12}{7}, \frac{27}{7}, \frac{52}{7}\right)
$
$
|AP| = \sqrt{\left(-\frac{19}{7}\right)^2 + \left(\frac{13}{7}\right)^2 + \left(\frac{31}{7}\right)^2} = \frac{\sqrt{1611}}{7}
$

Alternative Formula:
$
d = \frac{|\overrightarrow{AQ} \times \vec{b}|}{|\vec{b}|}
$
where \( Q \) is any point on the line.

3. Distance Between Parallel Lines

For lines \( L_1: \vec{r} = \vec{a}_1 + \lambda\vec{b} \) and \( L_2: \vec{r} = \vec{a}_2 + \mu\vec{b} \):

1. Pick point \( A \) on \( L_1 \)
2. Find distance from \( A \) to \( L_2 \) (as above)

4. Distance Between Skew Lines

Given non-parallel, non-intersecting lines:

$
L_1: \vec{r} = \vec{a}_1 + \lambda\vec{b}_1
$
$
L_2: \vec{r} = \vec{a}_2 + \mu\vec{b}_2
$

Method:

1. Find vector \( \overrightarrow{A_1A_2} = \vec{a}_2 – \vec{a}_1 \)
2. Compute \( \vec{b}_1 \times \vec{b}_2 \)
3. Distance:
$
d = \frac{|(\vec{a}_2 – \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}
$

Example

For lines:

$
L_1: \begin{pmatrix}1\\2\\3\end{pmatrix} + \lambda\begin{pmatrix}4\\5\\0\end{pmatrix}, \quad L_2: \begin{pmatrix}5\\7\\9\end{pmatrix} + \mu\begin{pmatrix}3\\2\\1\end{pmatrix}
$

Find the distance between the,.

▶️Answer/Explanation

Solution:

1. \( \vec{a}_2 – \vec{a}_1 = \begin{pmatrix}4\\5\\6\end{pmatrix} \)
2. \( \vec{b}_1 \times \vec{b}_2 = \begin{pmatrix}5\\-4\\-7\end{pmatrix} \)
3. \( d = \frac{|4(5)+5(-4)+6(-7)|}{\sqrt{25+16+49}} = \frac{42}{\sqrt{90}} = \frac{7\sqrt{10}}{5} \)

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