IB Mathematics AA Vector equation of a line in planes Study Notes
IB Mathematics AA Vector equation of a line in planes Study Notes
IB Mathematics AA Vector equation of a line in planes Notes Offer a clear explanation of Vector equation of a line in planes, 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 Vector equation of a line in planes.
Vector Equation of a Line (General Form)
Vector Equation of a Line (General Form)
A line through point \( A \) with position vector \( \mathbf{a} \), and in the direction of vector \( \mathbf{d} \), is given by:
\( \mathbf{r} = \mathbf{a} + \lambda \mathbf{d} \)
where:
- \( \mathbf{r} \) is the position vector of a general point on the line
- \( \lambda \in \mathbb{R} \) is a scalar parameter
In 2D (plane)
\( \mathbf{r} = \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} + \lambda \begin{pmatrix} d_1 \\ d_2 \end{pmatrix} \)
In 3D (space)
\( \mathbf{r} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} + \lambda \begin{pmatrix} d_1 \\ d_2 \\ d_3 \end{pmatrix} \)
Parametric Form
From the vector form, we can write:
- 2D: \( x = a_1 + \lambda d_1 \), \( y = a_2 + \lambda d_2 \)
- 3D: \( x = a_1 + \lambda d_1 \), \( y = a_2 + \lambda d_2 \), \( z = a_3 + \lambda d_3 \)
Symmetric Form (3D)
\( \frac{x – a_1}{d_1} = \frac{y – a_2}{d_2} = \frac{z – a_3}{d_3} \)
where \( d_1, d_2, d_3 \ne 0 \).
Example
Find the vector equation of the line through point \( (1, 2, 3) \) and parallel to vector \( (4, -2, 5) \).
▶️ Answer/Explanation
position vector of point:
\( \mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \)
Direction vector:
\( \mathbf{d} = \begin{pmatrix} 4 \\ -2 \\ 5 \end{pmatrix} \)
Vector equation:
\( \mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ -2 \\ 5 \end{pmatrix} \)
Parametric form:
\( x = 1 + 4\lambda \), \( y = 2 – 2\lambda \), \( z = 3 + 5\lambda \)
Angle Between Two Lines
Angle Between Two Lines
If two lines have direction vectors \( \mathbf{d_1} \) and \( \mathbf{d_2} \), the angle \( \theta \) between the lines is the angle between these vectors:
\( \cos \theta = \frac{ \mathbf{d_1} \cdot \mathbf{d_2} }{ |\mathbf{d_1}| |\mathbf{d_2}| } \)
- \( \mathbf{d_1} \cdot \mathbf{d_2} \) is the scalar (dot) product of the vectors.
- \( |\mathbf{d_1}| \) and \( |\mathbf{d_2}| \) are the magnitudes of the direction vectors.
- \( 0 \le \theta \le 90^\circ \) or \( 0 \le \theta \le \frac{\pi}{2} \)
Magnitude of a vector:
\( |\mathbf{d}| = \sqrt{ d_1^2 + d_2^2 + d_3^2 } \)
Example
Find the angle between the lines:
- Line 1: \( \mathbf{r} = \mathbf{a_1} + \lambda \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} \)
- Line 2: \( \mathbf{r} = \mathbf{a_2} + \mu \begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix} \)
▶️ View Solution
Direction vectors:
\( \mathbf{d_1} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} \), \( \mathbf{d_2} = \begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix} \)
Dot product:
\( \mathbf{d_1} \cdot \mathbf{d_2} = (2)(1) + (-1)(4) + (3)(-2) = 2 – 4 – 6 = -8 \)
Magnitudes:
\( |\mathbf{d_1}| = \sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{4 + 1 + 9} = \sqrt{14} \)
\( |\mathbf{d_2}| = \sqrt{1^2 + 4^2 + (-2)^2} = \sqrt{1 + 16 + 4} = \sqrt{21} \)
Find angle:
\( \cos \theta = \frac{-8}{\sqrt{14} \sqrt{21}} \)
\( \cos \theta = \frac{-8}{\sqrt{294}} \approx -0.466 \)
\( \theta = \cos^{-1}(-0.466) \approx 118^\circ \)
Simple Applications to Kinematics
Simple Applications to Kinematics
In kinematics, the vector equation of a line:
\( \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} \)
can represent the position of a particle at time \( \lambda \):
- \( \mathbf{a} \): Initial position vector
- \( \mathbf{b} \): Velocity vector (constant)
- \( \lambda \): Time parameter
- \( |\mathbf{b}| \): Speed (magnitude of velocity)
The particle moves along a straight path at constant velocity. The position at any time \( \lambda \) is determined by substituting \( \lambda \) into the equation.
Example
A particle starts at position \( \mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \) and moves with constant velocity \( \mathbf{b} = \begin{pmatrix} 4 \\ -2 \\ 1 \end{pmatrix} \).
Find:
- The position at \( t = 3 \)
- The speed of the particle
▶️ Answer/Explanation
Position at \( t = 3 \):
\( \mathbf{r} = \mathbf{a} + 3 \mathbf{b} \)
\( = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + 3 \begin{pmatrix} 4 \\ -2 \\ 1 \end{pmatrix} \)
\( = \begin{pmatrix} 1 + 12 \\ 2 – 6 \\ 3 + 3 \end{pmatrix} \)
\( = \begin{pmatrix} 13 \\ -4 \\ 6 \end{pmatrix} \)
Speed:
\( |\mathbf{b}| = \sqrt{4^2 + (-2)^2 + 1^2} \)
\( = \sqrt{16 + 4 + 1} = \sqrt{21} \)
\( \approx 4.58 \) units per time unit