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IB Mathematics AA Different types of lines Study Notes

IB Mathematics AA Different types of lines Study Notes

IB Mathematics AA Different types of lines Study Notes

IB Mathematics AA Different types of lines Notes Offer a clear explanation of Different types of lines, 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 Different types of lines.

Different Types of Lines

Introduction

In geometry, understanding the relationships between different types of lines is fundamental for analyzing their properties and interactions. Lines can be classified based on their relative positions and whether they intersect in two-dimensional or three-dimensional space.

Types of Lines

1. Coincident Lines

Coincident lines lie exactly on top of each other. They have the same slope and the same y-intercept in a two-dimensional plane. In equations, two lines \( y = mx + c \) and \( y = mx + c \) are coincident if they are identical.

Example: \( y = 2x + 3 \) and \( y = 2x + 3 \) represent coincident lines.

2. Parallel Lines

Parallel lines have the same slope but different y-intercepts. These lines never intersect and remain equidistant from each other at all points.

 

Mathematical Property: Two lines \( y = m_1x + c_1 \) and \( y = m_2x + c_2 \) are parallel if \( m_1 = m_2 \) and \( c_1 \neq c_2 \).

Example: \( y = 3x + 2 \) and \( y = 3x – 4 \) are parallel lines.

3. Intersecting Lines

Intersecting lines cross each other at a single point in the same plane. At the point of intersection, the coordinates satisfy the equations of both lines.

Finding the Point of Intersection: Solve the system of equations representing the lines:

\[ \begin{aligned} & y = m_1x + c_1 \\ & y = m_2x + c_2 \end{aligned} \]

Example: The lines \( y = 2x + 1 \) and \( y = -x + 4 \) intersect at \( (1, 3) \).

4. Skew Lines

Skew lines exist only in three-dimensional space. They are non-parallel and do not intersect because they lie in different planes.

Example: In 3D space, the lines:

  • \( L_1: \vec{r} = (1, 2, 3) + t(2, -1, 0) \)
  • \( L_2: \vec{r} = (4, 5, 6) + s(0, 1, -1) \)

are skew lines because they do not lie on the same plane and do not intersect.

Key Differences Between the Types

Type of LineKey Characteristics
CoincidentIdentical lines; overlap completely.
ParallelSame slope; never intersect; equidistant.
IntersectingCross at a single point; different slopes in 2D.
SkewNon-parallel and non-intersecting; exist only in 3D space.

Additional Notes on Skew Lines

  • Skew lines do not share a common plane (they are non-coplanar).
  • They cannot be visualized in 2D space.
  • Applications of skew lines appear in 3D geometry, robotics, and engineering.

Points of Intersection

The point of intersection is the coordinate pair (or triplet in 3D) where two lines meet. For two lines represented by equations, the point of intersection can be found by solving the equations simultaneously:

\[ \begin{aligned} & y = m_1x + c_1 \\ & y = m_2x + c_2 \end{aligned} \]

Substituting the value of \( x \) obtained into either equation gives the corresponding \( y \) value.

Example: For \( y = 3x + 2 \) and \( y = -2x + 8 \):

\[ \begin{aligned}
& 3x + 2 = -2x + 8 \\
& 3x + 2x = 8 – 2 \\
& 5x = 6 \\
& x = \frac{6}{5}
\end{aligned} \]

Substitute \( x = \frac{6}{5} \) into \( y = 3x + 2 \):

\[ y = 3\left(\frac{6}{5}\right) + 2 = \frac{18}{5} + 2 = \frac{28}{5} \]

The point of intersection is \( \left(\frac{6}{5}, \frac{28}{5}\right) \).

Connections to Other Topics

  • Unit Vectors: Directional components of lines.
  • Vectors: Representing lines in vector form aids in distinguishing skew lines.
  • Geometry in 3D: Understanding skew lines is crucial for analyzing spatial relationships.

IB Mathematics AA SL Different types of lines Exam Style Worked Out Questions

Question

Consider the plane \({\mathit{\Pi} _1}\), parallel to both lines \({L_1}\) and \({L_2}\). Point C lies in the plane \({\mathit{\Pi} _1}\).

The line \({L_3}\) has vector equation \(\boldsymbol{r} = \left( \begin{array}{l}3\\0\\1\end{array} \right) + \lambda \left( \begin{array}{c}k\\1\\ – 1\end{array} \right)\).

The plane \({\mathit{\Pi} _2}\) has Cartesian equation \(x + y = 12\).

The angle between the line \({L_3}\) and the plane \({\mathit{\Pi} _2}\) is 60°.

Given the points A(1, 0, 4), B(2, 3, −1) and C(0, 1, − 2) , find the vector equation of the line \({L_1}\) passing through the points A and B.[2]

a.

The line \({L_2}\) has Cartesian equation \(\frac{{x – 1}}{3} = \frac{{y + 2}}{1} = \frac{{z – 1}}{{ – 2}}\).

Show that \({L_1}\) and \({L_2}\) are skew lines.[5]

b.

Find the Cartesian equation of the plane \({\Pi _1}\).[4]

c.

(i)     Find the value of \(k\).

(ii)     Find the point of intersection P of the line \({L_3}\) and the plane \({\mathit{\Pi} _2}\).[7]

d.
▶️Answer/Explanation

Markscheme

direction vector \(\overrightarrow {{\rm{AB}}}  = \left( \begin{array}{c}1\\3\\ – 5\end{array} \right)\) or \(\overrightarrow {{\rm{BA}}}  = \left( \begin{array}{c} – 1\\ – 3\\5\end{array} \right)\)     A1

\(\boldsymbol{r} = \left( \begin{array}{l}1\\0\\4\end{array} \right) + t\left( \begin{array}{c}1\\3\\ – 5\end{array} \right)\) or \(\boldsymbol{r} = \left( \begin{array}{c}2\\3\\ – 1\end{array} \right) + t\left( \begin{array}{c}1\\3\\ – 5\end{array} \right)\) or equivalent     A1

Note:     Do not award final A1 unless ‘\(\boldsymbol{r} = {\text{K}}\)’ (or equivalent) seen.

     Allow FT on direction vector for final A1.

[2 marks]

a.

both lines expressed in parametric form:

\({L_1}\):

\(x = 1 + t\)

\(y = 3t\)

\(z = 4 – 5t\)

\({L_2}\):

\(x = 1 + 3s\)

\(y =  – 2 + s\)     M1A1

\(z =  – 2s + 1\)

Notes:     Award M1 for an attempt to convert \({L_2}\) from Cartesian to parametric form.

     Award A1 for correct parametric equations for \({L_1}\) and \({L_2}\).

     Allow M1A1 at this stage if same parameter is used in both lines.

attempt to solve simultaneously for x and y:     M1

\(1 + t = 1 + 3s\)

\(3t =  – 2 + s\)

\(t =  – \frac{3}{4},{\text{ }}s =  – \frac{1}{4}\)     A1

substituting both values back into z values respectively gives \(z = \frac{{31}}{4}\)

and \(z = \frac{3}{2}\) so a contradiction     R1

therefore \({L_1}\) and \({L_1}\) are skew lines     AG

[5 marks]

b.

finding the cross product:

\(\left( \begin{array}{c}1\\3\\ – 5\end{array} \right) \times \left( \begin{array}{c}3\\1\\ – 2\end{array} \right)\)     (M1)

= – i  – 13j  – 8k     A1

Note:     Accept i  + 13j  + 8k

\( – 1(0) – 13(1) – 8( – 2) = 3\)     (M1)

\( \Rightarrow  – x – 13y – 8z = 3\) or equivalent     A1

[4 marks]

c.

(i)     \((\cos \theta  = )\frac{{\left( \begin{array}{c}k\\1\\ – 1\end{array} \right) \bullet \left( \begin{array}{l}1\\1\\0\end{array} \right)}}{{\sqrt {{k^2} + 1 + 1}  \times \sqrt {1 + 1} }}\)     M1

Note:     Award M1 for an attempt to use angle between two vectors formula.

\(\frac{{\sqrt 3 }}{2} = \frac{{k + 1}}{{\sqrt {2({k^2} + 2)} }}\)     A1

obtaining the quadratic equation

\(4{(k + 1)^2} = 6({k^2} + 2)\)     M1

\({k^2} – 4k + 4 = 0\)

\({(k – 2)^2} = 0\)

\(k = 2\)     A1

Note:     Award M1A0M1A0 if \(\cos 60^\circ \) is used \((k = 0{\text{ or }}k =  – 4)\).

(ii)     \(r = \left( \begin{array}{l}3\\0\\1\end{array} \right) + \lambda \left( \begin{array}{c}2\\1\\ – 1\end{array} \right)\)

substituting into the equation of the plane \({\Pi _2}\):

\(3 + 2\lambda  + \lambda  = 12\)     M1

\(\lambda  = 3\)     A1

point P has the coordinates:

(9, 3, –2)   A1

Notes:     Accept 9i  + 3j  – 2k and \(\left( \begin{array}{l}9\\3\\- 2\end{array} \right)\).

     Do not allow FT if two values found for k.

[7 marks]

d.

Question

Two boats, A and B , move so that at time t hours, their position vectors, in kilometres, are r\(_A\) = (9t)i + (3 – 6t)j and r\(_B\) = (7 – 4t)i + (7t – 6)j .

a.Find the coordinates of the common point of the paths of the two boats.[4]

 

b.Show that the boats do not collide.[2]

 
▶️Answer/Explanation

Markscheme

METHOD 1

\(9{t_A} = 7 – 4{t_B}\) and

\(3 – 6{t_A} = – 6 + 7{t_B}\)     M1A1

solve simultaneously

\({t_A} = \frac{1}{3},{\text{ }}{t_B} = 1\)     A1 

Note: Only need to see one time for the A1.

therefore meet at (3, 1)     A1

[4 marks]

METHOD 2

path of A is a straight line: \(y = – \frac{2}{3}x + 3\)     M1A1 

Note: Award M1 for an attempt at simultaneous equations.

path of B is a straight line: \(y = – \frac{7}{4}x + \frac{{25}}{4}\)     A1

\( – \frac{2}{3}x + 3 = – \frac{7}{4}x + \frac{{25}}{4}{\text{ }}( \Rightarrow x = 3)\)

so the common point is (3, 1)     A1

[4 marks]

a.

METHOD 1

boats do not collide because the two times \(\left( {{t_A} = \frac{1}{3},{\text{ }}{t_B} = 1} \right)\)     (A1)

are different     R1

[2 marks]

METHOD 2

for boat A, \(9t = 3 \Rightarrow t = \frac{1}{3}\) and for boat B, \(7 – 4t = 3 \Rightarrow t = 1\)

times are different so boats do not collide     R1AG

[2 marks]

b.

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