IB Mathematics AA Intersections of a line with a planes Study Notes - New Syllabus
IB Mathematics AA Intersections of a line with a planes Study Notes
IB Mathematics AA Intersections of a line with a planes Notes Offer a clear explanation of Intersections of a line with a 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 Intersections of a line with a planes.
Intersections of a Line with a Plane
Introduction
The intersection of a line and a plane is a fundamental concept in 3D geometry. Determining the point of intersection (if it exists) involves solving equations derived from their respective representations. This topic also extends to analyzing the relationships between two or more planes and the angles between geometric objects in space.
Intersection of a Line and a Plane
The general vector equation of a line is:
\[ \vec{r} = \vec{a} + \lambda \vec{b} \]
Here:
- \( \vec{r} \): Position vector of any point on the line.
- \( \vec{a} \): Position vector of a known point on the line.
- \( \vec{b} \): Direction vector of the line.
- \( \lambda \): Scalar parameter.
The vector equation of a plane is:
\[ \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n} \]
or equivalently in Cartesian form:
\[ ax + by + cz = d \]
Here:
- \( \vec{n} \): Normal vector to the plane.
- \( a, b, c \): Components of \( \vec{n} \).
- \( d \): A constant representing \( \vec{a} \cdot \vec{n} \), where \( \vec{a} \) is a known point on the plane.
Finding the Intersection
To find the point of intersection of the line and plane:
- Substitute the parametric equation of the line \( \vec{r} = \vec{a} + \lambda \vec{b} \) into the equation of the plane \( ax + by + cz = d \).
- Solve for the parameter \( \lambda \).
- Use \( \lambda \) to calculate the coordinates of the intersection point.
Geometric Interpretation
- If a solution exists, the line intersects the plane at a single point.
- If no solution exists, the line is parallel to the plane.
- If the line lies entirely within the plane, the equations will be dependent.
Intersection of Two or More Planes
- Two Planes: The intersection of two planes is either a line (if they are not parallel) or they do not intersect (if they are parallel).
- Three Planes: Three planes may intersect at a point, form a line, or have no intersection (depending on their orientations).
Solving Intersections
To find intersections involving planes, solve the system of linear equations derived from their Cartesian forms:
\[ a_1x + b_1y + c_1z = d_1 \]
\[ a_2x + b_2y + c_2z = d_2 \]
\[ a_3x + b_3y + c_3z = d_3 \]
Angle Between a Line and a Plane
The angle \( \theta \) between a line (direction vector \( \vec{b} \)) and a plane (normal vector \( \vec{n} \)) is given by:
\[ \sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|} \]
Angle Between Two Planes
The angle \( \phi \) between two planes is determined by the angle between their normal vectors \( \vec{n_1} \) and \( \vec{n_2} \):
\[ \cos \phi = \frac{\vec{n_1} \cdot \vec{n_2}}{|\vec{n_1}| |\vec{n_2}|} \]
Examples
Example 1: Intersection of a Line and a Plane
Given the line \( \vec{r} = [1, 2, 3] + \lambda [2, 1, -1] \) and the plane \( 2x + y – z = 5 \):
- Substitute \( x = 1 + 2\lambda \), \( y = 2 + \lambda \), \( z = 3 – \lambda \) into \( 2x + y – z = 5 \):
\[ 2(1 + 2\lambda) + (2 + \lambda) – (3 – \lambda) = 5 \]
- Simplify to find \( \lambda = 1 \).
- Substitute \( \lambda = 1 \) into the parametric equations to find the intersection point \( [3, 3, 2] \).
Example 2: Angle Between a Line and a Plane
For a line with direction vector \( \vec{b} = [2, 1, -1] \) and a plane with normal vector \( \vec{n} = [2, 1, 1] \):
\[ \sin \theta = \frac{|(2)(2) + (1)(1) + (-1)(1)|}{\sqrt{(2^2 + 1^2 + (-1)^2)} \sqrt{(2^2 + 1^2 + 1^2)}} \]
\[ \sin \theta = \frac{|4 + 1 – 1|}{\sqrt{6} \sqrt{6}} = \frac{4}{6} \]
\[ \theta = \arcsin\left(\frac{2}{3}\right) \]
Applications
- Engineering: Designing intersections in 3D structures.
- Physics: Analyzing forces and trajectories in space.
- Computer Graphics: Rendering and collision detection.
IB Mathematics AA SL Intersections of a line with a planes Exam Style Worked Out Questions
Question
Consider the three planes
∏1 : 2x – y + z = 4
∏2 : x – 2y + 3z = 5
∏3 : -9x + 3y – 2z = 32
(a) Show that the three planes do not intersect.
(b) (i) Verify that the point P(1 , -2 , 0) lies on both ∏1 and ∏2 .
(ii) Find a vector equation of L, the line of intersection of ∏1 and ∏2 .
(c) Find the distance between L and ∏3 .
▶️Answer/Explanation
Ans:
(a) METHOD 1
attempt to eliminate a variable
obtain a pair of equations in two variables
EITHER
-3x + z = -3 and
-3x + z = 44
OR
-5x + y = -7 and
-5x + y = 40
OR
3x – z = 3 and
3x – z = \(\frac{79}{5}\)
THEN
the two lines are parallel ( – 3 ≠ 44 or – 7 ≠ 40 or 3 ≠ -\(\frac{79}{5}\)
Note: There are other possible pairs of equations in two variables.
To obtain the final R1, at least the initial M1 must have been awarded.
hence the three planes do not intersect
METHOD 2
vector product of the two normals = \(\begin{pmatrix}-1\\-5\\-3\end{pmatrix}\) (or equivalent)
\(r = \begin{pmatrix}1\\-2 \\0 \end{pmatrix} + \lambda \begin{pmatrix}1\\5 \\3 \end{pmatrix}\) (or equivalent)
Note: Award A0 if “r =” is missing. Subsequent marks may still be awarded.
Attempt to substitute (1 + λ, -2 + 5λ, 3λ) in ∏3
-9(1 + λ) + 3(-2+5λ) – 2(3λ) = 32
− 15 = 32, a contradiction
hence the three planes do not intersect
(b) (i) ∏1 : 2 + 2 + 0 = 4 and ∏2 : 1 + 4 + 0 = 5
(ii) METHOD 1
attempt to find the vector product of the two normals
\(\begin{pmatrix}2\\-1 \\1 \end{pmatrix} \times \begin{pmatrix}1\\-2 \\3 \end{pmatrix}\)
\(=\begin{pmatrix}-1\\-5 \\-3 \end{pmatrix}\)
\(r = \begin{pmatrix}1\\-2 \\0 \end{pmatrix} + \lambda \begin{pmatrix}1\\5 \\3 \end{pmatrix}\)
Note: Award A1A0 if “r =” is missing.
Accept any multiple of the direction vector.
Working for (b)(ii) may be seen in part (a) Method 2. In this case penalize
lack of “r =” only once.
METHOD 2
attempt to eliminate a variable from ∏1 and ∏2
3x – z = 3 OR 3y – 5z = -6 OR 5x – y = 7
Let x = t
substituting x = t in 3x – z = 3 to obtain
z = -3 + 3t and y = 5t – 7 (for all three variables in parametric form)
\(r = \begin{pmatrix}1\\-2 \\0 \end{pmatrix} + \lambda \begin{pmatrix}1\\5 \\3 \end{pmatrix}\)
Note: Award A1A0 if “r =” is missing.
Accept any multiple of the direction vector. Accept other position vectors
which satisfy both the planes ∏1 and ∏2 .
(c) METHOD 1
the line connecting L and ∏3 is given by L1
attempt to substitute position and direction vector to form L1
\(s = \begin{pmatrix}1\\-2 \\0 \end{pmatrix} + t \begin{pmatrix}-9\\3 \\-2 \end{pmatrix}\)
substitute (1 – 9t, – 2 + 3t, – 2t) in ∏3
-9(1-9t) + 3(-2+3t) – 2(-2t) = 32
\(94t = 47\Rightarrow t=\frac{1}{2}\)
attempt to find distance between (1, -2,0) and their point \(\begin{pmatrix}-\frac{7}{2}, & -\frac{1}{2}, & -1 \end{pmatrix}\)
\(=\left | \begin{pmatrix}1\\-2 \\0 \end{pmatrix} +\frac{1}{2}\begin{pmatrix}-9\\3 \\-2 \end{pmatrix}-\begin{pmatrix}1\\-2 \\0 \end{pmatrix}\right | = \frac{1}{2}\sqrt{(-9)^{2}+3^{2}+(-2)^{2}}\)
\(=\frac{\sqrt{94}}{2}\)
METHOD 2
unit normal vector equation of ∏3 is given by
\(=\frac{32}{\sqrt{94}}\)
Question
Consider the three planes
∏1 : 2x – y + z = 4
∏2 : x – 2y + 3z = 5
∏3 : -9x + 3y – 2z = 32
(a) Show that the three planes do not intersect.
(b) (i) Verify that the point P(1 , -2 , 0) lies on both ∏1 and ∏2 .
(ii) Find a vector equation of L, the line of intersection of ∏1 and ∏2 .
(c) Find the distance between L and ∏3 .
▶️Answer/Explanation
Ans:
(a) METHOD 1
attempt to eliminate a variable
obtain a pair of equations in two variables
EITHER
-3x + z = -3 and
-3x + z = 44
OR
-5x + y = -7 and
-5x + y = 40
OR
3x – z = 3 and
3x – z = \(\frac{79}{5}\)
THEN
the two lines are parallel ( – 3 ≠ 44 or – 7 ≠ 40 or 3 ≠ -\(\frac{79}{5}\)
Note: There are other possible pairs of equations in two variables.
To obtain the final R1, at least the initial M1 must have been awarded.
hence the three planes do not intersect
METHOD 2
vector product of the two normals = \(\begin{pmatrix}-1\\-5\\-3\end{pmatrix}\) (or equivalent)
\(r = \begin{pmatrix}1\\-2 \\0 \end{pmatrix} + \lambda \begin{pmatrix}1\\5 \\3 \end{pmatrix}\) (or equivalent)
Note: Award A0 if “r =” is missing. Subsequent marks may still be awarded.
Attempt to substitute (1 + λ, -2 + 5λ, 3λ) in ∏3
-9(1 + λ) + 3(-2+5λ) – 2(3λ) = 32
− 15 = 32, a contradiction
hence the three planes do not intersect
(b) (i) ∏1 : 2 + 2 + 0 = 4 and ∏2 : 1 + 4 + 0 = 5
(ii) METHOD 1
attempt to find the vector product of the two normals
\(\begin{pmatrix}2\\-1 \\1 \end{pmatrix} \times \begin{pmatrix}1\\-2 \\3 \end{pmatrix}\)
\(=\begin{pmatrix}-1\\-5 \\-3 \end{pmatrix}\)
\(r = \begin{pmatrix}1\\-2 \\0 \end{pmatrix} + \lambda \begin{pmatrix}1\\5 \\3 \end{pmatrix}\)
Note: Award A1A0 if “r =” is missing.
Accept any multiple of the direction vector.
Working for (b)(ii) may be seen in part (a) Method 2. In this case penalize
lack of “r =” only once.
METHOD 2
attempt to eliminate a variable from ∏1 and ∏2
3x – z = 3 OR 3y – 5z = -6 OR 5x – y = 7
Let x = t
substituting x = t in 3x – z = 3 to obtain
z = -3 + 3t and y = 5t – 7 (for all three variables in parametric form)
\(r = \begin{pmatrix}1\\-2 \\0 \end{pmatrix} + \lambda \begin{pmatrix}1\\5 \\3 \end{pmatrix}\)
Note: Award A1A0 if “r =” is missing.
Accept any multiple of the direction vector. Accept other position vectors
which satisfy both the planes ∏1 and ∏2 .
(c) METHOD 1
the line connecting L and ∏3 is given by L1
attempt to substitute position and direction vector to form L1
\(s = \begin{pmatrix}1\\-2 \\0 \end{pmatrix} + t \begin{pmatrix}-9\\3 \\-2 \end{pmatrix}\)
substitute (1 – 9t, – 2 + 3t, – 2t) in ∏3
-9(1-9t) + 3(-2+3t) – 2(-2t) = 32
\(94t = 47\Rightarrow t=\frac{1}{2}\)
attempt to find distance between (1, -2,0) and their point \(\begin{pmatrix}-\frac{7}{2}, & -\frac{1}{2}, & -1 \end{pmatrix}\)
\(=\left | \begin{pmatrix}1\\-2 \\0 \end{pmatrix} +\frac{1}{2}\begin{pmatrix}-9\\3 \\-2 \end{pmatrix}-\begin{pmatrix}1\\-2 \\0 \end{pmatrix}\right | = \frac{1}{2}\sqrt{(-9)^{2}+3^{2}+(-2)^{2}}\)
\(=\frac{\sqrt{94}}{2}\)
METHOD 2
unit normal vector equation of ∏3 is given by
\(=\frac{32}{\sqrt{94}}\)