IB Mathematics AA Vector equation of a plane Study Notes
IB Mathematics AA Vector equation of a plane Study Notes
IB Mathematics AA Vector equation of a plane Notes Offer a clear explanation of Vector equation of a plane, 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 plane.
Vector Equations of a Plane
Introduction
The vector equation of a plane is a mathematical representation of a plane in 3D space. This equation is derived using vectors and can also be converted into its Cartesian form. Planes are fundamental in geometry and are widely used in physics, computer graphics, and engineering.
Vector Equation of a Plane
The vector equation of a plane is given by:
\[ \vec{r} = \vec{a} + \lambda \vec{b} + \mu \vec{c} \]
Here:
- \( \vec{r} \): The position vector of any point on the plane.
- \( \vec{a} \): The position vector of a known point on the plane.
- \( \vec{b} \) and \( \vec{c} \): Non-parallel vectors lying within the plane, also known as direction vectors.
- \( \lambda \) and \( \mu \): Scalars that allow the combination of \( \vec{b} \) and \( \vec{c} \) to span the plane.
Alternative Form: Normal Vector Equation
The equation of a plane can also be expressed in terms of a normal vector \( \vec{n} \), which is perpendicular to the plane:
\[ \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n} \]
Here:
- \( \vec{r} \): The position vector of any point on the plane.
- \( \vec{a} \): The position vector of a known point on the plane.
- \( \vec{n} \): The normal vector to the plane.
This form is particularly useful for determining whether a point lies on the plane by substituting the point’s coordinates into the equation.
Cartesian Equation of a Plane
The vector equations can be converted into the Cartesian form of a plane:
\[ ax + by + cz = d \]
Here:
- \( a \), \( b \), \( c \): Components of the normal vector \( \vec{n} = [a, b, c] \).
- \( x, y, z \): Coordinates of a point on the plane.
- \( d \): A constant representing \( \vec{a} \cdot \vec{n} \), where \( \vec{a} \) is a known point on the plane.
Geometric Interpretation
- Normal Vector: The normal vector \( \vec{n} \) is perpendicular to the plane and uniquely defines its orientation.
- Direction Vectors: The vectors \( \vec{b} \) and \( \vec{c} \) span the plane and determine its extent.
Example Calculations
Example 1: Vector Equation of a Plane
Let \( \vec{a} = [1, 2, 3] \), \( \vec{b} = [2, 1, 0] \), and \( \vec{c} = [0, 1, -1] \). The vector equation of the plane is:
\[ \vec{r} = [1, 2, 3] + \lambda [2, 1, 0] + \mu [0, 1, -1] \]
Example 2: Normal Vector Equation
Given \( \vec{a} = [1, 2, 3] \) and \( \vec{n} = [2, -1, 1] \), the equation of the plane is:
\[ \vec{r} \cdot [2, -1, 1] = [1, 2, 3] \cdot [2, -1, 1] \]
\[ 2x – y + z = 2(1) – 1(2) + 1(3) = 3 \]
Thus, the Cartesian form is:
\[ 2x – y + z = 3 \]
Example 3: Cartesian Equation to Vector Form
Convert \( 3x + 2y – z = 6 \) into vector form:
- The normal vector is \( \vec{n} = [3, 2, -1] \).
- A point on the plane is \( [2, 0, 0] \), where \( x = 2 \), \( y = 0 \), and \( z = 0 \) satisfy the equation.
The vector equation is:
\[ \vec{r} = [2, 0, 0] + \lambda [2, -1, 0] + \mu [1, 1, 1] \]
Applications of Plane Equations
Equations of planes have wide-ranging applications, including:
- 3D Geometry: Analyzing shapes and intersections in space.
- Physics: Calculating forces, torques, and equilibrium conditions.
- Computer Graphics: Rendering and collision detection.
IB Mathematics AA SL Vector equation of a plane 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
\(=\frac{32}{\sqrt{94}}\)