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 Equation of a Plane
Vector Equation of a Plane
A plane in three-dimensional space can be defined by a point through which it passes and two non-parallel direction vectors lying in the plane.
General form:
\( \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} + \mu \mathbf{c} \)
- \( \mathbf{r} \): position vector of a general point on the plane
- \( \mathbf{a} \): position vector of a fixed point on the plane
- \( \mathbf{b}, \mathbf{c} \): two non-parallel direction vectors lying in the plane
- \( \lambda, \mu \): scalar parameters (real numbers)
The vectors \( \mathbf{b} \) and \( \mathbf{c} \) define the orientation of the plane. Any point \( \mathbf{r} \) on the plane can be reached by starting at point \( \mathbf{a} \) and moving along combinations of \( \mathbf{b} \) and \( \mathbf{c} \).
Example
Find the vector equation of the plane passing through point \( \mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \) with direction vectors \( \mathbf{b} = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} \) and \( \mathbf{c} = \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix} \).
▶️ Answer/Explanation
The equation is:
\( \mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} + \mu \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix} \)
where \( \lambda, \mu \in \mathbb{R} \).
Plane Equation Using a Normal Vector
Plane Equation Using a Normal Vector
If \( \mathbf{n} \) is a normal vector to the plane, and \( \mathbf{a} \) is the position vector of a point on the plane, then the plane’s equation can be written as:
\( \mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n} \)
- \( \mathbf{r} \): position vector of any point \( (x, y, z) \) on the plane
- \( \mathbf{a} \): position vector of a fixed point on the plane
- \( \mathbf{n} \): normal vector to the plane
Interpretation: This equation states that any point \( \mathbf{r} \) on the plane satisfies the condition that the dot product of its position vector with the normal vector is constant (equal to \( \mathbf{a} \cdot \mathbf{n} \)).
Example
Find the plane equation passing through point \( \mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \) with normal vector \( \mathbf{n} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} \).
▶️ Answer/Explanation
First compute \( \mathbf{a} \cdot \mathbf{n} \):
\( (1)(2) + (2)(-1) + (3)(4) = 2 – 2 + 12 = 12 \)
So the plane equation is:
\( \mathbf{r} \cdot \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} = 12 \)
Cartesian Equation of a Plane
Cartesian Equation of a Plane
The equation of a plane in Cartesian form is:
\( ax + by + cz = d \)
- \( (x, y, z) \) — coordinates of any point on the plane
- \( (a, b, c) \) — components of a normal vector to the plane
- \( d \) — a constant determined by substituting a known point on the plane
Interpretation: The plane consists of all points \( (x, y, z) \) whose position vectors satisfy this linear equation.
Example
Find the Cartesian equation of the plane that passes through the point \( (1, 2, -1) \) and has normal vector \( \mathbf{n} = \begin{pmatrix} 3 \\ -2 \\ 4 \end{pmatrix} \).
▶️ Answer/Explanation
We use the formula \( 3(x – 1) – 2(y – 2) + 4(z + 1) = 0 \)
\( 3x – 3 – 2y + 4 + 4z + 4 = 0 \)
\( 3x – 2y + 4z + 5 = 0 \)
So the Cartesian equation of the plane is:
\( 3x – 2y + 4z = -5 \)
Summary : Vector and Cartesian Equations
| Vector Equation | Cartesian Equation |
|---|---|
| \( \vec{r} \cdot \vec{n} = d \) | \( lx + ny + nz = d \) |
| \( (\vec{r} – \vec{a}) \cdot \vec{N} = 0 \) | \( A(x – x_1) + B(y – y_1) + C(z – z_1) = 0 \) |
| \( (\vec{r} – \vec{a}) \cdot [(\vec{b} – \vec{a}) \times (\vec{c} – \vec{a})] = 0 \) | \[ \begin{vmatrix} x – x_1 & y – y_1 & z – z_1 \\ x_2 – x_1 & y_2 – y_1 & z_2 – z_1 \\ x_3 – x_1 & y_3 – y_1 & z_3 – z_1 \end{vmatrix} = 0 \] |
| \( \vec{r} = (\vec{n}_1 + \lambda \vec{n}_2) = d_1 + \lambda d_2 \) | \( (A_1x + B_1y + C_1z – d_1) + \lambda(A_2x + B_2y + C_2z – d_2) = 0 \) |