Question

The vector equation of the plane passing through two fixed points A(a bar),B(b bar) and parallel to the vector c bar is:-(where c bar is not parallel to AB bar)

Answer 1

$(\vec{r} - \vec{a}) \cdot ((\vec{b} - \vec{a}) \times \vec{c}) = 0$

Answer 2

To find the vector equation of a plane, we need a point on the plane and a normal vector to the plane, or a point on the plane and two non-parallel vectors lying in the plane.

Given:

1. The plane passes through two fixed points A and B with position vectors $\vec{a}$ and $\vec{b}$ respectively.
2. The plane is parallel to vector $\vec{c}$.
3. Vector $\vec{c}$ is not parallel to $\vec{AB}$.

Method 1: Using a point and two non-parallel vectors in the plane (Parametric form)

• Since points A and B lie on the plane, the vector $\vec{AB} = \vec{b} - \vec{a}$ lies in the plane.
• Since the plane is parallel to vector $\vec{c}$, the vector $\vec{c}$ also lies in the plane (or can be translated to lie in the plane).
• We are given that $\vec{c}$ is not parallel to $\vec{AB}$ (i.e., $\vec{c}$ is not parallel to $\vec{b} - \vec{a}$). This ensures that $\vec{b} - \vec{a}$ and $\vec{c}$ are two non-collinear vectors lying in the plane.
• Let $\vec{r}$ be the position vector of an arbitrary point P on the plane.
• The vector $\vec{AP} = \vec{r} - \vec{a}$ also lies in the plane.
• Since $\vec{AP}$, $\vec{AB}$, and $\vec{c}$ are all vectors lying in the same plane, $\vec{AP}$ must be a linear combination of $\vec{AB}$ and $\vec{c}$. Therefore, $\vec{r} - \vec{a} = \lambda (\vec{b} - \vec{a}) + \mu \vec{c}$, where $\lambda$ and $\mu$ are scalar parameters.

The parametric vector equation of the plane is: $\vec{r} = \vec{a} + \lambda (\vec{b} - \vec{a}) + \mu \vec{c}$

Method 2: Using a point and a normal vector (Non-parametric form)

• The normal vector $\vec{n}$ to the plane must be perpendicular to any two non-parallel vectors lying in the plane.
• We have identified two such vectors: $\vec{AB} = \vec{b} - \vec{a}$ and $\vec{c}$.
• Therefore, the normal vector $\vec{n}$ can be found by taking the cross product of these two vectors: $\vec{n} = (\vec{b} - \vec{a}) \times \vec{c}$ Since $\vec{c}$ is not parallel to $\vec{b} - \vec{a}$, $\vec{n}$ will be a non-zero vector.
• The equation of a plane passing through a point with position vector $\vec{a}$ and having a normal vector $\vec{n}$ is given by: $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$
• Substitute the expression for $\vec{n}$: $(\vec{r} - \vec{a}) \cdot ((\vec{b} - \vec{a}) \times \vec{c}) = 0$ This equation can also be written in scalar triple product notation as: $[\vec{r} - \vec{a}, \vec{b} - \vec{a}, \vec{c}] = 0$

Both forms are valid vector equations. The non-parametric form is generally preferred as "the" vector equation of a plane as it does not involve arbitrary parameters.

The plane contains points A and B, so the vector $\vec{AB} = \vec{b} - \vec{a}$ lies in the plane. The plane is also parallel to vector $\vec{c}$, meaning $\vec{c}$ lies in the plane. Since $\vec{c}$ is not parallel to $\vec{AB}$, the vectors $\vec{b} - \vec{a}$ and $\vec{c}$ are non-collinear vectors lying in the plane. The normal vector to the plane is perpendicular to both these vectors, hence $\vec{n} = (\vec{b} - \vec{a}) \times \vec{c}$. The equation of a plane passing through a point $\vec{a}$ with normal $\vec{n}$ is $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$. Substituting $\vec{n}$ gives the required equation.