Understanding the Problem: Equation of a Plane

The problem asks us to find the equation of a plane that contains three specific non-collinear points. These points are given by their position vectors: $\vec{p}$, $\vec{q}$, and $\vec{r}$.

• Point A (from $\vec{p}$): $(2, 3, 4)$
• Point B (from $\vec{q}$): $(1, 4, -1)$
• Point C (from $\vec{r}$): $(2, 3, 1)$

To define the equation of a plane, we typically need two key pieces of information:

1. A point that lies on the plane.
2. A normal vector (a vector perpendicular) to the plane.

Since we have three points on the plane, we can easily select one as our reference point. To find the normal vector, we can form two vectors lying within the plane using these points and then calculate their cross product.

Finding Vectors within the Plane

Let's define two vectors that lie on the plane using the given position vectors. We will use point A as our reference point.

• Vector $\vec{AB}$ (from A to B): $\vec{AB} = \vec{q} - \vec{p} = (1\hat{i} + 4\hat{j} - 1\hat{k}) - (2\hat{i} + 3\hat{j} + 4\hat{k})$ $\vec{AB} = (1-2)\hat{i} + (4-3)\hat{j} + (-1-4)\hat{k}$ $\vec{AB} = -1\hat{i} + 1\hat{j} - 5\hat{k}$
• Vector $\vec{AC}$ (from A to C): $\vec{AC} = \vec{r} - \vec{p} = (2\hat{i} + 3\hat{j} + 1\hat{k}) - (2\hat{i} + 3\hat{j} + 4\hat{k})$ $\vec{AC} = (2-2)\hat{i} + (3-3)\hat{j} + (1-4)\hat{k}$ $\vec{AC} = 0\hat{i} + 0\hat{j} - 3\hat{k}$

Calculating the Normal Vector to the Plane

The normal vector ($\vec{n}$) to the plane is perpendicular to any vector lying in the plane. Therefore, we can find it by taking the cross product of the two vectors we just calculated, $\vec{AB}$ and $\vec{AC}$.

$\vec{n} = \vec{AB} \times \vec{AC}$

The cross product is calculated as the determinant of a matrix:

Expanding the determinant:

$\vec{n} = \hat{i}((1)(-3) - (-5)(0)) - \hat{j}((-1)(-3) - (-5)(0)) + \hat{k}((-1)(0) - (1)(0))$

$\vec{n} = \hat{i}(-3 - 0) - \hat{j}(3 - 0) + \hat{k}(0 - 0)$

$\vec{n} = -3\hat{i} - 3\hat{j} + 0\hat{k}$

We can simplify this normal vector by dividing by a scalar, for instance, by $-3$. A simplified normal vector $\vec{n}'$ would be:

$\vec{n}' = \frac{-3\hat{i} - 3\hat{j}}{-3} = 1\hat{i} + 1\hat{j} + 0\hat{k}$

So, the components of our simplified normal vector are $(1, 1, 0)$.

Deriving the Equation of the Plane

The general equation of a plane with a normal vector $(A, B, C)$ passing through a point $(x_0, y_0, z_0)$ is given by:

$A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$

We will use the simplified normal vector $(A, B, C) = (1, 1, 0)$ and point A $(x_0, y_0, z_0) = (2, 3, 4)$ which lies on the plane.

Substituting these values into the equation:

$1(x - 2) + 1(y - 3) + 0(z - 4) = 0$

$x - 2 + y - 3 + 0 = 0$

Combining the constant terms:

$x + y - 5 = 0$

This is the equation of the plane containing the given non-collinear points.

Alternative Method: Determinant Form for Equation of Plane

The equation of a plane passing through three points $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, and $(x_3, y_3, z_3)$ can also be found directly using a determinant:

$\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix} = 0$

Using our points A(2, 3, 4), B(1, 4, -1), and C(2, 3, 1):

This simplifies to:

Expanding the determinant along the first row:

$(x-2)((1)(-3) - (-5)(0)) - (y-3)((-1)(-3) - (-5)(0)) + (z-4)((-1)(0) - (1)(0)) = 0$

$(x-2)(-3 - 0) - (y-3)(3 - 0) + (z-4)(0 - 0) = 0$

$-3(x-2) - 3(y-3) = 0$

$-3x + 6 - 3y + 9 = 0$

$-3x - 3y + 15 = 0$

Dividing the entire equation by $-3$:

$x + y - 5 = 0$

Both methods confirm that the equation of the plane is $x + y - 5 = 0$.