Problem 25 - Algebra and Trigonometry

Question

Find an equation of the plane that passes through the points \(P, Q,\) and \(R .\) $$ P(6,1,1), \quad Q(3,2,0), \quad R(0,0,0) $$

Step-by-Step Solution

Answer

The equation of the plane is \\(-2x + 3y - 3z = -12\\).

Step 1: Identify Vectors

From the given points, we first need to determine two vectors that lie on the plane. Using points \(P(6,1,1)\), \(Q(3,2,0)\), and \(R(0,0,0)\), we can form the vectors \(\overrightarrow{PQ}\) and \(\overrightarrow{PR}\): - \(\overrightarrow{PQ} = Q - P = (3 - 6, 2 - 1, 0 - 1) = (-3, 1, -1)\) - \(\overrightarrow{PR} = R - P = (0 - 6, 0 - 1, 0 - 1) = (-6, -1, -1)\)

Step 2: Find the Normal Vector

Next, we compute the cross product of \(\overrightarrow{PQ}\) and \(\overrightarrow{PR}\) to find the normal vector \(\mathbf{n}\) to the plane: \[ \mathbf{n} = \overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -3 & 1 & -1 \ -6 & -1 & -1 \end{vmatrix} \ = \mathbf{i}(1 \cdot (-1) - (-1) \cdot (-1)) - \mathbf{j}(-3 \cdot (-1) - (-1) \cdot (-6)) + \mathbf{k}(-3 \cdot (-1) - 1 \cdot (-6)) \ = \mathbf{i}(-1 - 1) - \mathbf{j}(3 - 6) + \mathbf{k}(3 - 6) = \mathbf{i}(-2) + \mathbf{j}(3) + \mathbf{k}(-3) = (-2, 3, -3) \]

Step 3: Use the Point-Normal Form of the Plane

With the normal vector \((-2, 3, -3)\) obtained, and a point \((6, 1, 1)\) from the plane (point P), we use the point-normal form of the equation of a plane: \ \[-2(x - 6) + 3(y - 1) - 3(z - 1) = 0\]

Step 4: Simplify the Equation

Expand the equation: \[-2(x - 6) + 3(y - 1) - 3(z - 1) = 0\] \[-2x + 12 + 3y - 3 - 3z + 3 = 0\] \[-2x + 3y - 3z + 12 = 0\] Simplified equation of the plane is \[-2x + 3y - 3z = -12\]

Key Concepts

Vectors in 3D Space

Vectors are fundamental building blocks in 3D space. They help represent points and directions. In three-dimensional geometry, a vector is defined by three components:

• An x-component
• A y-component
• An z-component

Each of these components corresponds to one of the three spatial dimensions. Consider a vector \(\overrightarrow{AB}\), formed from point A to point B. If A has coordinates \( (x_1, y_1, z_1) \) and B has \( (x_2, y_2, z_2) \), the vector \(\overrightarrow{AB}\) can be calculated as \((x_2-x_1, y_2-y_1, z_2-z_1)\).
For example, to find vector \(\overrightarrow{PQ}\) from P to Q in a plane, you'd use the given points in the original exercise. Given point P as \( (6,1,1) \) and Q as \( (3,2,0) \), the vector \(\overrightarrow{PQ} = (3-6, 2-1, 0-1) = (-3,1,-1)\). This formalizes the idea of moving from one point to another in 3D space. Understanding vectors is crucial because they allow us to define directions and magnitudes of movement in this multi-dimensional setting.

Cross Product

The cross product is a powerful operation in vector algebra for finding a vector that is perpendicular to two given vectors. This is particularly useful in establishing the normal, or orthogonal, vector to a plane.
Given two vectors \(\overrightarrow{u}\) and \(\overrightarrow{v}\) in 3D, their cross product \((\overrightarrow{u} \times \overrightarrow{v})\) is a new vector, denoted by: \[\overrightarrow{u} \times \overrightarrow{v} = \begin{vmatrix}i & \mathbf{j} & \mathbf{k} \u_1 & u_2 & u_3 \v_1 & v_2 & v_3\end{vmatrix}\]
This determinant calculates to: \[(u_2v_3 - u_3v_2)\mathbf{i} - (u_1v_3 - u_3v_1)\mathbf{j} + (u_1v_2 - u_2v_1)\mathbf{k}.\]
In the context of the original exercise, finding the normal vector of the plane involved computing the cross product of vectors \(\overrightarrow{PQ}\) and \(\overrightarrow{PR}\). The result \((-2, 3, -3)\) reflects a direction perpendicular to both input vectors, thus serving as the normal vector to the plane through the points P, Q, and R.

Point-Normal Form

To define a plane in 3D space, we often use the point-normal form of the equation of a plane. This form requires a known point on the plane and a normal vector, which is perpendicular to the plane.
The general equation of a plane in point-normal form is \[a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\] where \(a, b,\) and \(c\) are the components of the normal vector and \( (x_0, y_0, z_0) \) are the coordinates of the point on the plane.
For the plane containing points \(P(6,1,1), Q(3,2,0),\) and \(R(0,0,0)\), we used the normal vector \((-2, 3, -3)\). Taking point \(P(6,1,1)\), the equation of the plane becomes:
\[-2(x - 6) + 3(y - 1) - 3(z - 1) = 0\]
Expanding and simplifying gives us the final plane equation \(-2x + 3y - 3z = -12\). This tells us that any point \( (x, y, z) \) on this plane satisfies this equation, aligning perfectly with the derived normal vector.