Problem 22 - Algebra and Trigonometry

Question

Find an equation of the plane that passes through the points \(P, Q,\) and \(R .\) $$ P(3,4,5), \quad Q(1,2,3), \quad R(4,7,6) $$

Step-by-Step Solution

Answer

The equation of the plane is \( x - z = 2 \).

Step 1: Determine Direction Vectors

To find the plane equation, calculate two direction vectors using the given points. Let vector \( \vec{PQ} \) be from \( P \) to \( Q \), and \( \vec{PR} \) be from \( P \) to \( R \). The components are calculated as follows:\[ \vec{PQ} = Q - P = (1 - 3, 2 - 4, 3 - 5) = (-2, -2, -2) \]\[ \vec{PR} = R - P = (4 - 3, 7 - 4, 6 - 5) = (1, 3, 1) \]

Step 2: Calculate the Normal Vector

The normal vector of the plane is found by taking the cross product of the two direction vectors \( \vec{PQ} \) and \( \vec{PR} \). The cross product \( \vec{n} = \vec{PQ} \times \vec{PR} \) is given by:\[\vec{n} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \-2 & -2 & -2 \1 & 3 & 1\end{vmatrix} = (\mathbf{i}(\{-2\cdot1 - (-2)\cdot3\}) - \mathbf{j}(\{-2\cdot1 - (-2)\cdot1\}) + \mathbf{k}(\{-2\cdot3 - (-2)\cdot1\}))\]This simplifies to:\[\vec{n} = (4\mathbf{i} + 0\mathbf{j} - 4\mathbf{k}) = (4, 0, -4)\]

Step 3: Form the Equation of the Plane

With the normal vector \( \vec{n} = (4, 0, -4) \) and a point \( P(3, 4, 5) \) on the plane, the equation of the plane is given by:\[ 4(x-3) + 0(y-4) - 4(z-5) = 0 \]This simplifies to:\[ 4x - 4z = 8 \]

Step 4: Simplify the Plane Equation

Simplify the equation by dividing through by 4:\[ x - z = 2 \]This is the final equation of the plane.

Key Concepts

Direction Vectors

In the study of geometry, direction vectors are essential when solving problems involving planes. Given a set of points, direction vectors help us understand how these points are positioned relative to one another. To find a direction vector between two points, you subtract the coordinates of one point from the other. For instance, with points \( P(3,4,5) \) and \( Q(1,2,3) \), you obtain the direction vector \( \vec{PQ} = (-2, -2, -2) \). Similarly, for points \( P(3,4,5) \) and \( R(4,7,6) \), the direction vector is \( \vec{PR} = (1, 3, 1) \).
Direction vectors show the relative change in position between points. They are not unique; any scalar multiple of a direction vector is also a valid direction vector.

Normal Vector

In the context of planes, a normal vector is a vector that is perpendicular to the plane. This vector is vital because it uniquely determines the plane's orientation. You can find it using the cross product of two direction vectors that lie in the plane. From the previous section, we use \( \vec{PQ} = (-2, -2, -2) \) and \( \vec{PR} = (1, 3, 1) \) to calculate the normal vector.
Using the cross product formula:

• The first component is \((-2\cdot1 - (-2)\cdot3) = 4\).
• The second component is \(-((-2)\cdot1 - (-2)\cdot1) = 0\).
• The third component is \((-2\cdot3 - (-2)\cdot1) = -4\).

Thus, the normal vector is \( \vec{n} = (4, 0, -4) \). This normal vector is essential for writing the plane's equation.

Cross Product

The cross product is a mathematical operation that takes two vectors and produces a third vector perpendicular to both. This is a crucial operation for finding a normal vector to the plane. The formula for the cross product \( \vec{a} \times \vec{b} \) of vectors \( \vec{a} = (a_1, a_2, a_3) \) and \( \vec{b} = (b_1, b_2, b_3) \) is expressed by \[ \vec{a} \times \vec{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} \] This determinant calculates the components as follows:

• The \( \mathbf{i} \) component as \( a_2b_3 - a_3b_2 \)
• The \( \mathbf{j} \) component as \( a_3b_1 - a_1b_3 \)
• The \( \mathbf{k} \) component as \( a_1b_2 - a_2b_1 \)

When you apply this to our direction vectors, you get \( (4, 0, -4) \) as the normal vector, confirming the earlier result.

Equation of a Plane

An equation of a plane provides a formula for all the points on that plane. Given a normal vector \( \vec{n} = (A, B, C) \) and a point \( (x_0, y_0, z_0) \) on the plane, the general equation of a plane is \[ A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 \]For the present case, we utilize the normal vector \( (4, 0, -4) \) and the point \( P(3, 4, 5) \). Substituting these into the formula yields: \[ 4(x - 3) + 0(y - 4) - 4(z - 5) = 0 \]
This simplifies further to \( 4x - 4z = 8 \), which, when divided by 4, becomes \( x - z = 2 \).
This is the final, simplified equation representing the plane through points \( P, Q, \) and \( R \).