Problem 28 - Precalculus Mathematics for Calculus

Question

Find the area of \(\triangle P Q R\). $$P(3,-2,6), Q(-1,-4,-6), R(3,4,6)$$

Step-by-Step Solution

Answer

The area of \(\triangle PQR\) is \(12\sqrt{10}\) square units.

Step 1: Understand the Problem

We need to find the area of the triangle formed by the points \(P(3,-2,6)\), \(Q(-1,-4,-6)\), and \(R(3,4,6)\) in 3-dimensional space. This can be achieved using the cross product method of vectors.

Step 2: Create Vectors PQ and PR

Vector \(\overrightarrow{PQ}\) is found by subtracting the coordinates of \(P\) from \(Q\):\(\overrightarrow{PQ} = (-1 - 3, -4 + 2, -6 - 6) = (-4, -2, -12)\).Vector \(\overrightarrow{PR}\) is found by subtracting the coordinates of \(P\) from \(R\):\(\overrightarrow{PR} = (3 - 3, 4 + 2, 6 - 6) = (0, 6, 0)\).

Step 3: Calculate the Cross Product

The cross product of \(\overrightarrow{PQ}\) and \(\overrightarrow{PR}\) is calculated using determinant form:\(\overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -4 & -2 & -12 \ 0 & 6 & 0 \end{vmatrix} = (72\hat{i} + 0\hat{j} - 24\hat{k}) = (72, 0, -24)\).

Step 4: Find the Magnitude of the Cross Product

The magnitude of the cross product vector \((72, 0, -24)\) is found using:\[\|\overrightarrow{PQ} \times \overrightarrow{PR}\| = \sqrt{72^2 + 0^2 + (-24)^2} = \sqrt{5184 + 576} = \sqrt{5760}.\]

Step 5: Calculate the Area of the Triangle

The area \(A\) of \(\triangle PQR\) is half the magnitude of the cross product:\[A = \frac{1}{2} \times \sqrt{5760} = \frac{1}{2} \times 24\sqrt{10} = 12\sqrt{10} \approx 37.95\] square units.

Key Concepts

Cross Product Method

The cross product method is an essential tool for finding the area of a triangle in 3D space. This technique involves using vectors formulated from the vertices of the triangle. In our exercise, the cross product was computed between vectors \(\overrightarrow{PQ}\) and \(\overrightarrow{PR}\).

This method helps to determine a new vector that is perpendicular, or orthogonal, to the plane containing the original two vectors. The magnitude of this cross product vector corresponds to the area of a parallelogram formed by the two vectors.

• It’s crucial to remember that the area of a triangle is half of this vector's magnitude.
• By employing the cross product method, we can bypass some complex geometry, taking advantage of vector properties.
• This is particularly helpful in three dimensions where visualizing shapes can be challenging.

Vector Subtraction

Vector subtraction is a simple yet fundamental operation in vector algebra used to create vectors that are needed for further calculations like the cross product. In this problem, vectors \(\overrightarrow{PQ}\) and \(\overrightarrow{PR}\) were formulated by performing vector subtraction between the given points.

To subtract vectors, you subtract the corresponding components of one vector from another. For example, to find \(\overrightarrow{PQ}\), we subtracted each component of point \(P\) from point \(Q\).

• This approach translates the problem of finding the triangle area in a 3D coordinate system into dealing with simpler vector components.
• Vector subtraction helps in determining the relative position between two points.
• It's a foundational step that allows us to advance to more complex calculations, like determining a meaningful cross product.

Magnitude of Vector

The magnitude of a vector is essentially its length and in context, it represents a distance between points or a physical quantity such as force. After calculating the cross product vector, its magnitude was found to solve the problem at hand.

To find the magnitude of a vector, you use the formula \(\sqrt{x^2 + y^2 + z^2}\) for a 3-dimensional vector \((x, y, z)\). This calculation gives a scalar value, which we use to understand the size of the vector in real terms.

• Finding the magnitude is crucial because it translates the abstract vector information into a quantifiable measure of area.
• This step combines all vector components into one meaningful number.
• Magnitude is used extensively in physics to find the strength and direction of vectors.

Determinant Form

The determinant form is a structured mathematical tool used for computing cross products in 3D. It's a compact way to perform calculations necessary for deriving a perpendicular vector from two initial vectors.

In the exercise, the determinant of matrices was used to compute \(\overrightarrow{PQ} \times \overrightarrow{PR}\). The determinants enable evaluating complex expressions involving vectors quickly and efficiently.

• This form organizes calculations so that they become more manageable.
• It provides an organized framework to handle terms involving \(\hat{i}, \hat{j},\) and \(\hat{k}\).
• Using determinants here streamlines the process of finding a vector orthogonal to the plane defined by \(\overrightarrow{PQ}\) and \(\overrightarrow{PR}\).