Problem 26 - Precalculus Mathematics for Calculus

Question

Find the area of \(\triangle P Q R\). $$P(2,1,0), Q(0,0,-1), R(-4,2,0)$$

Step-by-Step Solution

Answer

The area of \( \triangle PQR \) is \( \frac{\sqrt{101}}{2} \).

Step 1: Find Vectors PQ and PR

To find the area of \( \triangle PQR \), we first need to determine vectors \( \overrightarrow{PQ} \) and \( \overrightarrow{PR} \). Calculate \( \overrightarrow{PQ} = Q - P = (0 - 2, 0 - 1, -1 - 0) = (-2, -1, -1) \). Similarly, \( \overrightarrow{PR} = R - P = (-4 - 2, 2 - 1, 0 - 0) = (-6, 1, 0) \).

Step 2: Compute the Cross Product of PQ and PR

The area of the triangle can be found using the magnitude of the cross product of vectors \( \overrightarrow{PQ} \) and \( \overrightarrow{PR} \). Calculate the cross product: \( \overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -2 & -1 & -1 \ -6 & 1 & 0 \end{vmatrix} \).

Step 3: Calculate the Determinant for the Cross Product

Compute the determinant: \[ \overrightarrow{PQ} \times \overrightarrow{PR} = \mathbf{i}((-1)(0) - (-1)(1)) - \mathbf{j}((-2)(0) - (-1)(-6)) + \mathbf{k}((-2)(1) -(-1)(-6)) \]. Simplifying, we find this equals \( \mathbf{i}(1) - \mathbf{j}(6) - \mathbf{k}(8) = (1, -6, -8) \).

Step 4: Find The Magnitude of The Cross Product

The area of \( \triangle PQR \) is \( \frac{1}{2} \times \text{magnitude of } \overrightarrow{PQ} \times \overrightarrow{PR} \). Find the magnitude: \[ \sqrt{1^2 + (-6)^2 + (-8)^2} = \sqrt{1 + 36 + 64} = \sqrt{101} \].

Step 5: Compute the Area

Finally, compute the area of the triangle: \( \frac{1}{2} \times \sqrt{101} \). Thus, the area of \( \triangle PQR \) is \( \frac{\sqrt{101}}{2} \).

Key Concepts

Triangle Area

Triangles are fascinating shapes as they are the simplest polygons. They consist of three sides and three angles. To calculate the area of a triangle in a three-dimensional space, such as using points like P, Q, and R, you can use vector calculus. This method is particularly valuable when dealing with vertices that have three-dimensional coordinates, as it avoids the need to break the triangle down into simpler components.

You can determine the area of a triangle using the formula involving the cross product of vectors. The steps involve:

• Finding two vectors using the coordinates of the triangle's vertices.
• Calculating the cross product of these two vectors.
• Finding the magnitude of this cross product.
• The area of the triangle is half of this magnitude.

This approach is handy when the vertices do not lie in a single plane parallel to a base axis, making conventional area calculations challenging.

Applying vector calculus is a universal approach that simplifies the problem, regardless of the orientation of the triangle in the space.

Cross Product

The cross product is a fundamental operation in vector calculus that takes two vectors and returns a third vector that is perpendicular to the plane formed by the first two. If you have two vectors, \( \overrightarrow{A} \) and \( \overrightarrow{B} \), their cross product \( \overrightarrow{A} \times \overrightarrow{B} \) is a vector whose direction follows the right-hand rule.

To compute the cross product of two vectors in three-dimensional space:

• Set up a 3x3 determinant using the unit vectors \( \mathbf{i} \), \( \mathbf{j} \), \( \mathbf{k} \) along with the components of the given vectors.
• Calculate the determinant to find the resulting vector.

This method is beneficial because it provides a vector that not only helps in calculating areas of triangles but also finds applications in physics and engineering for problems involving torque and rotational forces.

The cross product has a magnitude that corresponds to the area of the parallelogram formed by the original two vectors, which is why it is so useful in computations like those involved in determining triangle area.

Vector Magnitude

Vectors represent quantities that have both direction and magnitude; the latter is the "size" or length of the vector. Calculating vector magnitude is crucial when determining results like the size of a resultant vector from a cross product.

For any three-dimensional vector \( \overrightarrow{V} = (x, y, z) \), the magnitude, denoted \( \| \overrightarrow{V} \| \), is calculated using the formula:\[ \| \overrightarrow{V} \| = \sqrt{x^2 + y^2 + z^2} \]

This formula involves squaring each component of the vector, summing them, and then taking the square root of the result. It effectively provides the "length" of the vector in three-dimensional space.

Having the magnitude of a vector is often a final step in many vector calculus problems, including determining areas and volumes, as it gives a real number value, which is often more interpretable in practical applications. In the context of triangle area, once the cross product is found, its magnitude directly contributes to our understanding of the spatial relationships and distances in geometry.