Question

Find the area of the following triangles \(T\). (The area of a triangle is half the area of the corresponding parallelogram.) The vertices of \(T\) are \(O(0,0,0), P(1,2,3),\) and \(Q(6,5,4)\)

Short answer

Answer: The area of the triangle T is 1/2√(294) units squared.

Step-by-step solution

Defining Vectors for Sides of the Triangle

We will represent the positions of points O, P, and Q using coordinates. Let OP and OQ be the two sides of the triangle T. We can find the vectors representing OP and OQ by subtracting the coordinates of O from P and Q respectively. OP = P - O = (1, 2, 3) - (0, 0, 0) = (1, 2, 3) OQ = Q - O = (6, 5, 4) - (0, 0, 0) = (6, 5, 4)

Finding Cross Product of the Vectors

The cross product of two vectors is a new vector that is perpendicular to both input vectors and has a magnitude equal to the area of the parallelogram formed by the input vectors. To find the cross product, we perform the following operation: (OP x OQ) = (OP_y * OQ_z - OP_z * OQ_y, OP_z * OQ_x - OP_x * OQ_z, OP_x * OQ_y - OP_y * OQ_x) Here, the subscripts x, y, and z refer to the coordinates of the corresponding vectors. Plugging the values of OP and OQ, we get: (OP x OQ) = (2 * 4 - 3 * 5, 3 * 6 - 1 * 4, 1 * 5 - 2 * 6) = (-7, 14, -7)

Finding the Magnitude of the Cross Product

Next, we will find the magnitude of the cross product vector, which is equal to the area of the parallelogram formed by the input vectors. We can use the Euclidean norm formula to calculate the magnitude: ||OP x OQ|| = √((-7)^2 + (14)^2 + (-7)^2) = √(49 + 196 + 49) = √(294)

Finding the Area of the Triangle

To find the area of the triangle T, we can simply divide the area of the parallelogram (magnitude of the cross product) by 2: Area(T) = 1/2 ||OP x OQ|| = 1/2 * √(294) So, the area of the triangle T is 1/2√(294) units squared.

Key concepts

Understanding Vectors

Vectors are an essential concept in mathematics and physics. They are quantities that have both magnitude and direction, unlike scalar quantities that only have magnitude. In the context of our triangle exercise, vectors are used to represent the sides of the triangle. The points given, such as O(0,0,0), P(1,2,3), and Q(6,5,4), help us define these vectors easily.

• We found vector OP by taking point P and subtracting point O, resulting in the vector \(OP = (1, 2, 3)\).
• Similarly, the vector OQ is the subtraction of O from Q, giving us \(OQ = (6, 5, 4)\).

Vectors allow us to perform operations like addition, subtraction, and multiplication using vector algebra. In this problem, understanding how to define and manipulate these vectors is key to finding the solution.

The Cross Product and Its Significance

The cross product is a vector multiplication operation that results in a new vector perpendicular to the original vectors. It's particularly useful in three-dimensional space calculations. When you take the cross product of two vectors, it gives a vector whose magnitude represents the area of the parallelogram formed by the initial vectors.
To compute the cross product \(OP \times OQ\), we perform the operation: \((OP_y \cdot OQ_z - OP_z \cdot OQ_y, OP_z \cdot OQ_x - OP_x \cdot OQ_z, OP_x \cdot OQ_y - OP_y \cdot OQ_x)\). By substituting the values of \(OP\) and \(OQ\), we find that \(OP \times OQ = (-7, 14, -7)\).
The resultant vector points in the direction that is orthogonal to both OP and OQ, which is critical when we're seeking the area the two sides encompass in space. The magnitude of this cross product vector is what will help us determine the area of the parallelogram, consequently aiding us in discovering the area of the triangle.

Calculating Magnitudes Using the Euclidean Norm

The Euclidean norm, also known as the magnitude of a vector, is a measure of the length or size of that vector in space. It's calculated using the formula: \[ ||\mathbf{v}|| = \sqrt{v_1^2 + v_2^2 + v_3^2} \]where \(v_1, v_2, \) and \(_3\) are the components of the vector.
In our exercise, we applied this formula to the cross product vector \( (-7, 14, -7) \). By substituting into the Euclidean norm, we get: \[ ||-7, 14, -7|| = \sqrt{(-7)^2 + (14)^2 + (-7)^2} = \sqrt{294} \]This magnitude represents the area of the parallelogram formed by vectors OP and OQ. Since the area of a triangle is half the area of this parallelogram, we divided the magnitude by two to finally get the area of triangle T as \( \frac{1}{2} \sqrt{294} \).
Understanding the Euclidean norm allows efficient computation of this kind of vector problem, making it crucial for students to grasp for various applications in physics and engineering.