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Compute the area of the triangle. T = = ||OP x oQ|| = |(36, -36, 36) || = < 18, – 18,18 > II ト

Compute the area of the triangle. T = = ||OP x oQ|| = |(36, -36, 36) || = < 18, – 18,18 > II ト

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Compute the Area of the Triangle** To find the area of a triangle, we can use the following formula: \[ T = \frac{1}{2} A \] Where \(T\) is the area of the triangle. ### Step-by-Step Calculation 1. We first represent the area \(A\) in terms of the cross product of two vectors. Let’s denote the vectors originating from point \(O\) to points \(P\) and \(Q\) as \(\overrightarrow{OP}\) and \(\overrightarrow{OQ}\) respectively. The area \(A\) of the parallelogram formed by these vectors can be computed as: \[ A = \left\lVert \overrightarrow{OP} \times \overrightarrow{OQ} \right\lVert \] Thus, the area of the triangle can be written as: \[ T = \frac{1}{2} \left\lVert \overrightarrow{OP} \times \overrightarrow{OQ} \right\lVert \] 2. Next, we compute the cross product \(( \overrightarrow{OP} \times \overrightarrow{OQ} )\). Suppose \(\overrightarrow{OP} = \left< a_1, a_2, a_3 \right>\) and \(\overrightarrow{OQ} = \left< b_1, b_2, b_3 \right>\). For illustration, let's say the result after computation is: \[ \overrightarrow{OP} \times \overrightarrow{OQ} = \left< 36, -36, 36 \right> \] 3. Then, compute the magnitude of the resulting vector: \[ \left\lVert \left< 36, -36, 36 \right> \right\lVert = \sqrt{36^2 + (-36)^2 + 36^2} = \sqrt{3 \times 36^2} = \sqrt{3 \times 1296} = \sqrt{3888} \] 4. Therefore, the area \(T\) is: \[ T = \frac{1}{2} \left\lVert \left< 36, -36, 36 \right> \right\lVert = \frac{1}{2} \times \sqrt{3888} = \frac{
Transcribed Image Text:**Compute the Area of the Triangle** To find the area of a triangle, we can use the following formula: \[ T = \frac{1}{2} A \] Where \(T\) is the area of the triangle. ### Step-by-Step Calculation 1. We first represent the area \(A\) in terms of the cross product of two vectors. Let’s denote the vectors originating from point \(O\) to points \(P\) and \(Q\) as \(\overrightarrow{OP}\) and \(\overrightarrow{OQ}\) respectively. The area \(A\) of the parallelogram formed by these vectors can be computed as: \[ A = \left\lVert \overrightarrow{OP} \times \overrightarrow{OQ} \right\lVert \] Thus, the area of the triangle can be written as: \[ T = \frac{1}{2} \left\lVert \overrightarrow{OP} \times \overrightarrow{OQ} \right\lVert \] 2. Next, we compute the cross product \(( \overrightarrow{OP} \times \overrightarrow{OQ} )\). Suppose \(\overrightarrow{OP} = \left< a_1, a_2, a_3 \right>\) and \(\overrightarrow{OQ} = \left< b_1, b_2, b_3 \right>\). For illustration, let's say the result after computation is: \[ \overrightarrow{OP} \times \overrightarrow{OQ} = \left< 36, -36, 36 \right> \] 3. Then, compute the magnitude of the resulting vector: \[ \left\lVert \left< 36, -36, 36 \right> \right\lVert = \sqrt{36^2 + (-36)^2 + 36^2} = \sqrt{3 \times 36^2} = \sqrt{3 \times 1296} = \sqrt{3888} \] 4. Therefore, the area \(T\) is: \[ T = \frac{1}{2} \left\lVert \left< 36, -36, 36 \right> \right\lVert = \frac{1}{2} \times \sqrt{3888} = \frac{
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