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Find the area of triangle with vertices Q C5, 1, 32, R (2,4,6), SC1,-1,9).

Find the area of triangle with vertices Q C5, 1, 32, R (2,4,6), SC1,-1,9).

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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**Finding the Area of a Triangle Given Vertices** To find the area of a triangle with vertices at the points \( Q(5, 1, 3) \), \( R(2, 4, 6) \), and \( S(1, -1, 9) \), you can use the formula for the area of a triangle in 3D space: 1. **Define the vectors**: Compute vectors \( \overrightarrow{QR} \) and \( \overrightarrow{QS} \). - \( \overrightarrow{QR} = R - Q = (2-5, 4-1, 6-3) = (-3, 3, 3) \) - \( \overrightarrow{QS} = S - Q = (1-5, -1-1, 9-3) = (-4, -2, 6) \) 2. **Compute the cross product** \( \overrightarrow{QR} \times \overrightarrow{QS} \): - \(\overrightarrow{QR} \times \overrightarrow{QS} = \left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -3 & 3 & 3 \\ -4 & -2 & 6 \end{matrix} \right|\) - Calculate the determinant: - \( = \mathbf{i}(3 \cdot 6 - 3 \cdot (-2)) - \mathbf{j}(-3 \cdot 6 - 3 \cdot (-4)) + \mathbf{k}(-3 \cdot (-2) - 3 \cdot (-4)) \) - \( = \mathbf{i}(18 + 6) - \mathbf{j}(-18 + 12) + \mathbf{k}(6 + 12) \) - \( = \mathbf{i}(24) - \mathbf{j}(-6) + \mathbf{k}(18) \) - \( = (24, 6, 18) \) 3. **Magnitude of the cross product**: - \( \text{Magnitude} = \sqrt{(24)^2 + (6)^2 + (18)^2} = \sqrt{576 + 36 + 324} = \sqrt{
Transcribed Image Text:**Finding the Area of a Triangle Given Vertices** To find the area of a triangle with vertices at the points \( Q(5, 1, 3) \), \( R(2, 4, 6) \), and \( S(1, -1, 9) \), you can use the formula for the area of a triangle in 3D space: 1. **Define the vectors**: Compute vectors \( \overrightarrow{QR} \) and \( \overrightarrow{QS} \). - \( \overrightarrow{QR} = R - Q = (2-5, 4-1, 6-3) = (-3, 3, 3) \) - \( \overrightarrow{QS} = S - Q = (1-5, -1-1, 9-3) = (-4, -2, 6) \) 2. **Compute the cross product** \( \overrightarrow{QR} \times \overrightarrow{QS} \): - \(\overrightarrow{QR} \times \overrightarrow{QS} = \left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -3 & 3 & 3 \\ -4 & -2 & 6 \end{matrix} \right|\) - Calculate the determinant: - \( = \mathbf{i}(3 \cdot 6 - 3 \cdot (-2)) - \mathbf{j}(-3 \cdot 6 - 3 \cdot (-4)) + \mathbf{k}(-3 \cdot (-2) - 3 \cdot (-4)) \) - \( = \mathbf{i}(18 + 6) - \mathbf{j}(-18 + 12) + \mathbf{k}(6 + 12) \) - \( = \mathbf{i}(24) - \mathbf{j}(-6) + \mathbf{k}(18) \) - \( = (24, 6, 18) \) 3. **Magnitude of the cross product**: - \( \text{Magnitude} = \sqrt{(24)^2 + (6)^2 + (18)^2} = \sqrt{576 + 36 + 324} = \sqrt{
# Find the Area of a Triangle with Given Vertices ## Problem Statement Calculate the area of a triangle with the following vertices: - Q(5, 13) - R(2, 4) - S(11, 19) ## Solution Approach To find the area of a triangle given the coordinates of its vertices, we use the formula: \[ \text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right| \] where (x₁, y₁), (x₂, y₂), and (x₃, y₃) are the coordinates of the vertices. ## Summary - Assign Q(5, 13) as (x₁, y₁), R(2, 4) as (x₂, y₂), and S(11, 19) as (x₃, y₃). - Substitute these coordinates into the formula and solve for the area. This method provides a straightforward means of calculating the area using determinant-like computations.
Transcribed Image Text:# Find the Area of a Triangle with Given Vertices ## Problem Statement Calculate the area of a triangle with the following vertices: - Q(5, 13) - R(2, 4) - S(11, 19) ## Solution Approach To find the area of a triangle given the coordinates of its vertices, we use the formula: \[ \text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right| \] where (x₁, y₁), (x₂, y₂), and (x₃, y₃) are the coordinates of the vertices. ## Summary - Assign Q(5, 13) as (x₁, y₁), R(2, 4) as (x₂, y₂), and S(11, 19) as (x₃, y₃). - Substitute these coordinates into the formula and solve for the area. This method provides a straightforward means of calculating the area using determinant-like computations.
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