Chapter 13: Problem 35
Find the area of the following triangles \(T\). The vertices of \(T\) are \(A(5,6,2), B(7,16,4),\) and \(C(6,7,3)\)
Short Answer
Expert verified
Answer: The area of the triangle is 6 square units.
Step by step solution
01
Finding the vectors AB and AC
We can find the vector \(\vec{AB}\) by subtracting the coordinates of points \(A\) and \(B\). Similarly, we can find \(\vec{AC}\) by subtracting the coordinates of points \(A\) and \(C\).
\(\vec{AB} = B - A = (7-5, 16-6, 4-2) = (2,10,2)\)
\(\vec{AC} = C - A = (6-5, 7-6, 3-2) = (1,1,1)\)
So, \(\vec{AB} = (2,10,2)\) and \(\vec{AC} = (1,1,1)\).
02
Finding the cross product of AB and AC
Now that we have the vectors \(\vec{AB}\) and \(\vec{AC}\), we can find their cross product, denoted as \(\vec{AB} \times \vec{AC}\), using the formula:
\(\vec{AB} \times \vec{AC} = \begin{pmatrix} i & j & k \\ 2 & 10 & 2 \\ 1 & 1 & 1 \end{pmatrix}\)
\(\vec{AB} \times \vec{AC} = i \begin{pmatrix} 10 & 2 \\ 1 & 1 \end{pmatrix} - j\begin{pmatrix}2 & 2 \\ 1 & 1\end{pmatrix} + k \begin{pmatrix}2 & 10 \\ 1 & 1\end{pmatrix} = (8, -2, 8)\)
03
Finding the magnitude of the cross product
Next, we need to find the magnitude of the cross product vector. We can do this using the formula for the magnitude of a vector:
\(||\vec{AB} \times \vec{AC}|| = \sqrt{(8)^2 + (-2)^2 + (8)^2} = \sqrt{144} = 12\)
04
Calculate the area of the triangle
Finally, we can find the area of the triangle using the formula:
\(Area = \frac{1}{2}||\vec{AB} \times \vec{AC}|| = \frac{1}{2}(12) = 6\)
The area of the triangle with vertices \(A(5,6,2), B(7,16,4),\) and \(C(6,7,3)\) is 6 square units.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Area of a Triangle
To find the area of a triangle in three-dimensional space, you can use the concept of the cross product. When you have a triangle with vertices given by three points in 3D space, like points \( A \), \( B \), and \( C \), you can form two vectors from these points, say \( \vec{AB} \) and \( \vec{AC} \). The cross product of these two vectors gives you a new vector perpendicular to the plane containing \( \vec{AB} \) and \( \vec{AC} \).
The magnitude of this cross product vector is directly related to the area of the parallelogram formed by the vectors \( \vec{AB} \) and \( \vec{AC} \). Since a triangle is half of a parallelogram, you can find the area of the triangle by taking half of the magnitude of the cross product:
The magnitude of this cross product vector is directly related to the area of the parallelogram formed by the vectors \( \vec{AB} \) and \( \vec{AC} \). Since a triangle is half of a parallelogram, you can find the area of the triangle by taking half of the magnitude of the cross product:
- Compute the vectors \( \vec{AB} \) and \( \vec{AC} \).
- Find the cross product \( \vec{AB} \times \vec{AC} \).
- Calculate the magnitude \( || \vec{AB} \times \vec{AC} || \).
- Divide by 2 to find the triangle's area: \( \frac{1}{2} || \vec{AB} \times \vec{AC} || \).
Magnitude of a Vector
The magnitude of a vector, often called its "length" or "norm," is a measure of how long the vector is. In three-dimensional space, a vector is represented as \( \vec{v} = (x, y, z) \). To find its magnitude, you apply the formula:
\[||\vec{v}|| = \sqrt{x^2 + y^2 + z^2}\]
This formula reflects the Pythagorean Theorem, where you square each component of the vector, sum these squares, and then take the square root. The magnitude essentially "lengthens" the vector into a simple scalar distance measurement.
For example, if you have a vector \( \vec{v} = (8, -2, 8) \), as obtained from a cross product, you can compute its magnitude like so:
\[||\vec{v}|| = \sqrt{x^2 + y^2 + z^2}\]
This formula reflects the Pythagorean Theorem, where you square each component of the vector, sum these squares, and then take the square root. The magnitude essentially "lengthens" the vector into a simple scalar distance measurement.
For example, if you have a vector \( \vec{v} = (8, -2, 8) \), as obtained from a cross product, you can compute its magnitude like so:
- Square each component: \( 8^2 = 64 \), \( (-2)^2 = 4 \), and \( 8^2 = 64 \).
- Sum these squares: \( 64 + 4 + 64 = 132 \).
- Take the square root: \( \sqrt{132} \approx 11.48 \).
Three-Dimensional Vectors
Three-dimensional vectors are mathematical objects that have both direction and magnitude in a three-dimensional space. They can be used to describe positions, forces, and more within that space. A vector in 3D is typically represented in Cartesian coordinates as \( \vec{v} = (x, y, z) \), where \( x \), \( y \), and \( z \) are its components along the respective axes.
In the context of calculating areas, three-dimensional vectors can define edges of a triangle formed by points in 3D. If you take the points \( A \), \( B \), and \( C \) as vertices, you can form vectors \( \vec{AB} \) and \( \vec{AC} \), which are found by subtracting the coordinates of two points, e.g., \( \vec{AB} = B - A \).
In the context of calculating areas, three-dimensional vectors can define edges of a triangle formed by points in 3D. If you take the points \( A \), \( B \), and \( C \) as vertices, you can form vectors \( \vec{AB} \) and \( \vec{AC} \), which are found by subtracting the coordinates of two points, e.g., \( \vec{AB} = B - A \).
- Vectors are added by simply adding their corresponding components.
- Subtraction works similarly, leading to vectors like \( \vec{AB} \) = (7-5, 16-6, 4-2).
- These operations allow vectors to describe displacement and orientation in space effectively.