Problem 50 - Advanced Engineering Mathematics

Question

Find the area of the triangle determined by the given points. $$ P_{1}(1,0,3), P_{2}(0,0,6), P_{3}(2,4,5) $$

Step-by-Step Solution

Answer

The area of the triangle is \(\frac{\sqrt{185}}{2}\).

Step 1: Understanding the Problem

We are given three points in three-dimensional space and need to find the area of the triangle formed by these points. Let the points be \(P_1(1,0,3)\), \(P_2(0,0,6)\), and \(P_3(2,4,5)\).

Step 2: Calculate Vectors From Points

First, we find the vectors representing the sides of the triangle. We need vectors \(\vec{a} = \overrightarrow{P_1P_2}\) and \(\vec{b} = \overrightarrow{P_1P_3}\). Use the formula \(\overrightarrow{AB} = (x_2-x_1, y_2-y_1, z_2-z_1)\). Compute:\[\vec{a} = P_2 - P_1 = (0-1, 0-0, 6-3) = (-1, 0, 3)\],\[\vec{b} = P_3 - P_1 = (2-1, 4-0, 5-3) = (1, 4, 2)\].

Step 3: Compute the Cross Product

The area of the triangle is half the magnitude of the cross product of vectors \(\vec{a}\) and \(\vec{b}\). Compute the cross product \(\vec{a} \times \vec{b}\) using the determinant method:\[\vec{a} \times \vec{b} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \-1 & 0 & 3 \1 & 4 & 2\end{vmatrix}\]This results in:\[= \mathbf{i}(0\cdot2 - 3\cdot4) - \mathbf{j}(-1\cdot2 - 3\cdot1) + \mathbf{k}(-1\cdot4 - 0\cdot1)\]\[= \mathbf{i}(-12) - \mathbf{j}(-2 - 3) + \mathbf{k}(-4)\]\[= (-12, 5, -4)\].

Step 4: Find the Magnitude of the Cross Product

The magnitude of \(\vec{a} \times \vec{b}\) gives us twice the area of the triangle. Compute the magnitude:\[|\vec{a} \times \vec{b}| = \sqrt{(-12)^2 + 5^2 + (-4)^2}\]\[= \sqrt{144 + 25 + 16}\]\[= \sqrt{185}\].

Step 5: Compute the Area of the Triangle

The area of the triangle, \(A\), is half of the magnitude of the cross product:\[A = \frac{1}{2} |\vec{a} \times \vec{b}| = \frac{1}{2} \sqrt{185}\]\[= \frac{\sqrt{185}}{2}\].

Key Concepts

Cross Product

The cross product is a crucial operation in vector geometry, especially when working in three-dimensional space. It takes two vectors and produces a third vector that is perpendicular to both. This results in a vector that represents a plane area.
The formula for the cross product of two vectors \( \vec{a} = (a_1, a_2, a_3) \) and \( \vec{b} = (b_1, b_2, b_3) \) is:
\[\vec{a} \times \vec{b} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3\end{vmatrix}\]This can be expanded to:- \( \mathbf{i}(a_2b_3 - a_3b_2) \)- \( -\mathbf{j}(a_1b_3 - a_3b_1) \)- \( \mathbf{k}(a_1b_2 - a_2b_1) \)
In our exercise, we find the cross product of vectors \( \vec{a} = (-1, 0, 3) \) and \( \vec{b} = (1, 4, 2) \). This results in the vector \( \vec{a} \times \vec{b} = (-12, 5, -4) \). It provides a direction that is perpendicular to the plane formed by the original vectors.

Triangle Area in 3D

In three-dimensional vector geometry, the area of a triangle formed by three points can be determined using vectors and the cross product.
First, identify the two sides of the triangle as vectors. For this, take any two of the three given points to form vectors along the triangle's sides. In our problem, the points are:

• \(P_1(1, 0, 3)\)
• \(P_2(0, 0, 6)\)
• \(P_3(2, 4, 5)\)

Next, compute vectors \( \vec{a} = \overrightarrow{P_1P_2} \) and \( \vec{b} = \overrightarrow{P_1P_3} \) using the formula \( \overrightarrow{AB} = (x_2-x_1, y_2-y_1, z_2-z_1) \).
The area of the triangle is half the magnitude of the vector resulting from the cross product of these two vectors:\[A = \frac{1}{2} |\vec{a} \times \vec{b}|\]This gives a practical method to find the area of a triangle in 3D space without resorting to more complicated geometry.

Magnitude of Vectors

The magnitude of a vector provides a measure of its length or size. Calculating the magnitude is essential in many applications of vector geometry.
For a vector \( \vec{v} = (v_1, v_2, v_3) \), the magnitude is obtained through the formula:\[|\vec{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}\]
In our exercise, after finding the cross product \( \vec{a} \times \vec{b} = (-12, 5, -4) \), we compute its magnitude to determine how large or impactful this vector is in space. This calculation is essential because the magnitude of the cross product directly relates to the area of the triangle. Here, the magnitude is:\[|\vec{a} \times \vec{b}| = \sqrt{(-12)^2 + 5^2 + (-4)^2} = \sqrt{185}\]
This magnitude, when divided by 2, gives us the exact area of the triangle formed by the points in three-dimensional space.