Problem 9 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the area of the triangle with \((3,2,1),(2,4,6)\), and \((-1,2,5)\) as vertices.

Step-by-Step Solution

Answer

The area of the triangle is \(2\sqrt{13}\) square units.

Step 1: Identify the Coordinates

Consider the given vertices: - Point A: (3, 2, 1) - Point B: (2, 4, 6) - Point C: (-1, 2, 5). These represent the points in 3-dimensional space.

Step 2: Calculate the Vectors Formed by Two Edges

We need two vectors originating from the same point, so let's choose A as the origin: - Vector AB: B - A = (2-3, 4-2, 6-1) = (-1, 2, 5) - Vector AC: C - A = (-1-3, 2-2, 5-1) = (-4, 0, 4).

Step 3: Find the Cross Product of the Vectors

The cross product of vectors AB and AC gives a vector perpendicular to the plane containing the triangle: \[\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \-1 & 2 & 5 \-4 & 0 & 4\end{vmatrix}\]Calculate: \[ (-1)(0)-(2)(4) \mathbf{i} - [(5)(-4)-(4)(5)] \mathbf{j} + [(-1)(4)-(-4)(2)] \mathbf{k} \]\[= -8 \mathbf{i} + 0 \mathbf{j} + 12 \mathbf{k}\]

Step 4: Calculate the Magnitude of the Cross Product

The magnitude of the cross product vector gives twice the area of the triangle:\[ | -8 \mathbf{i} + 0 \mathbf{j} + 12 \mathbf{k} | = \sqrt{(-8)^2 + 0^2 + 12^2} \]\[ = \sqrt{64 + 0 + 144} = \sqrt{208} = 4\sqrt{13}.\]

Step 5: Calculate the Area of the Triangle

Divide the magnitude of the cross product by 2 to find the area:\[ \text{Area} = \frac{1}{2} \times 4\sqrt{13} = 2\sqrt{13}. \]Thus, the area of the triangle is \(2\sqrt{13}\) square units.

Key Concepts

Vector Cross Product

When dealing with geometry in three dimensions, understanding the vector cross product is crucial. The cross product of two vectors results in a third vector that is perpendicular to both input vectors. This cross product is useful in various applications, especially when you're finding areas or checking perpendicularity between planes.

For example, in our problem, we had two vectors from the triangle's vertices, AB and AC. To find a vector perpendicular to these in three-dimensional space, we perform the cross product. This uses a determinant that involves the unit vectors \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\), and the components of the vectors.

• Use the determinant: \[\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -1 & 2 & 5 \ -4 & 0 & 4 \end{vmatrix}\]
• Calculate each component of the resulting vector.
• The result from our exercise gives the vector \(-8\mathbf{i} + 0\mathbf{j} + 12\mathbf{k}\).

This vector tells us about the orientation of the plane on which the triangle lies.

Area of a Triangle

In three-dimensional space, finding the area of a triangle requires some unique strategies. Unlike two-dimensional geometry, you can't just use the simple base times height formula. Instead, one efficient method is using vectors and the cross product.

If you have vectors derived from the triangle's vertices, you can find the area by first computing the cross product of these vectors to get a vector perpendicular to the plane. The magnitude of this vector is crucial, as it directly relates to the triangle's area.

• The area of the triangle is half the magnitude of the cross product vector.
• This is because the cross product's magnitude is analogous to twice the area of the parallelogram formed by the two vectors.
• Our calculated vector \(-8\mathbf{i} + 0\mathbf{j} + 12\mathbf{k}\) has a magnitude of \(4\sqrt{13}\).

Thus, the area of the triangle can be found by taking \(\frac{1}{2}\) of this magnitude, which in this case is \(2\sqrt{13}\).

Calculating Magnitudes

Calculating the magnitude of a vector is a fundamental aspect of vector mathematics. It allows you to determine the size or length of the vector, essential for numerous applications in physics and engineering.

For a 3-dimensional vector \(\mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}\), the magnitude is calculated using the formula \(\sqrt{a^2 + b^2 + c^2}\). This is derived directly from the Euclidean distance formula.

• Write down the components of the vector. For instance, \(-8\) for \(\mathbf{i}\), \(0\) for \(\mathbf{j}\), and \(12\) for \(\mathbf{k}\).
• Square each component: \((-8)^2 = 64\), \(0^2 = 0\), and \(12^2 = 144\).
• Add these squares together and take the square root to find the magnitude: \(\sqrt{64 + 0 + 144} = \sqrt{208} = 4\sqrt{13}\).

This method applies no matter the dimensions and holds true for any vectors encountered in geometric problems.