Question

Find the area of the following triangles \(T\). (The area of a triangle is half the area of the corresponding parallelogram.) The sides of \(T\) are \(\mathbf{u}=\langle 3,3,3\rangle, \mathbf{v}=\langle 6,0,6\rangle,\) and \(\mathbf{u}-\mathbf{v}\)

Short answer

Answer: The area of the triangle is 4.5.

Step-by-step solution

Find Cross Product of \(\mathbf{u}\) and \(\mathbf{v}\)

Firstly, we calculate the cross product of vectors \(\mathbf{u}\) and \(\mathbf{v}\). The cross product is defined as: \(\mathbf{u} \times \mathbf{v} = \langle u_2v_3-u_3v_2, u_3v_1-u_1v_3, u_1v_2-u_2v_1 \rangle\) where \(\mathbf{u}=\langle u_1,u_2,u_3 \rangle\) and \(\mathbf{v}=\langle v_1,v_2,v_3 \rangle.\) We plug in the given values: \(\mathbf{u} \times \mathbf{v} = \langle (3)(6)- (3)(6),(3)(6)-(3)(6),(3)(0)-(3)(3)\rangle = \langle 0, 0 , -9 \rangle\)

Magnitude of Cross Product

Now, we need to find the magnitude of the cross product: \(|\mathbf{u} \times \mathbf{v}| = \sqrt{0^2 + 0^2 + (-9)^2} = \sqrt{81} = 9\)

Find the Area of the Triangle

As the area of a triangle is half the area of the corresponding parallelogram, we divide the magnitude of the cross product by 2: \(Area~of~Triangle =\frac{|\mathbf{u} \times \mathbf{v}|}{2}=\frac{9}{2}=4.5\) So, the area of the triangle \(T\) is 4.5.

Key concepts

Cross Product

The cross product is a crucial concept when dealing with vectors in three-dimensional space. It provides a way to find a vector that is orthogonal to two given vectors. In simpler terms, the cross product of two vectors results in another vector that points perpendicular to both.
This is particularly useful in physics and engineering, for instance, when determining the direction of the force exerted by a torque.To compute the cross product for vectors \(\mathbf{u}\) and \(\mathbf{v}\), you can use the determinant-like structure:

• \(\mathbf{u} \times \mathbf{v} = \langle u_2v_3-u_3v_2, u_3v_1-u_1v_3, u_1v_2-u_2v_1 \rangle\)

In our exercise, with \(\mathbf{u}=\langle 3,3,3\rangle\) and \(\mathbf{v}=\langle 6,0,6\rangle\), the cross product calculation yields \(\langle 0, 0, -9 \rangle\).
Notice how each component is calculated using a small part of each vector's coordinates, emphasizing the perpendicular nature of the result.

Parallelogram Area

The area of a parallelogram is closely related to vectors and their cross product. When two vectors are positioned such that they originate from the same point, they form a parallelogram.
The area of this parallelogram can be found by calculating the magnitude of the cross product of these two vectors.The steps to find the area using vectors \(\mathbf{u}\) and \(\mathbf{v}\) can be summarized as:

• Calculate the cross product: \(\mathbf{u} \times \mathbf{v}\).
• Find the magnitude of the resulting vector to determine the area of the parallelogram.

In our scenario, the resulting vector from the cross product was \(\langle 0, 0, -9 \rangle\), and hence, the magnitude \(|\mathbf{u} \times \mathbf{v}| = \sqrt{0^2 + 0^2 + (-9)^2} = 9\).
Thus, the area of the parallelogram is 9. Remember, for a triangle that shares the same base and height, the area is half of this.

Vector Magnitude

The magnitude of a vector essentially measures its length. In a context where vectors represent physical quantities like force or displacement, understanding magnitude helps us gauge their intensity or distance.The magnitude of a vector \(\mathbf{a} = \langle a_1, a_2, a_3 \rangle\) is calculated using the formula:

• \(|\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}\)

This formula is a natural extension into three dimensions of the Pythagorean theorem.In the exercise, we calculated the magnitude of the cross product vector \(\mathbf{u} \times \mathbf{v} = \langle 0, 0, -9 \rangle\). By applying the magnitude formula, we found \(|\mathbf{u} \times \mathbf{v}| = \sqrt{81} = 9\).
This step is vital as it underpins the calculation of both the area of the parallelogram and subsequently the triangle.