Question

Find the area of the parallelogram defined by the given vectors. \(\vec{u}=\langle-2,1,5\rangle, \quad \vec{v}=\langle-1,3,1\rangle\)

Short answer

The area of the parallelogram is \( \sqrt{270} \).

Step-by-step solution

Understanding the Formula

To find the area of a parallelogram defined by two vectors, you use the cross product of the vectors. The magnitude of the cross product vector gives the area. If \( \vec{a} = \langle a_1, a_2, a_3 \rangle \) and \( \vec{b} = \langle b_1, b_2, b_3 \rangle \), then the cross product \( \vec{a} \times \vec{b} \) is calculated using the determinant of a 3x3 matrix.

Calculate the Cross Product

Calculate the cross product \( \vec{u} \times \vec{v} \). \[\vec{u} \times \vec{v} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \-2 & 1 & 5 \-1 & 3 & 1 \\end{vmatrix}\]This expands to: \( \mathbf{i}(1 \cdot 1 - 5 \cdot 3) - \mathbf{j}(-2 \cdot 1 - 5 \cdot (-1)) + \mathbf{k}(-2 \cdot 3 - 1 \cdot (-1)) \).Simplify this to get:\( \mathbf{i}(-14) + \mathbf{j}(7) + \mathbf{k}(-5) \).Thus, \( \vec{u} \times \vec{v} = \langle -14, 7, -5 \rangle \).

Calculate the Magnitude

The magnitude of a vector \( \vec{a} = \langle a_1, a_2, a_3 \rangle \) is given by the formula \[ \|\vec{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2} \].Apply this to the cross product vector \( \langle -14, 7, -5 \rangle \):\[ \|\vec{u} \times \vec{v}\| = \sqrt{(-14)^2 + 7^2 + (-5)^2} \]. Simplify the calculation:\[ \|\vec{u} \times \vec{v}\| = \sqrt{196 + 49 + 25} = \sqrt{270} \].

Key concepts

Cross Product

The cross product of two vectors is a fundamental concept in vector calculus, especially within three-dimensional space. It results in a vector that is perpendicular to the plane formed by the original vectors. This unique feature can be immensely helpful when trying to find areas or solve geometric problems.
When you compute the cross product of vectors such as \( \vec{a} = \langle a_1, a_2, a_3 \rangle \) and \( \vec{b} = \langle b_1, b_2, b_3 \rangle \), you use determinants of a matrix to find the resulting vector:
\[\vec{a} \times \vec{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix}\]
- Using this expansion, you calculate each component:
- **i-component:** \( a_2 \times b_3 - a_3 \times b_2 \)
- **j-component:** \( -(a_1 \times b_3 - a_3 \times b_1) \)
- **k-component:** \( a_1 \times b_2 - a_2 \times b_1 \)
For example, the cross product of vectors \( \vec{u}=\langle -2, 1, 5 \rangle \) and \( \vec{v}=\langle -1, 3, 1 \rangle \) results in a new vector \( \langle -14, 7, -5 \rangle \). This vector is orthogonal to both \( \vec{u} \) and \( \vec{v} \).
The cross product plays a vital role, especially for defining areas like those of parallelograms and determining torque in physics.

Magnitude of a Vector

The magnitude of a vector gives an idea of its length or size. Understanding how to calculate it helps to measure the size or extent of the vector in space.
For a vector \( \vec{a}=\langle a_1, a_2, a_3 \rangle \), the magnitude is calculated using the formula:
\[ \|\vec{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2} \]
This formula effectively extends the Pythagorean theorem into three dimensions.
When working with vectors like the cross product result \( \langle -14, 7, -5 \rangle \), you find:
- Compute each component squared: \((-14)^2 = 196\), \(7^2 = 49\), \((-5)^2 = 25\)
- Add them together: \(196 + 49 + 25 = 270\)
- Take the square root: \(\sqrt{270}\)
Evaluating this gives an approximate value of 16.43.
This magnitude equates to the length of the resulting vector—the cross product, which is crucial for geometric interpretations like area measurements.

Area of a Parallelogram

The formula for the area of a parallelogram created by two vectors involves using the cross product. Intuitively, the parallelogram's area corresponds to the magnitude of the cross product vector.
The key steps involve:
- Calculate the cross product of the given pair of vectors.
- Find the magnitude of this cross product to determine the parallelogram's area.
For example, given vectors \( \vec{u} = \langle -2, 1, 5 \rangle \) and \( \vec{v} = \langle -1, 3, 1 \rangle \), the resultant vector from the cross product is \( \langle -14, 7, -5 \rangle \). The magnitude of this vector, \( \sqrt{270} \), provides the area enclosed by the parallelogram.
The measured area, therefore, quantifies the space within the parallelogram, flourishing from the richness of vector calculus.