Problem 7 - Thomas Calculus

Question

In Exercises \(1-8,\) find the length and direction (when defined) of \(\mathbf{u} \times \mathbf{v}\) and \(\mathbf{v} \times \mathbf{u} .\) $$ \mathbf{u}=-8 \mathbf{i}-2 \mathbf{j}-4 \mathbf{k}, \quad \mathbf{v}=2 \mathbf{i}+2 \mathbf{j}+\mathbf{k} $$

Step-by-Step Solution

Answer

Lengths are both \(6\sqrt{5}\). Directions are opposite: \(\frac{1}{\sqrt{5}}\mathbf{i} - \frac{2}{\sqrt{5}}\mathbf{k}\) and \(-\frac{1}{\sqrt{5}}\mathbf{i} + \frac{2}{\sqrt{5}}\mathbf{k}\).

Step 1: Find the Cross Product \( \mathbf{u} \times \mathbf{v} \)

The cross product \( \mathbf{u} \times \mathbf{v} \) is given by the determinant of the matrix: \[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -8 & -2 & -4 \ 2 & 2 & 1 \end{vmatrix} \] To evaluate this determinant, use the formula: \[ \mathbf{u} \times \mathbf{v} = \mathbf{i}((-2)(1) - (-4)(2)) - \mathbf{j}((-8)(1) - (-4)(2)) + \mathbf{k}((-8)(2) - (-2)(2)). \] Calculating: - For \( \mathbf{i} \): \((-2)(1) - (-4)(2) = -2 + 8 = 6\)- For \( \mathbf{j} \): \((-8)(1) - (-4)(2) = -8 + 8 = 0\)- For \( \mathbf{k} \): \((-8)(2) - (-2)(2) = -16 + 4 = -12\) Thus, \( \mathbf{u} \times \mathbf{v} = 6\mathbf{i} + 0\mathbf{j} - 12\mathbf{k} = 6\mathbf{i} - 12\mathbf{k} \).

Step 2: Calculate the Length of \( \mathbf{u} \times \mathbf{v} \)

The length of a vector \( \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} \) is given by: \[ \|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2} \] So, for \( \mathbf{u} \times \mathbf{v} = 6\mathbf{i} + 0\mathbf{j} - 12\mathbf{k} \), the length is: \[ \|\mathbf{u} \times \mathbf{v}\| = \sqrt{6^2 + 0^2 + (-12)^2} = \sqrt{36 + 0 + 144} = \sqrt{180} = \sqrt{36 \times 5} = 6\sqrt{5} \].

Step 3: Find the Direction of \( \mathbf{u} \times \mathbf{v} \)

The direction of vector \( \mathbf{u} \times \mathbf{v} \) is given by the unit vector: \[ \mathbf{a}_{\text{unit}} = \frac{\mathbf{a}}{\|\mathbf{a}\|} \] For \( \mathbf{u} \times \mathbf{v} = 6\mathbf{i} + 0\mathbf{j} - 12\mathbf{k} \), \[ \mathbf{u} \times \mathbf{v}_{\text{unit}} = \frac{6\mathbf{i} + 0\mathbf{j} - 12\mathbf{k}}{6\sqrt{5}} = \frac{1}{\sqrt{5}}\mathbf{i} - \frac{2}{\sqrt{5}}\mathbf{k} \].

Step 4: Find the Cross Product \( \mathbf{v} \times \mathbf{u} \)

Using the property \( \mathbf{v} \times \mathbf{u} = - (\mathbf{u} \times \mathbf{v}) \), therefore \( \mathbf{v} \times \mathbf{u} = -(6\mathbf{i} - 12\mathbf{k}) = -6\mathbf{i} + 12\mathbf{k} \).

Step 5: Calculate the Length of \( \mathbf{v} \times \mathbf{u} \)

The length of \( \mathbf{v} \times \mathbf{u} \) is the same as \( \mathbf{u} \times \mathbf{v} \) since flipping the cross product only changes direction, not length. Thus, \[ \|\mathbf{v} \times \mathbf{u}\| = 6\sqrt{5} \].

Step 6: Find the Direction of \( \mathbf{v} \times \mathbf{u} \)

Similarly, the direction is just the negative of the direction of \( \mathbf{u} \times \mathbf{v} \). So, the unit vector is: \[ \mathbf{v} \times \mathbf{u}_{\text{unit}} = -\left(\frac{1}{\sqrt{5}}\mathbf{i} - \frac{2}{\sqrt{5}}\mathbf{k}\right) = -\frac{1}{\sqrt{5}}\mathbf{i} + \frac{2}{\sqrt{5}}\mathbf{k} \].

Key Concepts

Determinants in Linear Algebra

When we are working with vectors and need to calculate the cross product, we often use determinants. In linear algebra, a determinant is a scalar value that provides important information about a matrix. For vectors in three-dimensional space, we use a 3x3 matrix to calculate the cross product using the format:

• First row: Unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \)
• Second row: Components of the first vector
• Third row: Components of the second vector

To solve the determinant, you expand it by minors, which involves finding determinants of smaller 2x2 matrices inside the larger matrix. This might sound a bit complex, but it simply requires focusing on one row to eliminate one variable at a time, multiplying and summing the resulting products.

Vector Length

The length or magnitude of a vector is a measure of how long the vector is, which we calculate using the Pythagorean theorem. If a vector can be represented as \( \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} \), its length is given by:\[ \|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2} \]For instance, in our exercise, the cross product \( \mathbf{u} \times \mathbf{v} = 6\mathbf{i} + 0\mathbf{j} - 12\mathbf{k} \) has a length \( \sqrt{180} = 6\sqrt{5} \). Understanding vector length is crucial because it tells us about the size of the vector regardless of its direction.

Unit Vector Calculation

A unit vector is a vector that has a length of one, pointing in a specific direction. To convert any vector into a unit vector, divide the vector by its magnitude. The formula is:\[ \mathbf{a}_{\text{unit}} = \frac{\mathbf{a}}{\|\mathbf{a}\|} \]This process, known as normalization, scales the original vector to a unit vector without changing its direction. In our example, the unit vector for \( \mathbf{u} \times \mathbf{v} \) is \( \frac{1}{\sqrt{5}}\mathbf{i} - \frac{2}{\sqrt{5}}\mathbf{k} \). Unit vectors are significant in physics and engineering because they simplify directional expressions and calculations.

Direction of Vectors

The direction of a vector determines where it points in space. It's typically represented by a unit vector. This ensures the vector shows direction without being influenced by magnitude. For cross products like \( \mathbf{u} \times \mathbf{v} \), the direction is perpendicular to the plane formed by the original two vectors.
Vector direction plays a crucial role in applications such as finding torques and forces, where direction indicates the axis and sense of rotation or movement.