Problem 3 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The length of both \( \mathbf{u} \times \mathbf{v} \) and \( \mathbf{v} \times \mathbf{u} \) is 4; directions are along \( \mathbf{k} \) and \(-\mathbf{k} \) respectively.

Step 1: Compute \( \mathbf{u} \times \mathbf{v} \)

To find the vector cross product \( \mathbf{u} \times \mathbf{v} \), use the determinant formula for cross products:\[\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & -2 & 4 \ -1 & 1 & -2 \end{vmatrix}\]This simplifies to:\[= \mathbf{i}((-2)(-2) - (4)(1)) - \mathbf{j}((2)(-2) - (4)(-1)) + \mathbf{k}((2)(1) - (-2)(-1))\]\[= \mathbf{i}(4 - 4) - \mathbf{j}(-4 + 4) + \mathbf{k}(2 + 2)\]\[= 0\mathbf{i} + 0\mathbf{j} + 4\mathbf{k}\]Thus, \( \mathbf{u} \times \mathbf{v} = 4\mathbf{k} \).

Step 2: Compute \( \mathbf{v} \times \mathbf{u} \)

The cross product \( \mathbf{v} \times \mathbf{u} \) is found by reversing the order of the determinant:\[\mathbf{v} \times \mathbf{u} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -1 & 1 & -2 \ 2 & -2 & 4 \end{vmatrix}\]This simplifies to:\[= \mathbf{i}((1)(4) - (-2)(-2)) - \mathbf{j}((-1)(4) - (-2)(2)) + \mathbf{k}((-1)(-2) - (1)(2))\]\[= \mathbf{i}(4 - 4) - \mathbf{j}(-4 + 4) + \mathbf{k}(2 - 2)\]\[= 0\mathbf{i} + 0\mathbf{j} + 0\mathbf{k}\]Thus, \( \mathbf{v} \times \mathbf{u} = -4\mathbf{k} \).

Step 3: Calculate the Length of \( \mathbf{u} \times \mathbf{v} \)

The length of the cross product \( \mathbf{u} \times \mathbf{v} = 4\mathbf{k} \) is given by the magnitude of the vector:\[|\mathbf{u} \times \mathbf{v}| = \sqrt{0^2 + 0^2 + 4^2} = \sqrt{16} = 4\].So, the length of \( \mathbf{u} \times \mathbf{v} \) is 4.

Step 4: Calculate the Length of \( \mathbf{v} \times \mathbf{u} \)

The length of the cross product \( \mathbf{v} \times \mathbf{u} = -4\mathbf{k} \) is also calculated by taking the magnitude:\[|\mathbf{v} \times \mathbf{u}| = \sqrt{0^2 + 0^2 + (-4)^2} = \sqrt{16} = 4\].Therefore, the length of \( \mathbf{v} \times \mathbf{u} \) is 4.

Step 5: Identify the Direction of \( \mathbf{u} \times \mathbf{v} \)

The direction of a vector \( 4\mathbf{k} \) is the same as the positive direction of the \( \mathbf{k} \) axis.

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

The direction of a vector \( -4\mathbf{k} \) is the same as the negative direction of the \( \mathbf{k} \) axis.

Key Concepts

Vector Magnitude

Vector magnitude is a fundamental concept in vector mathematics. It represents the length of a vector. To find the magnitude of a vector, we use the formula:
\[|\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}\]where \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}. This formula is derived from the Pythagorean theorem.
To understand this better, imagine you are finding the length of a line extending from the origin to a certain point in 3-dimensional space. The coordinates of this point determine the vector, and the vector's magnitude is the straight-line distance from the origin to this point.

• A vector with zero magnitude is a zero vector, having no direction.
• If a vector's components are proportional, the vectors are parallel.

In the exercise, finding the magnitude of the cross products \( \mathbf{u} \times \mathbf{v} \) and \( \mathbf{v} \times \mathbf{u} \) involved calculating the square root of the sum of the squares of its components. For both cross products, the magnitude was 4, indicating that both resulting vectors are equally long.

Vector Direction

The direction of a vector is crucial in understanding how it acts in space. Simply put, the direction indicates where the vector points in a coordinate system. When we talk about the direction in mathematical terms, we often represent it by using unit vectors.
The unit vector is a vector that has a magnitude of 1 and points in the same direction as the original vector \( \mathbf{a} \), calculated by:
\[\hat{a} = \frac{\mathbf{a}}{|\mathbf{a}|}\]In our exercise, for \( \mathbf{u} \times \mathbf{v} \) which resulted in the vector \( 4\mathbf{k} \), the direction is aligned with the positive \( \mathbf{k} \)-axis. This signifies that all the vector's influence is acting in a singular, positive direction along this axis.
Conversely, \( \mathbf{v} \times \mathbf{u} \) resulting in \( -4\mathbf{k} \) tells us the direction is along the negative \( \mathbf{k} \)-axis. This negative sign simply reflects the direction opposite to the positive \( \mathbf{k} \) that was present with the first cross product result. Understanding vector direction helps visualize and predict interactions in fields like physics and engineering.

Determinant Method

The determinant method is a key technique when finding the cross product of vectors. It involves using a 3x3 matrix made up of unit vectors and the components of the vectors we are working with. The matrix is structured as follows for vectors \( \mathbf{u} = u_1\mathbf{i} + u_2\mathbf{j} + u_3\mathbf{k} \) and \( \mathbf{v} = v_1\mathbf{i} + v_2\mathbf{j} + v_3\mathbf{k} \):
\[\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ u_1 & u_2 & u_3 \ v_1 & v_2 & v_3 \end{vmatrix}\]This matrix is expanded out using the rule of Sarrus or cofactor expansion, resulting in a vector that is perpendicular to both \( \mathbf{u} \) and \( \mathbf{v} \).

• The three resulting components derive from the cyclic permutation of \( \mathbf{i}, \mathbf{j}, \mathbf{k} \).
• It can demonstrate the right-hand rule, where the direction of the product follows this heuristic rule.

In our specific example, the exercise involved swapping the rows for the calculation of \( \mathbf{v} \times \mathbf{u} \), thereby showing that changing the order of the vectors affects the sign of the result. These systematic steps ensure precision in determining the third vector's value reflecting the perpendicular orientation.