Problem 1 - University Calculus: Early Transcendentals

Question

Find the length and direction (when defined) of \(\mathbf{u} \times \mathbf{v}\) and \(\mathbf{v} \times \mathbf{u}\). $$\mathbf{u}=2 \mathbf{i}-2 \mathbf{j}-\mathbf{k}, \quad \mathbf{v}=\mathbf{i}-\mathbf{k}$$

Step-by-Step Solution

Answer

The length is \(\sqrt{17}\), and \(\mathbf{u} \times \mathbf{v}\) and \(\mathbf{v} \times \mathbf{u}\) are anti-parallel.

Step 1: Calculate Cross Product

We will find the cross product \(\mathbf{u} \times \mathbf{v}\) using the determinant method. The formula for the cross product is given by \((\mathbf{u} \times \mathbf{v}) = |\begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & -2 & -1 \ 1 & 0 & -1 \end{array}|\). We calculate the determinant by expanding along the first row.

Step 2: Expand Determinant for Cross Product

Calculate the determinant: \( (\mathbf{u} \times \mathbf{v}) = \mathbf{i}((-2)(-1) - (0)(-1)) - \mathbf{j}(2(-1) - 1(-1)) + \mathbf{k}(2(0) - (-2)(1)) \). Simplifying these terms gives \( \mathbf{i}(2) - \mathbf{j}(-3) + \mathbf{k}(-2) \), which results in \(2\mathbf{i} + 3\mathbf{j} - 2\mathbf{k}\).

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

To find the length, use the formula \(\|\mathbf{u} \times \mathbf{v}\| = \sqrt{2^2 + 3^2 + (-2)^2}\). Calculating, \(\sqrt{4 + 9 + 4} = \sqrt{17}\).

Step 4: Calculation for \(\mathbf{v} \times \mathbf{u}\)

The operation \(\mathbf{v} \times \mathbf{u}\) involves switching \(\mathbf{u}\) and \(\mathbf{v}\), giving \(|\begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 0 & -1 \ 2 & -2 & -1 \end{array}|\). Calculate the determinant: \(1(-2 - 2) - \mathbf{j}(-2 \times -1 - (-1) \times 2) + (-1)(-2 - 0)\), resulting in \( -4\mathbf{i} + 0\mathbf{j} + 2\mathbf{k} \), simplified as \(-2\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}\).

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

Use the same method as before for \(\mathbf{v} \times \mathbf{u}\) with vector \(-2\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}\). Its length is \(\sqrt{(-2)^2 + (-3)^2 + 2^2} = \sqrt{4 + 9 + 4} = \sqrt{17}\).

Step 6: Conclude the Direction (If Any)

Both vectors \(\mathbf{u} \times \mathbf{v}\) and \(\mathbf{v} \times \mathbf{u}\) have the same magnitude, \(\sqrt{17}\), but opposite directions (note: their vectors are negatives of each other), demonstrating they are anti-parallel.

Key Concepts

Determinant Method

The determinant method is a powerful technique used to calculate the cross product of two vectors. When you perform a cross product, you're essentially determining the vector that's perpendicular to both original vectors. To use the determinant method, arrange the vectors and unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) in a 3x3 matrix. The cross product \( \mathbf{u} \times \mathbf{v} \) is expressed as the determinant of this matrix:\[(\mathbf{u} \times \mathbf{v}) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ u_1 & u_2 & u_3 \ v_1 & v_2 & v_3 \end{vmatrix}\]For example, in your exercise, vector \( \mathbf{u} = 2\mathbf{i} - 2\mathbf{j} - \mathbf{k} \) and vector \( \mathbf{v} = \mathbf{i} - \mathbf{k} \). You expand the determinant by picking a row and applying alternated signs to the corresponding minors. This expansion results in terms which you simplify to generate the final cross product vector.

Length of a Vector

The length of a vector, also known as its magnitude, is a measure of how long the vector is. To find it, you need to apply the Pythagorean theorem in a three-dimensional space. If you have a vector represented as \( a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \), the formula for its length \( \| \mathbf{a} \| \) is:\[\| \mathbf{a} \| = \sqrt{a^2 + b^2 + c^2}\]This method was applied in the exercise to both \( \mathbf{u} \times \mathbf{v} \) and \( \mathbf{v} \times \mathbf{u} \). For each, the length came out to be \( \sqrt{17} \), evidencing the symmetry of the length when vectors are anti-parallel.

Magnitude of Cross Product

The magnitude of a cross product represents the area of the parallelogram formed by the original vectors. It is calculated as:\[\| \mathbf{u} \times \mathbf{v} \| = \| \mathbf{u} \| \cdot \| \mathbf{v} \| \cdot \sin(\theta)\]Here, \( \theta \) is the angle between the two vectors. For the given vectors, since they are not explicitly parallel or perpendicular, calculating the magnitude by this formula gives the same outcome as the determinant-based approach used in your example. This is vital for understanding the geometry related to the vector positions and the space they occupy.

Anti-parallel Vectors

Anti-parallel vectors have the same magnitude but are in opposite directions. In terms of cross products, if \( \mathbf{u} \times \mathbf{v} \) results in a vector, then \( \mathbf{v} \times \mathbf{u} \) will be its negative. This was demonstrated in the exercise as \( \mathbf{u} \times \mathbf{v} = 2\mathbf{i} + 3\mathbf{j} - 2\mathbf{k} \) and \( \mathbf{v} \times \mathbf{u}= -2\mathbf{i} - 3\mathbf{j} + 2\mathbf{k} \). This underscores an important principle: cross products are not commutative, meaning the order in which you multiply vectors matters, resulting in opposite, or anti-parallel, vectors.