Math Problem Statement

Let u = (2,2,0) and v = (2,2,2√2). Compute |u × v|. Then sketch u, v, and u × v.

Solution

To solve the given problem, we need to compute the magnitude of the cross product ∣u×v∣|\mathbf{u} \times \mathbf{v}|, and identify the correct graph that represents vectors u\mathbf{u}, v\mathbf{v}, and u×v\mathbf{u} \times \mathbf{v}.

Given:

  • u=(2,2,0)\mathbf{u} = (2, 2, 0)
  • v=(2,2,22)\mathbf{v} = (2, 2, 2\sqrt{2})

Step 1: Compute the Cross Product u×v\mathbf{u} \times \mathbf{v}

The cross product of two vectors u=(u1,u2,u3)\mathbf{u} = (u_1, u_2, u_3) and v=(v1,v2,v3)\mathbf{v} = (v_1, v_2, v_3) is given by:

\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix}$$ Substitute the values of $$\mathbf{u}$$ and $$\mathbf{v}$$: $$\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 2 & 0 \\ 2 & 2 & 2\sqrt{2} \end{vmatrix}$$ Compute the determinant: $$\mathbf{u} \times \mathbf{v} = \mathbf{i} \left(2 \cdot 2\sqrt{2} - 0 \cdot 2\right) - \mathbf{j} \left(2 \cdot 2\sqrt{2} - 0 \cdot 2\right) + \mathbf{k} \left(2 \cdot 2 - 2 \cdot 2\right)$$ Simplify the expression: $$\mathbf{u} \times \mathbf{v} = \mathbf{i} \cdot 4\sqrt{2} - \mathbf{j} \cdot 4\sqrt{2} + \mathbf{k} \cdot 0$$ $$\mathbf{u} \times \mathbf{v} = (4\sqrt{2}, -4\sqrt{2}, 0)$$ ### Step 2: Compute the Magnitude of the Cross Product The magnitude of the cross product $$|\mathbf{u} \times \mathbf{v}|$$ is calculated as: $$|\mathbf{u} \times \mathbf{v}| = \sqrt{(4\sqrt{2})^2 + (-4\sqrt{2})^2 + 0^2}$$ Simplify further: $$|\mathbf{u} \times \mathbf{v}| = \sqrt{(16 \cdot 2) + (16 \cdot 2)}$$ $$|\mathbf{u} \times \mathbf{v}| = \sqrt{32 + 32}$$ $$|\mathbf{u} \times \mathbf{v}| = \sqrt{64} = 8$$ ### Step 3: Determine the Correct Graph Now that we know $$\mathbf{u} \times \mathbf{v} = (4\sqrt{2}, -4\sqrt{2}, 0)$$, we can see that the cross product lies in the xy-plane (since the z-component is 0). The correct graph should show $$\mathbf{u} \times \mathbf{v}$$ in the xy-plane. From the given choices: - Graph **A** shows $$\mathbf{u} \times \mathbf{v}$$ in the xy-plane. - Graph **B** and **C** do not show the correct orientation of $$\mathbf{u} \times \mathbf{v}$$. Thus, the correct graph is **A**. ### Conclusion: - $$|\mathbf{u} \times \mathbf{v}| = 8$$ - The correct graph is **A**. Would you like more details or have any questions? #### Further Related Questions: 1. What is the geometric meaning of the cross product of two vectors? 2. How do you determine the direction of the cross product using the right-hand rule? 3. Can the magnitude of a cross product ever be negative? Why or why not? 4. What are some real-world applications of the cross product in physics? 5. How does the cross product change if vectors $$\mathbf{u}$$ and $$\mathbf{v}$$ are parallel? #### Tip: When visualizing vector cross products, always use the right-hand rule to determine the direction of the resulting vector!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Vectors
Cross Product
Magnitude of Cross Product

Formulas

u × v = |i j k| |u1 u2 u3| |v1 v2 v3|
|u × v| = √((u × v)_x^2 + (u × v)_y^2 + (u × v)_z^2)

Theorems

Cross Product Theorem

Suitable Grade Level

Grades 10-12 or introductory college-level physics or vector calculus

Related Recommendation

Cross Product of Vectors u = (-1, 3, 2) and v = (1, 7, 0)

Cross Product of Vectors (1,2,0) and (1,4,2) - Step-by-Step Solution

Cross Product Calculation for Vectors u and v

Calculate Cross Product of Vectors u = [3, 0, -1] and v = [4, 0, -2]

Calculate the Cross Product and Direction of u x v for Given Vectors