Vector Algebra
Cross Product and Scalar Triple Product
Grade None
Question:
<p>Given vectors <strong>p</strong>, <strong>q</strong>, and <strong>r</strong> such that <strong>p</strong> = <math>\frac{2}{3}\mathbf{i} - \frac{1}{3}\mathbf{j} - \frac{1}{3}\mathbf{k}</math>, <strong>q</strong> = <math>-\frac{1}{3}\mathbf{i} + \frac{2}{3}\mathbf{j} - \frac{1}{3}\mathbf{k}</math>, and <strong>r</strong> = <math>-\frac{1}{3}\mathbf{i} - \frac{1}{3}\mathbf{j} + \frac{2}{3}\mathbf{k}</math>. If <math>3(\mathbf{p} \times \mathbf{q})^2 - \lambda|\mathbf{r} \times \mathbf{q}| = 0</math>, find the value of <math>\lambda</math>.</p>
Step-by-Step Solution
Key Concept: Compute the cross product of vectors and use the given algebraic condition to solve for the unknown scalar λ.
Step 1: Calculate <math>(\mathbf{p} \times \mathbf{q})^2</math>. From the cross product computation: <math>(\mathbf{p} \times \mathbf{q})^2 = \frac{1}{9}</math> Step 2: Calculate <math>|\mathbf{r} \times \mathbf{q}|</math>. <math>\mathbf{r} \times \mathbf{q} = -\frac{1}{3}\mathbf{i} - \frac{1}{3}\mathbf{j} - \frac{1}{3}\mathbf{k}</math> <math>|\mathbf{r} \times \mathbf{q}| = \sqrt{\frac{1}{9} + \frac{1}{9} + \frac{1}{9}} = \frac{\sqrt{3}}{3}</math> Step 3: Substitute into the given equation. <math>3 \cdot \frac{1}{9} - \lambda \cdot \frac{\sqrt{3}}{3} = 0</math> <math>\frac{1}{3} = \lambda \cdot \frac{\sqrt{3}}{3}</math> <math>\lambda = 1</math> ∴ Answer is 1 .
Correct Answer: 1