Vector Algebra
Cross product and scalar product
Grade 12
Question:
<p>Given <strong>59.</strong>
\(\vec{a} = \hat{i} + \hat{j} + \hat{k},\ \vec{c} = \hat{j} - \hat{k},\ \vec{a} \cdot \vec{b} = 3\) and \(\vec{a} \times \vec{b} = \vec{c}\), find \(|\vec{b}|\).</p>
<p>\(\sqrt{\dfrac{11}{3}}\)</p>
<p>\(\sqrt{\dfrac{10}{3}}\)</p>
<p>\(\sqrt{\dfrac{11}{3}}\)</p>
<p>\(\sqrt{\dfrac{13}{3}}\)</p>
Step-by-Step Solution
Key Concept: Use the identity |a⃗ × b⃗|² = |a⃗|²|b⃗|² - (a⃗ · b⃗)² to find |b⃗|, combined with the given dot and cross product constraints.
Step 1: Calculate |a⃗| and |c⃗|. |a⃗| = √(1^2 + 1^2 + 1^2) = √3 |c⃗| = √(0^2 + 1^2 + (-1)^2) = √2 Step 2: Use the vector identity: |a⃗ × b⃗|^2 = |a⃗|^2|b⃗|^2 - (a⃗ · b⃗)^2 Since a⃗ × b⃗ = c⃗, we have |a⃗ × b⃗| = |c⃗| = √2 Therefore: (√2)^2 = (√3)^2|b⃗|^2 - (3)^2 Step 3: Solve for |b⃗|. 2 = 3|b⃗|^2 - 9 3|b⃗|^2 = 11 |b⃗|^2 = 11/3 |b⃗| = √(11/3) = √33/3 ∴ Answer: A
Correct Answer: A