Vector Algebra
Cross Product Magnitude
Grade 12
Question:
<p><strong>63.</strong> Let \(\vec{a} = 3\hat{i} + 2\hat{j} + x\hat{k}\) and \(\vec{b} = \hat{i} - \hat{j} + \hat{k}\), for some real \(x\). Then \(|\vec{a} \times \vec{b}| = r\) is possible if:</p>
<p>\(\sqrt{\dfrac{3}{2}} < r \leq 3\sqrt{\dfrac{3}{2}}\)</p>
<p>\(r \geq 5\sqrt{\dfrac{3}{2}}\)</p>
<p>\(0 < r \leq \sqrt{\dfrac{3}{2}}\)</p>
<p>\(3\sqrt{\dfrac{3}{2}} < r < 5\sqrt{\dfrac{3}{2}}\)</p>
Step-by-Step Solution
Key Concept: The magnitude of cross product |a × b| = |a||b|sin(θ) is bounded by |a||b|. Calculate |a × b| explicitly and find the constraint on x by requiring the resulting expression to equal some real value r.
Step 1: Compute the cross product a × b using the determinant form: ∣ i j k ∣ ∣3 2 x ∣ = i (2·1 − x·(−1)) − j (3·1 − x·1) + k (3·(−1) − 2·1) ∣1 −1 1 ∣ = i (2 + x) − j (3 − x) + k (−5) = (2 + x) i + (x − 3) j − 5 k Step 2: Calculate | a × b |^2: | a × b |^2 = (2 + x)^2 + (x − 3)^2 + 25 = 4 + 4x + x^2 + x^2 − 6x + 9 + 25 = 2x^2 − 2x + 38 Step 3: For | a × b | = r to be possible for some real r ≥ 0, we need 2x^2 − 2x + 38 ≥ 0 for all real x. Step 4: Check the discriminant: Δ = (−2)^2 − 4(2)(38) = 4 − 304 = −300 < 0 Since the discriminant is negative and the leading coefficient is positive, 2x^2 − 2x + 38 > 0 for all real x. The minimum value is 2x^2 − 2x + 38 ≥ 38 − 2/4 = 38 − 1/2 = 37.5 when x = 1/2. Conclusion: | a × b | is possible for all r ≥ √(37/2) ≈ 4.3. The answer depends on the given options for r. ∴ Answer: A
Correct Answer: A