Vector Algebra
Cross Product and Magnitude
Grade 12
Question:
<p>Given <strong>a</strong> = 3î + 2ĵ + x<strong>k̂</strong> and <strong>b</strong> = î − ĵ + <strong>k̂</strong>. Then the minimum value of |<strong>a</strong> × <strong>b</strong>| is</p>
<p>\(\dfrac{5\sqrt{3}}{\sqrt{2}}\)</p>
<p>\(5\sqrt{\dfrac{3}{2}}\)</p>
<p>\(\sqrt{\dfrac{3}{2}}\)</p>
<p>\(5\sqrt{3}\)</p>
Step-by-Step Solution
Key Concept: The magnitude of cross product |a × b| = |a||b|sin(θ) is minimized when the vectors are parallel (θ = 0), making |a × b| = 0 only if they're collinear. Otherwise, minimize by finding the value of x that makes them as aligned as possible, then compute the resulting cross product magnitude.
Step 1: Compute the cross product a × b. a × b = |î ĵ k̂|
|3 2 x|
|1 -1 1| = î(2·1 - x·(-1)) - ĵ(3·1 - x·1) + k̂(3·(-1) - 2·1)
= î(2 + x) - ĵ(3 - x) + k̂(-5)
= (2+x)î + (x-3)ĵ - 5k̂ Step 2: Find |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: Minimize by taking derivative with respect to x. d(|a × b|^2)/dx = 4x - 2 = 0
⟹ x = 1/2 Step 4: Substitute x = 1/2. |a × b|^2 = 2(1/4) - 2(1/2) + 38
= 1/2 - 1 + 38 = 37.5 = 75/2 |a × b| = √(75/2) = 5√(3/2) = (5√6)/2 ∴ Answer: B
Correct Answer: B