Vector Algebra
Magnitude of Vector
Grade 12
Question:
<p>The vector <span>\(\hat{i} + x\hat{j} + 3\hat{k}\)</span> is rotated through an angle <span>\(\theta\)</span> and doubled in magnitude, then it becomes <span>\(4\hat{i} + (4x-2)\hat{j} + 2\hat{k}\)</span>. The values of <span>\(x\)</span> are</p>
<p>\(\left\{-\dfrac{2}{3}, 2\right\}\)</p>
<p>\(\left\{\dfrac{1}{3}, 2\right\}\)</p>
<p>\(\left\{\dfrac{2}{3}, 0\right\}\)</p>
<p>\(\{2, 7\}\)</p>
Step-by-Step Solution
Key Concept: Use the constraint that rotating a vector preserves its magnitude relationship: if vector **a** is rotated and scaled by factor k, then |**a'**| = k|**a**|. Also, the dot product relationship |**a**|·|**a'**|·cos(θ) = **a**·**a'** provides the rotation constraint.
Step 1: Write the original vector a = î + x ĵ + 3 k̂ and rotated/scaled vector a' = 4 î + (4x-2) ĵ + 2 k̂ Step 2: Calculate | a |^2 = 1 + x^2 + 9 = 10 + x^2 Step 3: Calculate | a' |^2 = 16 + (4x-2)^2 + 4 = 16 + 16x^2 - 16x + 4 + 4 = 16x^2 - 16x + 24 Step 4: Apply the magnitude doubling condition: | a' |^2 = 4| a |^2 16x^2 - 16x + 24 = 4(10 + x^2) 16x^2 - 16x + 24 = 40 + 4x^2 12x^2 - 16x - 16 = 0 3x^2 - 4x - 4 = 0 Step 5: Factor or use quadratic formula: (3x + 2)(x - 2) = 0 x = 2 or x = -2/3 Step 6: Verify both values satisfy the rotation constraint using **a**· a' = | a || a' |cos(θ) where 0 ≤ θ ≤ π ∴ Answer: A
Correct Answer: A