Vector Algebra
Scalar Triple Product
Grade 12
Question:
<p>Let \(\vec{a}, \vec{b}, \vec{c}\) be three vectors of magnitude 2, 3, 5 respectively, satisfying \(|[\vec{a},\, \vec{b},\, \vec{c}]| = 30\). If \((2\vec{a} + \vec{b} + \vec{c}) \cdot ((\vec{a} \times \vec{c}) \times (\vec{a} - \vec{c}) + \vec{b}) = k\), then the value of \(\left\lfloor \dfrac{k}{103} \right\rfloor\) is:</p>
<p>(a) 1</p>
<p>(b) 2</p>
<p>(c) 3</p>
<p>(d) 4</p>
Step-by-Step Solution
Key Concept: The scalar triple product |[a,b,c]| = 30 with magnitudes 2,3,5 means the vectors form a right-handed orthogonal system (since 2×3×5=30). Use this orthogonality to simplify the complex vector expression by setting up coordinates.
Step 1: Interpret the scalar triple product condition. Given |[a,b,c]| = |a·(b×c)| = 30 with |a|=2, |b|=3, |c|=5. Since 2×3×5 = 30, we have sin(θ) = 1 for the angle between vectors, meaning a, b, c are mutually orthogonal . Step 2: Set up orthogonal coordinate system. Let a = 2î, b = 3ĵ, c = 5k̂ Step 3: Compute a×c. a×c = 2î×5k̂ = -10ĵ Step 4: Compute a-c. a-c = 2î - 5k̂ Step 5: Compute (a×c)×(a-c). (-10ĵ)×(2î-5k̂) = -10ĵ×2î - (-10ĵ)×5k̂ = -20k̂ - 50î Step 6: Form (a×c)×(a-c) + b. -50î - 20k̂ + 3ĵ Step 7: Compute 2a+b+c. 4î + 3ĵ + 5k̂ Step 8: Calculate the dot product. (4î + 3ĵ + 5k̂)·(-50î + 3ĵ - 20k̂) = (4)(-50) + (3)(3) + (5)(-20) = -200 + 9 - 100 = -291 So k = -291 Step 9: Find ⌊k/103⌋. ⌊-291/103⌋ = ⌊-2.825...⌋ = -3 ∴ Answer: A
Correct Answer: A