Vector Algebra
Orthogonal Vectors
Grade 12
Question:
<p>If the vectors <strong>a</strong> = <strong>i</strong> − <strong>j</strong> + 2<strong>k</strong>, <strong>b</strong> = 2<strong>i</strong> + 4<strong>j</strong> + <strong>k</strong> and <strong>c</strong> = λ<strong>i</strong> + <strong>j</strong> + μ<strong>k</strong> are mutually orthogonal, then (λ, μ) is equal to</p>
<p>(a) (−3, 2)</p>
<p>(b) (2, −3)</p>
<p>(c) (−2, 3)</p>
<p>(d) (3, −2)</p>
Step-by-Step Solution
Key Concept: Two vectors are orthogonal if and only if their dot product equals zero. Since a, b, and c are mutually orthogonal, we need a·b = 0, a·c = 0, and b·c = 0 to find λ and μ.
Step 1: Check orthogonality of a and b (should equal 0) a·b = (i − j + 2k)·(2i + 4j + k) = (1)(2) + (−1)(4) + (2)(1) = 2 − 4 + 2 = 0 ✓ This confirms a ⊥ b, so we need to find λ and μ such that a ⊥ c and b ⊥ c. Step 2: Use condition a·c = 0 a·c = (i − j + 2k)·(λi + j + μk) = (1)(λ) + (−1)(1) + (2)(μ) = λ − 1 + 2μ = 0 Therefore: λ + 2μ = 1 ... (Equation 1) Step 3: Use condition b·c = 0 b·c = (2i + 4j + k)·(λi + j + μk) = (2)(λ) + (4)(1) + (1)(μ) = 2λ + 4 + μ = 0 Therefore: 2λ + μ = −4 ... (Equation 2) Step 4: Solve the system of equations From Equation 1: λ + 2μ = 1 From Equation 2: 2λ + μ = −4 Multiply Equation 1 by 2: 2λ + 4μ = 2 Subtract Equation 2: (2λ + 4μ) − (2λ + μ) = 2 − (−4) 3μ = 6 ⟹ μ = 2 Step 5: Find λ Substitute μ = 2 into Equation 1: λ + 2(2) = 1 λ + 4 = 1 ⟹ λ = −3 Step 6: Verify the solution a·c = (−3) − 1 + 2(2) = −3 − 1 + 4 = 0 ✓ b·c = 2(−3) + 4 + 2 = −6 + 4 + 2 = 0 ✓ ∴ Answer: A
Correct Answer: A