Vector Algebra
Coplanarity of Vectors
Grade 12
Question:
<p>If <span>\(\vec{a}, \vec{b}, \vec{c}\)</span> are non-coplanar vectors and <span>\(\lambda\)</span> is a real number, then the vectors <span>\(\vec{a} + 2\vec{b} + 3\vec{c}\)</span>, <span>\(\lambda\vec{b} + 4\vec{c}\)</span> and <span>\((2\lambda - 1)\vec{c}\)</span> are non-coplanar for</p>
<p>all values of <span>\(\lambda\)</span>.</p>
<p>all except one value of <span>\(\lambda\)</span>.</p>
<p>all except two values of <span>\(\lambda\)</span>.</p>
<p>no value of <span>\(\lambda\)</span>.</p>
Step-by-Step Solution
Key Concept: Three vectors are coplanar if and only if their scalar triple product equals zero. For the given vectors to be non-coplanar, we need [v₁ v₂ v₃] ≠ 0, which requires the coefficient of the scalar triple product [a b c] to be non-zero.
Step 1: Three vectors are coplanar iff their scalar triple product is zero. We need to find when [v_1 v_2 v_3] = 0 for the given vectors to become coplanar. Step 2: Let v_1 = a + 2b + 3c, v_2 = λb + 4c, v_3 = (2λ - 1)c Compute [v_1 v_2 v_3] = (a + 2b + 3c)·[(λb + 4c) × (2λ - 1)c] Step 3: Since (2λ - 1)c is parallel to c, we have (λb + 4c) × (2λ - 1)c = λb × (2λ - 1)c = λ(2λ - 1)(b × c) Step 4: Therefore [v_1 v_2 v_3] = (a + 2b + 3c)·[λ(2λ - 1)(b × c)] = λ(2λ - 1)[a·(b × c)] = λ(2λ - 1)[a b c] Step 5: Since a, b, c are non-coplanar, [a b c] ≠ 0. For non-coplanarity of v_1, v_2, v_3, we need λ(2λ - 1) ≠ 0 Step 6: This gives λ ≠ 0 and λ ≠ 1/2 ∴ The vectors are non-coplanar for all λ except λ = 0 and λ = 1/2
Correct Answer: B