Vector Algebra
Scalar Triple Product
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 <span>\([\lambda(\vec{a}+\vec{b})\ \lambda^2\vec{b}\ \lambda\vec{c}] = [\vec{a}\ \vec{b}+\vec{c}\ \vec{b}]\)</span> for</p>
<p>exactly one value of <span>\(\lambda\)</span>.</p>
<p>no value of <span>\(\lambda\)</span>.</p>
<p>exactly three values of <span>\(\lambda\)</span>.</p>
<p>exactly two values of <span>\(\lambda\)</span>.</p>
Step-by-Step Solution
Key Concept: The scalar triple product is linear in each vector position and a scalar factor multiplying any vector can be extracted as a power outside the bracket. Use the distributive property of scalar triple product and factor out scalar multiples systematically.
Step 1: Expand the left side using linearity of scalar triple product in first position: [λ(a+b) λ^2b λc] = λ[((a+b) λ^2b λc] = λ·λ^2[(a+b) b c] = λ^3[(a+b) b c] Step 2: Expand [(a+b) b c] using linearity in first position: [(a+b) b c] = [a b c] + [b b c] = [a b c] (since [b b c] = 0) Step 3: So LHS = λ^3[a b c] Step 4: Expand the right side [a b+c b] using linearity in second position: [a b+c b] = [a b b] + [a c b] = 0 + [a c b] Step 5: Note that [a c b] = -[a b c] (swapping two vectors changes sign) Step 6: So RHS = -[a b c] Step 7: Set LHS = RHS: λ^3[a b c] = -[a b c] Step 8: Since a, b, c are non-coplanar, [a b c] ≠ 0, so: λ^3 = -1, which gives λ = -1 ∴ Answer: D
Correct Answer: D