Vector Algebra
Cross Product and Dot Product
Grade 12
Question:
<p>The vectors \(\vec{a}\) and \(\vec{b}\) are not perpendicular and \(\vec{c}\) and \(\vec{d}\) are two vectors satisfying: \(\vec{b} \times \vec{c} = \vec{b} \times \vec{d}\) and \(\vec{a} \cdot \vec{d} = 0\). Then the vector \(\vec{d}\) is equal to</p>
<p>\(\vec{c} + \left(\dfrac{\vec{a} \cdot \vec{c}}{\vec{a} \cdot \vec{b}}\right)\vec{b}\)</p>
<p>\(\vec{b} + \left(\dfrac{\vec{b} \cdot \vec{c}}{\vec{a} \cdot \vec{b}}\right)\vec{c}\)</p>
<p>\(\vec{c} - \left(\dfrac{\vec{a} \cdot \vec{c}}{\vec{a} \cdot \vec{b}}\right)\vec{b}\)</p>
<p>\(\vec{b} - \left(\dfrac{\vec{b} \cdot \vec{c}}{\vec{a} \cdot \vec{b}}\right)\vec{c}\)</p>
Step-by-Step Solution
Key Concept: Use the property that ⃗b × ⃗c = ⃗b × ⃗d implies ⃗b × (⃗c - ⃗d) = 0, so (⃗c - ⃗d) is parallel to ⃗b. Combined with ⃗a · ⃗d = 0, express ⃗d in terms of ⃗a and ⃗b using scalar parameters.
Step 1: From ⃗b × ⃗c = ⃗b × ⃗d, we get: ⃗b × (⃗c - ⃗d) = 0 This means (⃗c - ⃗d) is parallel to ⃗b, so: ⃗c - ⃗d = λ⃗b for some scalar λ Therefore: ⃗d = ⃗c - λ⃗b Step 2: Apply the condition ⃗a · ⃗d = 0: ⃗a · (⃗c - λ⃗b) = 0 ⃗a · ⃗c - λ(⃗a · ⃗b) = 0 Step 3: Solve for λ (since ⃗a and ⃗b are not perpendicular, ⃗a · ⃗b ≠ 0): λ = (⃗a · ⃗c)/(⃗a · ⃗b) Step 4: Substitute back: ⃗d = ⃗c - [(⃗a · ⃗c)/(⃗a · ⃗b)]⃗b ∴ Answer: ⃗d = ⃗c - [(⃗a · ⃗c)/(⃗a · ⃗b)]⃗b (or equivalent form depending on option C)
Correct Answer: C