Vector Algebra
Cross Product and Triple Cross Product
Grade 12
Question:
<p>Let \(\vec{a} = 2\hat{i}+\hat{j}-2\hat{k}\) and \(\vec{b} = \hat{i}+\hat{j}\). If \(\vec{c}\) is a vector such that \(\vec{a}\cdot\vec{c} = |\vec{c}|\), \(|\vec{c}-\vec{a}| = 2\sqrt{2}\) and the angle between \((\vec{a}\times\vec{b})\) and \(\vec{c}\) is 30°, then \(|(\vec{a}\times\vec{b})\times\vec{c}|\) is equal to ________.</p>
Step-by-Step Solution
Key Concept: Use the dot product constraint to express |c| in terms of projection onto a, then apply the distance constraint to find |c|, and finally use the angle condition with the triple cross product magnitude formula |u × v| = |u||v|sin(θ).
Step 1: Find |a| and a × b | a | = √(4+1+4) = 3 a × b = | i j k ; 2 1 -2; 1 1 0| = i (0+2) - j (0+2) + k (2-1) = 2 i - 2 j + k | a × b | = √(4+4+1) = 3 Step 2: Use constraint a·c = |c| This means: a · c = | a || c |cos(α), where α is angle between a and c So: | a || c |cos(α) = | c | ⟹ cos(α) = 1/3 Step 3: Use distance constraint |c - a| = 2√2 | c - a |^2 = 8 | c |^2 - 2 a · c + | a |^2 = 8 | c |^2 - 2| c | + 9 = 8 | c |^2 - 2| c + 1 = 0 (| c | - 1)^2 = 0 ⟹ | c | = 1 Step 4: Apply triple cross product formula |( a × b ) × c | = | a × b | · | c | · sin(30°) = 3 × 1 × (1/2) = 3/2 ∴ Answer: 3/2
Correct Answer: 3