Vector Algebra
Cross Product and Dot Product
Grade 12
Question:
<p>Let \(\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k}\) and \(\vec{b} = \hat{i} + \hat{j}\). Let \(\vec{c}\) be a vector such that \(|\vec{c} - \vec{a}| = 3\), \(|(\vec{a} \times \vec{b}) \times \vec{c}| = 3\) and the angle between \(\vec{c}\) and \(\vec{a} \times \vec{b}\) be \(30°\). Then \(\vec{a} \cdot \vec{c}\) is equal to</p>
<p>2</p>
<p>5</p>
<p>\(\dfrac{1}{8}\)</p>
<p>\(\dfrac{25}{8}\)</p>
Step-by-Step Solution
Key Concept: Use the scalar triple product identity |(**a** × **b**) × **c**| = |**a** × **b**| · |**c**| · sin(θ) to find |**c**|, then apply the distance constraint |**c** - **a**| = 3 with the dot product formula to isolate **a** · **c**.
Step 1: Calculate a × b a × b = (2 i + j - 2 k ) × ( i + j ) = i (1·0 - (-2)·1) - j (2·0 - (-2)·1) + k (2·1 - 1·1) = 2 i - 2 j + k | a × b | = √(4 + 4 + 1) = 3 Step 2: Use the vector triple product magnitude formula |( a × b ) × c | = | a × b | · | c | · sin(30°) 3 = 3 · | c | · (1/2) | c | = 2 Step 3: Apply the distance constraint | c - a |^2 = 9 | c |^2 + | a |^2 - 2( a · c ) = 9 | a |^2 = 4 + 1 + 4 = 9 4 + 9 - 2( a · c ) = 9 2( a · c ) = 4 a · c = 2 ∴ Answer: D
Correct Answer: D