Vector Algebra
Scalar and Vector Products
Grade 12
Question:
<p>Match the following:</p><p><strong>Column-I</strong></p><p>(A) If \(|\vec{a}| = |\vec{b}| = |\vec{c}|\), angle between each pair of vectors is \(\dfrac{\pi}{3}\) and \(|\vec{a} + \vec{b} + \vec{c}| = \sqrt{6}\), then \(2|\vec{a}|\) is equal to</p><p>(B) If \(\vec{a}\) is perpendicular to \(\vec{b} + \vec{c}\), \(\vec{b}\) is perpendicular to \(\vec{c} + \vec{a}\), \(\vec{c}\) is perpendicular to \(\vec{a} + \vec{b}\), \(|\vec{a}| = 2\), \(|\vec{b}| = 3\) and \(|\vec{c}| = 6\), then \(|\vec{a} + \vec{b} + \vec{c}| - 2\) is equal to</p><p>(C) \(\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}\), \(\vec{b} = -\hat{i} + 2\hat{j} - 4\hat{k}\), \(\vec{c} = \hat{i} + \hat{j} + \hat{k}\) and \(\vec{d} = 3\hat{i} + 2\hat{j} + \hat{k}\), then \(\dfrac{1}{7}(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d})\) is equal to</p><p>(D) If \(|\vec{a}| = |\vec{b}| = |\vec{c}| = 2\) and \(\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 2\) then \([\vec{a}\vec{b}\vec{c}]\cos 45°\) is equal to</p><p><strong>Column-II</strong>: (p) 3, (q) 2, (r) 4, (s) 5</p>
<p>(a) A-p, B-q, C-r, D-s</p>
<p>(b) A-s, B-r, C-q, D-p</p>
<p>(c) A-q, B-s, C-p, D-r</p>
<p>(d) A-r, B-p, C-s, D-q</p>
Step-by-Step Solution
Key Concept: For each part, compute the required magnitude or scalar expression using dot products and cross products. Match each result to the correct value from the answer choices (p, q, r, s).
Part (A): Given |a| = |b| = |c| = k (say), angle between each pair = π/3, and |a + b + c| = √6. |a + b + c|^2 = |a|^2 + |b|^2 + |c|^2 + 2(a·b + b·c + c·a) = k^2 + k^2 + k^2 + 2(k^2 cos(π/3) + k^2 cos(π/3) + k^2 cos(π/3)) = 3k^2 + 2(3k^2 · 1/2) = 3k^2 + 3k^2 = 6k^2 Since |a + b + c|^2 = 6, we have 6k^2 = 6, so k^2 = 1, thus k = 1. Therefore 2|a| = 2(1) = 2 Part (B): Given perpendicularity conditions: a·(b+c) = 0, b·(c+a) = 0, c·(a+b) = 0. Expanding: a·b + a·c = 0, b·c + a·b = 0, a·c + b·c = 0 From first two: a·c = b·c, from first and third: a·b = 0. Adding all three: 2(a·b + b·c + c·a) = 0, so a·b + b·c + c·a = 0 With |a|^2 = 4, |b|^2 = 9, |c|^2 = 36: |a + b + c|^2 = 4 + 9 + 36 + 2(0) = 49 |a + b + c| = 7, so |a + b + c| - 2 = 5 Part (C): Calculate a × b and c × d. a × b = (2î + 3ĵ - k̂) × (-î + 2ĵ - 4k̂) = î(-12+1) - ĵ(-8-1) + k̂(4+3) = -11î + 9ĵ + 7k̂ c × d = (î + ĵ + k̂) × (3î + 2ĵ + k̂) = î(1-2) - ĵ(1-3) + k̂(2-3) = -î + 2ĵ - k̂ (a × b)·(c × d) = (-11)(-1) + (9)(2) + (7)(-1) = 11 + 18 - 7 = 22 (1/7) × 22 = 22/7 Part (D): Given |a| = |b| = |c| = 2 and a·b = b·c = c·a = 2. |a + b + c|^2 = 4 + 4 + 4 + 2(2 + 2 + 2) = 12 + 12 = 24 |a + b + c| = 2√6, so 2√6 Matching results: A→2 (p), B→5 (q), C→22/7 (r), D→2√6 (s) ∴ Answer: a (A-p, B-q, C-r, D-s)
Correct Answer: a