By Fatskills Exam Guides Team — the exam nerds behind 28,500+ quizzes and 2.1M practice questions across 500+ global exams.
Mastering vectors unlocks 10-15 marks in IIT JEE (Main + Advanced)—enough to push you into the top 1%. Whether it’s finding angles between planes, proving collinearity, or solving 3D geometry problems, vectors are the secret weapon that separates AIR 1 from the rest.
Formula: a · b = |a||b|cosθ = a₁b₁ + a₂b₂ + a₃b₃ - a, b = vectors - θ = angle between them - MEMORISE THIS (used in projections, angles, work done)
Formula: a × b = |a||b|sinθ n̂ = (a₂b₃ – a₃b₂)i + (a₃b₁ – a₁b₃)j + (a₁b₂ – a₂b₁)k - n̂ = unit vector perpendicular to both a and b - MEMORISE THIS (used in area, torque, perpendicular vectors)
Formula: [a b c] = a · (b × c) = |a₁ a₂ a₃| |b₁ b₂ b₃| |c₁ c₂ c₃| - MEMORISE THIS (used in coplanarity, volume)
Formula: (a × b) × c = (a · c)b – (b · c)a - MEMORISE THIS (used in vector identities)
Formula: cosθ = (a · b) / (|a||b|) - MEMORISE THIS (used in geometry problems)
Formula: Projection = (a · b) / |b| - MEMORISE THIS (used in distance problems)
Formula: Area = |a × b| - MEMORISE THIS (used in geometry)
Formula: [a b c] = 0 - MEMORISE THIS (used in collinearity, plane problems)
Problem: Given vectors a = 2i + 3j – k and b = i – j + 2k, find: 1. a · b 2. a × b 3. The angle between a and b.
Already given.
1. Dot Product (a · b) = (2)(1) + (3)(-1) + (-1)(2) = 2 – 3 – 2 = -3
2. Cross Product (a × b) = [(3)(2) – (-1)(-1)]i + [(-1)(1) – (2)(2)]j + [(2)(-1) – (3)(1)]k = (6 – 1)i + (-1 – 4)j + (-2 – 3)k = 5i – 5j – 5k
3. Angle Between a and b First, find magnitudes: |a| = √(2² + 3² + (-1)²) = √(4 + 9 + 1) = √14 |b| = √(1² + (-1)² + 2²) = √(1 + 1 + 4) = √6
Now, cosθ = (a · b) / (|a||b|) = -3 / (√14 × √6) = -3 / √84 θ = cos⁻¹(-3/√84)
What we did and why: - Used dot product for angle and projection. - Used cross product for perpendicular vector and area. - Verified calculations step-by-step to avoid sign errors.
Problem: Find the angle between a = i + 2j – k and b = 2i – j + k.
Solution: 1. a · b = (1)(2) + (2)(-1) + (-1)(1) = 2 – 2 – 1 = -1 2. |a| = √(1 + 4 + 1) = √6 3. |b| = √(4 + 1 + 1) = √6 4. cosθ = -1 / (√6 × √6) = -1/6 5. θ = cos⁻¹(-1/6)
What we did and why: - Used dot product formula to find angle. - Calculated magnitudes separately to avoid mistakes.
Problem: Find the area of the parallelogram formed by a = 3i – j + 2k and b = i + 2j – k.
Solution: 1. a × b = [( -1)(-1) – (2)(2)]i + [(2)(1) – (3)(-1)]j + [(3)(2) – (-1)(1)]k = (1 – 4)i + (2 + 3)j + (6 + 1)k = -3i + 5j + 7k 2. Area = |a × b| = √((-3)² + 5² + 7²) = √(9 + 25 + 49) = √83
What we did and why: - Used cross product to find area. - Took magnitude of the resulting vector.
Problem: If a = i + 2j + 3k, b = 2i – j + k, c = 3i + j + 2k, check if they are coplanar.
Solution: 1. Scalar Triple Product = [a b c] = a · (b × c) 2. First, find b × c: = [( -1)(2) – (1)(1)]i + [(1)(3) – (2)(2)]j + [(2)(1) – (-1)(3)]k = (-2 – 1)i + (3 – 4)j + (2 + 3)k = -3i – j + 5k 3. Now, a · (b × c) = (1)(-3) + (2)(-1) + (3)(5) = -3 – 2 + 15 = 10 4. Since [a b c] ≠ 0, the vectors are not coplanar.
What we did and why: - Used scalar triple product to check coplanarity. - If [a b c] = 0, vectors are coplanar.
"Listen up—this is your 60-second vector survival guide for IIT JEE.
Memorise the formulas, practice 3 problems tonight, and you’ll own this topic tomorrow. Go crush it!
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