Vector Algebraसदिश बीजगणितVector Algebra

• a·b = |a||b| cos θ (θ = angle between them, 0 ≤ θ ≤ π). The result is a scalar.
• In components: a·b = a₁b₁ + a₂b₂ + a₃b₃. So cos θ = (a₁b₁ + a₂b₂ + a₃b₃)/(|a||b|).
• î·î = ĵ·ĵ = k̂·k̂ = 1 and î·ĵ = ĵ·k̂ = k̂·î = 0. Also a·a = |a|².
• If a + b + c = 0: square it (dot with itself): |a|² + |b|² + |c|² + 2(a·b + b·c + c·a) = 0, so a·b + b·c + c·a = −(|a|² + |b|² + |c|²)/2.
• Right-angled triangle check: find the three side vectors; if the dot product of any two is 0, there is a right angle between them.
• MCQ: î·î − ĵ·ĵ + k̂×k̂: here 1 − 1 = 0 and k̂×k̂ = 0 (zero vector), so the value is 0.

• a·b = |a||b| cos θ (θ = उनके बीच का कोण, 0 ≤ θ ≤ π)। परिणाम एक अदिश है।
• घटकों में: a·b = a₁b₁ + a₂b₂ + a₃b₃। अतः cos θ = (a₁b₁ + a₂b₂ + a₃b₃)/(|a||b|)।
• î·î = ĵ·ĵ = k̂·k̂ = 1 तथा î·ĵ = ĵ·k̂ = k̂·î = 0। साथ ही a·a = |a|²।
• यदि a + b + c = 0: इसका स्वयं से अदिश गुणन कीजिए: |a|² + |b|² + |c|² + 2(a·b + b·c + c·a) = 0, अतः a·b + b·c + c·a = −(|a|² + |b|² + |c|²)/2।
• समकोण त्रिभुज की जाँच: तीनों भुजा-सदिश ज्ञात कीजिए; यदि किन्हीं दो का अदिश गुणनफल 0 है, तो उनके बीच समकोण है।
• MCQ: î·î − ĵ·ĵ + k̂×k̂: यहाँ 1 − 1 = 0 तथा k̂×k̂ = 0 (शून्य सदिश), अतः मान 0 है।

• a·b = |a||b| cos θ (θ = unke beech ka angle, 0 ≤ θ ≤ π). Result ek scalar hai.
• Components mein: a·b = a₁b₁ + a₂b₂ + a₃b₃. Isliye cos θ = (a₁b₁ + a₂b₂ + a₃b₃)/(|a||b|).
• î·î = ĵ·ĵ = k̂·k̂ = 1 aur î·ĵ = ĵ·k̂ = k̂·î = 0. Saath hi a·a = |a|².
• Agar a + b + c = 0: iska khud se dot product kijiye: |a|² + |b|² + |c|² + 2(a·b + b·c + c·a) = 0, isliye a·b + b·c + c·a = −(|a|² + |b|² + |c|²)/2.
• Right-angled triangle check: teeno side vectors nikaliye; agar kinhi do ka dot product 0 hai, toh unke beech right angle hai.
• MCQ: î·î − ĵ·ĵ + k̂×k̂: yahan 1 − 1 = 0 aur k̂×k̂ = 0 (zero vector), isliye value 0 hai.

💡 Write the dot product term by term — (2)(1) + (−1)(1) + … — so the examiner sees each step; most errors are sign errors here.अदिश गुणनफल पद-दर-पद लिखिए — (2)(1) + (−1)(1) + … — ताकि परीक्षक हर चरण देखे; अधिकतर गलतियाँ यहीं चिह्न की होती हैं।Dot product term by term likhiye — (2)(1) + (−1)(1) + … — taaki examiner har step dekhe; zyadatar galtiyan yahin sign ki hoti hain.

• a × b = |a||b| sin θ n̂, where n̂ is the unit vector perpendicular to both (right-hand rule). The result is a vector.
• Determinant form: a × b = | î ĵ k̂ ; a₁ a₂ a₃ ; b₁ b₂ b₃ | = (a₂b₃ − a₃b₂)î − (a₁b₃ − a₃b₁)ĵ + (a₁b₂ − a₂b₁)k̂.
• î × î = ĵ × ĵ = k̂ × k̂ = 0; î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ; reversing the order changes the sign (b × a = −a × b).
• Unit vector perpendicular to a and b: ±(a × b)/|a × b|.
• a × b = 0 ⇔ a ∥ b. If |a·b| = |a × b|, then |cos θ| = sin θ, so θ = π/4 (or 3π/4).
• MCQ: î·(ĵ × k̂) + ĵ·(î × k̂) + k̂·(î × ĵ) = 1 + ĵ·(−ĵ) + 1 = 1 − 1 + 1 = 1.

• a × b = |a||b| sin θ n̂, जहाँ n̂ दोनों पर लंबवत मात्रक सदिश है (दाएँ हाथ का नियम)। परिणाम एक सदिश है।
• सारणिक रूप: a × b = | î ĵ k̂ ; a₁ a₂ a₃ ; b₁ b₂ b₃ | = (a₂b₃ − a₃b₂)î − (a₁b₃ − a₃b₁)ĵ + (a₁b₂ − a₂b₁)k̂।
• î × î = ĵ × ĵ = k̂ × k̂ = 0; î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ; क्रम बदलने पर चिह्न बदलता है (b × a = −a × b)।
• a और b पर लंबवत मात्रक सदिश: ±(a × b)/|a × b|।
• a × b = 0 ⇔ a ∥ b। यदि |a·b| = |a × b|, तो |cos θ| = sin θ, अतः θ = π/4 (या 3π/4)।
• MCQ: î·(ĵ × k̂) + ĵ·(î × k̂) + k̂·(î × ĵ) = 1 + ĵ·(−ĵ) + 1 = 1 − 1 + 1 = 1।

• a × b = |a||b| sin θ n̂, jahan n̂ dono par perpendicular unit vector hai (right-hand rule). Result ek vector hai.
• Determinant form: a × b = | î ĵ k̂ ; a₁ a₂ a₃ ; b₁ b₂ b₃ | = (a₂b₃ − a₃b₂)î − (a₁b₃ − a₃b₁)ĵ + (a₁b₂ − a₂b₁)k̂.
• î × î = ĵ × ĵ = k̂ × k̂ = 0; î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ; order badalne par sign badalta hai (b × a = −a × b).
• a aur b par perpendicular unit vector: ±(a × b)/|a × b|.
• a × b = 0 ⇔ a ∥ b. Agar |a·b| = |a × b|, toh |cos θ| = sin θ, isliye θ = π/4 (ya 3π/4).
• MCQ: î·(ĵ × k̂) + ĵ·(î × k̂) + k̂·(î × ĵ) = 1 + ĵ·(−ĵ) + 1 = 1 − 1 + 1 = 1.

💡 In the determinant, the middle (ĵ) term has a MINUS sign in front. Most wrong answers come from forgetting it.सारणिक में बीच वाले (ĵ) पद के आगे ऋण चिह्न होता है। अधिकतर गलत उत्तर इसी को भूलने से आते हैं।Determinant mein beech wale (ĵ) term ke aage MINUS sign hota hai. Zyadatar galat answers ise bhoolne se aate hain.

• Two non-zero vectors are perpendicular ⇔ a·b = 0.
• Find λ: write a·b = 0 and solve the linear equation in λ.
• (a + b) ⊥ (a − b) ⇔ |a| = |b|, because (a + b)·(a − b) = |a|² − |b|².
• |a + b|² = |a|² + |b|² ⇔ a ⊥ b, because |a + b|² = |a|² + |b|² + 2a·b.
• Parallel vectors: b = λa, i.e. a₁/b₁ = a₂/b₂ = a₃/b₃.

• दो अशून्य सदिश लंबवत हैं ⇔ a·b = 0।
• λ ज्ञात करना: a·b = 0 लिखकर λ का रैखिक समीकरण हल कीजिए।
• (a + b) ⊥ (a − b) ⇔ |a| = |b|, क्योंकि (a + b)·(a − b) = |a|² − |b|²।
• |a + b|² = |a|² + |b|² ⇔ a ⊥ b, क्योंकि |a + b|² = |a|² + |b|² + 2a·b।
• समांतर सदिश: b = λa, अर्थात a₁/b₁ = a₂/b₂ = a₃/b₃।

• Do non-zero vectors perpendicular hain ⇔ a·b = 0.
• λ nikaalna: a·b = 0 likhkar λ ki linear equation solve kijiye.
• (a + b) ⊥ (a − b) ⇔ |a| = |b|, kyunki (a + b)·(a − b) = |a|² − |b|².
• |a + b|² = |a|² + |b|² ⇔ a ⊥ b, kyunki |a + b|² = |a|² + |b|² + 2a·b.
• Parallel vectors: b = λa, yaani a₁/b₁ = a₂/b₂ = a₃/b₃.

• Parallelogram with adjacent sides a, b: area = |a × b|.
• Parallelogram with diagonals d₁, d₂: area = ½|d₁ × d₂|.
• Triangle ABC: area = ½|AB × AC|. With position vectors a, b, c: area = ½|a × b + b × c + c × a|.
• Three points are collinear if the area of the triangle they make is 0.

• आसन्न भुजाओं a, b वाले समांतर चतुर्भुज का क्षेत्रफल = |a × b|।
• विकर्णों d₁, d₂ वाले समांतर चतुर्भुज का क्षेत्रफल = ½|d₁ × d₂|।
• त्रिभुज ABC: क्षेत्रफल = ½|AB × AC|। स्थिति सदिशों a, b, c से: क्षेत्रफल = ½|a × b + b × c + c × a|।
• तीन बिंदु संरेख हैं यदि उनसे बने त्रिभुज का क्षेत्रफल 0 है।

• Adjacent sides a, b wale parallelogram ka area = |a × b|.
• Diagonals d₁, d₂ wale parallelogram ka area = ½|d₁ × d₂|.
• Triangle ABC: area = ½|AB × AC|. Position vectors a, b, c se: area = ½|a × b + b × c + c × a|.
• Teen points collinear hain agar unse bane triangle ka area 0 hai.

💡 Do not forget the ½ for a triangle and for diagonals.त्रिभुज और विकर्णों के लिए ½ न भूलें।Triangle aur diagonals ke liye ½ mat bhooliye.

• Projection of a on b = a·b/|b| (a scalar; it can be negative).
• Divide by the magnitude of the vector you project ON (b), not of a.
• MCQ: projection of î − ĵ on î + ĵ = (1 − 1)/√2 = 0.

• a का b पर प्रक्षेप = a·b/|b| (एक अदिश; ऋणात्मक भी हो सकता है)।
• उस सदिश (b) के परिमाण से भाग दीजिए जिस पर प्रक्षेप ले रहे हैं, a के नहीं।
• MCQ: î − ĵ का î + ĵ पर प्रक्षेप = (1 − 1)/√2 = 0।

• a ka b par projection = a·b/|b| (ek scalar; negative bhi ho sakta hai).
• Us vector (b) ki magnitude se divide kijiye jis PAR projection le rahe hain, a ki nahi.
• MCQ: î − ĵ ka î + ĵ par projection = (1 − 1)/√2 = 0.