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1. Chapter 10 Class 12 Vector Algebra
2. Serial order wise
3. Miscellaneous

Transcript

Misc 8(Introduction) Show that the points A(1, – 2, – 8), B (5, 0, – 2) and C (11, 3, 7) are collinear, and find the ratio in which B divides AC. (1) Three points collinear i.e. AB + BC = AC (2) Three vectors. i.e. |(𝐴𝐵) ⃗ | + |(𝐵𝐶) ⃗ | = |(𝐴𝐶) ⃗ | Misc 8 Show that the points A(1, – 2, – 8), B (5, 0, – 2) and C (11, 3, 7) are collinear, and find the ratio in which B divides AC. 3 points A, B, C are collinear if i.e. |(𝑨𝑩) ⃗ | + |(𝑩𝑪) ⃗ | = |(𝑨𝑪) ⃗ | A (1, -2, −8) B (5, 0,−2) C (11, 3, 7) (𝐴𝐵) ⃗ = (5 − 1) 𝑖 ̂ + (0 − (−2)) 𝑗 ̂ + (−2−(−8)) 𝑘 ̂ = 4𝑖 ̂ + 2𝑗 ̂ + 6𝑘 ̂ (𝐵𝐶) ⃗ = (11 − 5) 𝑖 ̂ + (3 − 0) 𝑗 ̂ + (7−(−2)) 𝑘 ̂ = 6𝑖 ̂ + 3𝑗 ̂ + 9𝑘 ̂ (𝐴𝐶) ⃗ = (11 − 1) 𝑖 ̂ + (3 − (−2)) 𝑗 ̂ + (7−(−8)) 𝑘 ̂ = 10𝑖 ̂ + 5𝑗 ̂ + 15𝑘 ̂ Magnitude of (𝐴𝐵) ⃗ = √(42+22+62) (𝐴𝐵) ⃗ = √(16+4+36) = √56 = √(4×14 ) = 2√14 Magnitude of (𝐵𝐶) ⃗ = √(62+32+92) |(𝐵𝐶) ⃗ |= √(36+9+81) = √126 = √(9×14 ) = 3√14 Magnitude of (𝐴𝐶) ⃗ = √(102+52+152) |(𝐴𝐶) ⃗ |= √(100+25+225)= √350 = √(25×14 ) = 5√14 |(𝐴𝐵) ⃗ | + |(𝐵𝐶) ⃗ | = 2√(14 ) + 3√(14 ) = 5√(14 ) = |(𝐴𝐶) ⃗ | ∴ A, B and C are collinear. Finding the ratio in which B divides AC Let B divide AC in the ratio k : 1 (𝑂𝐴) ⃗ = 1𝑖 ̂ − 2𝑗 ̂ − 8𝑘 ̂ (𝑂𝐵) ⃗ = 5𝑖 ̂ + 0𝑗 ̂ − 2𝑘 ̂ and (𝑂𝐶) ⃗ = 11𝑖 ̂ + 3𝑗 ̂ + 7𝑘 ̂ Position vector of 𝐵 = (𝑘(𝑂𝐶) ⃗ + 1.(𝑂𝐴) ⃗)/(𝑘 + 1) (𝑂𝐵) ⃗ = (𝑘(11𝑖 ̂ + 3𝑗 ̂ + 7𝑘 ̂ ) + 1(1𝑖 ̂ − 2𝑗 ̂ − 8𝑘 ̂ ))/(𝑘 + 1) 5𝑖 ̂ + 0𝑗 ̂ − 2𝑘 ̂ = (11𝑘𝑖 ̂ + 3𝑘𝑗 ̂ + 7𝑘 𝑘 ̂ + 𝑖 ̂ − 2𝑗 ̂ − 8𝑘 ̂)/(𝑘 + 1) 5𝑖 ̂ + 0𝑗 ̂ − 2𝑘 ̂ = ((11𝑘 + 1) 𝑖 ̂ + (3𝑘 − 2) 𝑗 ̂ + (7𝑘 − 8) 𝑘 ̂)/(𝑘 + 1) ∴ 5𝑖 ̂ + 0𝑗 ̂ − 2𝑘 ̂ = ((11𝑘 + 1) )/((𝑘 + 1)) 𝑖 ̂ + ((3𝑘 − 2) )/((𝑘 + 1)) 𝑗 ̂ + ((7𝑘 − 8) )/((𝑘 + 1)) 𝑘 ̂ Since the two vectors are equal, corresponding components are also equal. So, (11𝑘 + 1)/(𝑘 + 1) = 5 11k + 1 = 5k + 5 11k – 5k = 5 − 1 6k = 4 k = 4/6 = 2/3 Thus, B divides AC in the ratio 2/3 : 1 or 2 : 3

Miscellaneous 