     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﷯ = 4 𝑖﷯ + 2 𝑗﷯ + 6 𝑘﷯ 𝐵𝐶﷯ = (11 − 5) 𝑖﷯ + (3 − 0) 𝑗﷯ + (7−(−2)) 𝑘﷯ = 6 𝑖﷯ + 3 𝑗﷯ + 9 𝑘﷯ 𝐴𝐶﷯ = (11 − 1) 𝑖﷯ + (3 − (−2)) 𝑗﷯ + (7−(−8)) 𝑘﷯ = 10 𝑖﷯ + 5 𝑗﷯ + 15 𝑘﷯ 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﷯ = 1 𝑖﷯ − 2 𝑗﷯ − 8 𝑘﷯ 𝑂𝐵﷯ = 5 𝑖﷯ + 0 𝑗﷯ − 2 𝑘﷯ and 𝑂𝐶﷯ = 11 𝑖﷯ + 3 𝑗﷯ + 7 𝑘﷯ Position vector of 𝐵 = 𝑘 𝑂𝐶﷯+1. 𝑂𝐴﷯﷮𝑘+1﷯ 𝑂𝐵﷯ = 𝑘 11 𝑖﷯ + 3 𝑗﷯ + 7 𝑘﷯﷯ + 1 1 𝑖﷯ −2 𝑗﷯ − 8 𝑘﷯﷯﷮𝑘 + 1﷯ 5 𝑖﷯ + 0 𝑗﷯ − 2 𝑘﷯ = 11𝑘 𝑖﷯ + 3𝑘 𝑗﷯ + 7𝑘 𝑘﷯ + 𝑖﷯ − 2 𝑗﷯ − 8 𝑘﷯﷮𝑘 + 1﷯ 5 𝑖﷯ + 0 𝑗﷯ − 2 𝑘﷯ = 11𝑘 + 1﷯ 𝑖﷯ + 3𝑘 − 2﷯ 𝑗﷯ + (7𝑘− 8) 𝑘﷯﷮𝑘 + 1﷯ ∴ 5 𝑖﷯ + 0 𝑗﷯ − 2 𝑘﷯ = (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 