# Misc 10

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Misc 10 The two adjacent sides of a parallelogram are 2 𝑖 − 4 𝑗 + 5 𝑘 and 𝑖 − 2 𝑗 − 3 𝑘 Find the unit vector parallel to its diagonal. Also, find its area. Let 𝑎 and 𝑏 are adjacent side of a parallelogram, where 𝑎 = 2 𝑖 − 4 𝑗 + 5 𝑘 𝑏 = 𝑖 − 2 𝑗 − 3 𝑘 Let diagonal be 𝑐 Hence, 𝒄 = 𝒂 + 𝒃 = (2 𝑖 − 4 𝑗 + 5 𝑘) + (1 𝑖 − 2 𝑗 − 3 𝑘) = (2 + 1) 𝑖 − (4 + 2) 𝑗 + (5 − 3) 𝑘 = 3 𝒊 − 6 𝒋 + 2 𝒌 Magnitude of 𝑐 = 32+ −62+ 22 = 9+36+4 = 49 = 7 Unit vector in direction of 𝑐 = 1𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑐 × 𝑐 𝒄 = 𝟏𝟕 3 𝒊 − 6 𝒋 + 2 𝒌 Area of parallelogram = 𝑎× 𝑏 So 𝑎× 𝑏 = 𝑖 𝑗 𝑘2−451−2−3 = 𝑖 −3×−4−(−2×5) − 𝑗 2×−3−(1×5) + 𝑘 −2×2−(−4×1) = 𝑖 (12 − (−10) − 𝑗 (−6 −5) + 𝑘 (−4 − (−4)) = 𝑖 (12 + 10) − 𝑗 (−11) + 𝑘 (−4 + 4) = 22 𝑖 + 11 𝑗 + 0 𝑘 = 22 𝑖 + 11 𝑗 So 𝒂× 𝒃 = 22 𝒊 + 11 𝒋 𝑎× 𝑏 = 222+ 112 = 22( 11)2+ 112 = 112( 22+1) = 11 5 Hence, Area of parallelogram = 𝑎× 𝑏= 𝟏𝟏 𝟓

About the Author

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .