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Misc 10 - Find unit vector parallel to parallelogram diagonal

Misc 10 - Chapter 10 Class 12 Vector Algebra - Part 2
Misc 10 - Chapter 10 Class 12 Vector Algebra - Part 3

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Misc 10 The two adjacent sides of a parallelogram are 2𝑖 ̂ − 4𝑗 ̂ + 5𝑘 ̂ and 𝑖 ̂ − 2𝑗 ̂ − 3𝑘 ̂ Find the unit vector parallel to its diagonal. Also, find its area. Let 𝑎 ⃗ and 𝑏 ⃗ are adjacent side of a parallelogram, where 𝒂 ⃗ = 2𝑖 ̂ − 4𝑗 ̂ + 5𝑘 ̂ 𝒃 ⃗ = 𝑖 ̂ − 2𝑗 ̂ − 3𝑘 ̂ Let diagonal be 𝒄 ⃗ Hence, 𝒄 ⃗ = 𝒂 ⃗ + 𝒃 ⃗ = (2𝑖 ̂ − 4𝑗 ̂ + 5𝑘 ̂) + (1𝑖 ̂ − 2𝑗 ̂ − 3𝑘 ̂) = (2 + 1) 𝑖 ̂ − (4 + 2) 𝑗 ̂ + (5 − 3)𝑘 ̂ = 3𝒊 ̂ − 6𝒋 ̂ + 2𝒌 ̂ Magnitude of ⌈𝑐 ⃗ ⌉ = √((3)^2+(−6)^2+(2)^2 ) = √(9+36+4) = √49 = 7 Unit vector in direction of 𝑐 ⃗ = 1/(𝑀𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑐 ⃗ ) × 𝑐 ⃗ 𝒄 ̂ = 𝟏/𝟕 ("3" 𝒊 ̂" − 6" 𝒋 ̂" + 2" 𝒌 ̂ ) Finding Area of parallelogram Area of parallelogram = |𝑎 ⃗ × 𝑏 ⃗ | Now, 𝑎 ⃗ × 𝑏 ⃗ = |■8(𝑖 ̂&𝑗 ̂&𝑘 ̂@2&−4&[email protected]&−2&−3)| = 𝑖 ̂ (12 − (−10) − 𝑗 ̂ (−6 −5) + 𝑘 ̂ (−4 − (−4)) = 22𝑖 ̂ + 11𝑗 ̂ + 0𝑘 ̂ = 22𝑖 ̂ + 11𝑗 ̂ So 𝒂 ⃗ × 𝒃 ⃗ = 22𝒊 ̂ + 11𝒋 ̂ Now, |𝑎 ⃗ × 𝑏 ⃗ | = √(22^2+11^2 ) = √(2^2 (〖11)〗^2+11^2 ) = √( 11^2 (2^2+1)) = 11√5 Hence, Area of parallelogram = |𝑎 ⃗ × 𝑏 ⃗ |= 𝟏𝟏√𝟓

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.