Question 10 (OR 1 st question)

Find the area of the parallelogram whose diagonals are represented by the vectors a = 2i  – 3j  + 4k and b = 2i – j + 2k

We use the Area of Parallelogram formula with Diagonals

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Question 10 (OR 1st question) Find the area of the parallelogram whose diagonals are represented by the vectors ๐‘Ž โƒ— = 2๐‘– ฬ‚ โ€“ 3๐‘— ฬ‚ + 4๐‘˜ ฬ‚ and ๐‘ โƒ— = 2๐‘– ฬ‚ โ€“ ๐‘— ฬ‚ + 2๐‘˜ ฬ‚ Area of parallelograms with diagonal Area = 1/2 |(๐‘‘_1 ) โƒ—ร—(๐‘‘_2 ) โƒ— | Given Diagonals of a parallelogram as ๐‘Ž โƒ— = 2๐‘– ฬ‚ โ€“ 3๐‘— ฬ‚ + 4๐‘˜ ฬ‚ and ๐‘ โƒ— = 2๐‘– ฬ‚ โ€“ ๐‘— ฬ‚ + 2๐‘˜ ฬ‚ Area of the parallelogram = 1/2 |๐‘Ž โƒ—ร—๐‘ โƒ— | Finding ๐’‚ โƒ— ร— ๐’ƒ โƒ— ๐‘Ž โƒ— ร— ๐‘ โƒ— = |โ– 8(๐‘– ฬ‚&๐‘— ฬ‚&๐‘˜ ฬ‚@2&โˆ’3&4@2&โˆ’1&2)| = ๐‘– ฬ‚ ((โ€“3) ร— 2 โ€“ (-1) ร— 4) โˆ’ ๐‘— ฬ‚ (2 ร— 2 โˆ’ 2 ร— 4) + ๐‘˜ ฬ‚ (2 ร— (-1) โˆ’ 2 ร— (-3)) = ๐‘– ฬ‚ (-6 + 4) โˆ’ ๐‘— ฬ‚ (4 โ€“ 8) + ๐‘˜ ฬ‚ (โ€“2 + 6) = ๐‘– ฬ‚ (โˆ’2) - ๐‘— ฬ‚ (โˆ’4) + ๐‘˜ ฬ‚ (4) = โ€“2๐‘– ฬ‚ + 4๐‘— ฬ‚ + 4๐‘˜ ฬ‚ So, Magnitude of ๐‘Ž โƒ— ร— ๐‘ โƒ— = โˆš((โˆ’2)2+(4)2+(4)2) |๐‘Ž โƒ—" ร— " ๐‘ โƒ— | = โˆš(4+16+16) |๐‘Ž โƒ—" ร— " ๐‘ โƒ— |= โˆš36 |๐‘Ž โƒ—" ร— " ๐‘ โƒ— |= 6 Thus, Area of the parallelogram = 1/2 |๐‘Ž โƒ—ร—๐‘ โƒ— | = 1/2 ร— 6 = 3 square units

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.