Ex 10.4, 10 - Find area of parallelogram whose adjacent sides are

Ex 10.4, 10 - Chapter 10 Class 12 Vector Algebra - Part 2

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Ex 10.4, 10 Find the area of the parallelogram whose adjacent sides are determined by the vectors š‘Ž āƒ— = š‘– Ģ‚ āˆ’ š‘— Ģ‚ + 3š‘˜ Ģ‚ and b = 2š‘– Ģ‚ āˆ’ 7š‘— Ģ‚ + š‘˜ Ģ‚ . š‘Ž āƒ— = š‘– Ģ‚ āˆ’ š‘— Ģ‚ + 3š‘˜ Ģ‚ = 1š‘– Ģ‚ āˆ’ 1š‘— Ģ‚ + 3k Ģ‚ š‘ āƒ— = 2š‘– Ģ‚ āˆ’ 7š‘— Ģ‚ + š‘˜ Ģ‚ = 2š‘– Ģ‚ āˆ’ 7š‘— Ģ‚ + 1k Ģ‚ Area of parallelogram ABCD = |š‘Ž āƒ—" Ɨ " š‘ āƒ— | š’‚ āƒ— Ɨ š’ƒ āƒ— = |ā– 8(š‘– Ģ‚&š‘— Ģ‚&š‘˜ Ģ‚@1&āˆ’1&3@2&āˆ’7&1)| = š‘– Ģ‚ (āˆ’1 Ɨ 1 āˆ’ (āˆ’7) Ɨ 3) āˆ’ š‘— Ģ‚ (1 Ɨ 1 āˆ’ 2 Ɨ 3) + š‘˜ Ģ‚ (1 Ɨ āˆ’7 āˆ’ 2 Ɨ āˆ’1) = š‘– Ģ‚ (āˆ’1āˆ’(āˆ’21)) āˆ’ š‘— Ģ‚ (1 āˆ’ 6) + š‘˜ Ģ‚ (āˆ’7 āˆ’(āˆ’2)) = š‘– Ģ‚ (āˆ’1 + 21) āˆ’ š‘— Ģ‚ (āˆ’5) + š‘˜ Ģ‚ (āˆ’7 + 2) = 20 š’Š Ģ‚ + 5š’‹ Ģ‚ āˆ’ 5š’Œ Ģ‚ Magnitude of š‘Ž āƒ— Ɨ š‘ āƒ— = √(202+52+(āˆ’5)2) |š‘Ž āƒ—" Ɨ " š‘ āƒ— | = √(400+25+25) = √450 = √(25Ɨ9Ɨ2) = 5 Ɨ 3 Ɨ √2 = 15 √2 Therefore, the area of parallelogram is 15āˆššŸ .

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