Ex 10.4, 1 Class 12 - NCERT Solutions Vector Algebra - Find |a x b|

Ex 10.4, 1 - Chapter 10 Class 12 Vector Algebra - Part 2

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Ex 10.4, 1 Find |š‘Ž āƒ—Ć—š‘ āƒ— |, if š‘Ž āƒ— = š‘– Ģ‚ āˆ’ 7š‘— Ģ‚ + 7š‘˜ Ģ‚ and š‘ āƒ— = 3š‘– Ģ‚ āˆ’ 2š‘— Ģ‚ + 2š‘˜ Ģ‚š‘Ž āƒ— = š‘– Ģ‚ āˆ’ 7š‘— Ģ‚ + 7š‘˜ Ģ‚ = 1š‘– Ģ‚ āˆ’ 7š‘— Ģ‚ + 7š‘˜ Ģ‚ š‘ āƒ— = 3š‘– Ģ‚ āˆ’ 2š‘— Ģ‚ + 2k Ģ‚ š‘Ž āƒ— Ɨ š‘ āƒ— = |ā– 8(š‘– Ģ‚&š‘— Ģ‚&š‘˜ Ģ‚@ā–ˆ(1@3)&ā–ˆ(āˆ’7@āˆ’2)&ā–ˆ(7@2))| = š‘– Ģ‚ |ā– 8(āˆ’7&7@āˆ’2&2)| āˆ’š‘— Ģ‚ |ā– 8(1&7@3&2)| + k Ģ‚ |ā– 8(1&āˆ’7@3&āˆ’2)| = š‘– Ģ‚ ("āˆ’7 Ɨ 2 – (āˆ’2 Ɨ 7)" ) āˆ’ š‘— Ģ‚((1Ɨ2 ) āˆ’ (3Ɨ7 )) + š‘˜ Ģ‚((1Ɨ2 ) āˆ’ (3 Ɨ āˆ’7)) = š‘– Ģ‚ (āˆ’14āˆ’(āˆ’14)) āˆ’ š‘— Ģ‚(2āˆ’21 ) + š‘˜ Ģ‚((āˆ’2āˆ’(āˆ’21)) = š‘– Ģ‚ (0) āˆ’ š‘— Ģ‚ (āˆ’19) + š‘˜ Ģ‚(19) = 0š‘– Ģ‚ + 19š‘— Ģ‚ + 19š‘˜ Ģ‚ ∓ š’‚ āƒ— Ɨ š’ƒ āƒ— = 0š’Š Ģ‚ + 19š’‹ Ģ‚ + 19š’Œ Ģ‚ Magnitude of š‘Ž āƒ— Ɨ š‘ āƒ— = √(02+192+192) |š’‚ āƒ—" Ɨ" š’ƒ āƒ— | = √(0+361+361) = √722 = √(19Ɨ19Ɨ2 ) = 19āˆššŸ Therefore, the magnitude of š‘Ž āƒ—" Ɨ" š‘ āƒ— is 19āˆššŸ.

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