If š‘Ž āƒ— ≠ 0 āƒ—, š‘Ž āƒ—. š‘ āƒ— = š‘Ž āƒ—. š‘ āƒ—, š‘Ž āƒ— Ɨ š‘ āƒ— = š‘Ž āƒ— Ɨ š‘ āƒ— , then show that š‘ āƒ— = š‘ āƒ—.

[Class 12] If Vector a ≠ 0, a .b = a.c, aƗb=aƗb, then show that b = c - CBSE Class 12 Sample Paper for 2022 Boards (For Term 2)

part 2 - Question 9 - CBSE Class 12 Sample Paper for 2022 Boards (For Term 2) - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12

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Question 9 If š‘Ž āƒ— ≠ 0 āƒ—, š‘Ž āƒ—. š‘ āƒ— = š‘Ž āƒ—. š‘ āƒ—, š‘Ž āƒ— Ɨ š‘ āƒ— = š‘Ž āƒ— Ɨ š‘ āƒ— , then show that š‘ āƒ— = š‘ āƒ—. Given that š’‚ āƒ—. š’ƒ āƒ— = š’‚ āƒ—. š’„ āƒ— Now, š‘Ž āƒ—. š‘ āƒ— = š‘Ž āƒ—. š‘ āƒ— š‘Ž āƒ—. š‘ āƒ— āˆ’ š‘Ž āƒ—. š‘ āƒ— = 0 š’‚ āƒ—. (š’ƒ āƒ— āˆ’ š’„ āƒ—) = 0 Thus, either (š’ƒ āƒ— āˆ’ š’„ āƒ—) = šŸŽ āƒ— Or š’‚ āƒ— is perpendicular to (š’ƒ āƒ— āˆ’ š’„ āƒ—) š’‚ āƒ— Ɨ š’ƒ āƒ— = š’‚ āƒ— Ɨ š’„ āƒ— Now, š‘Ž āƒ— Ɨ š‘ āƒ— = š‘Ž āƒ— Ɨ š‘ āƒ— š‘Ž āƒ— Ɨ š‘ āƒ— āˆ’ š‘Ž āƒ— Ɨ š‘ āƒ— = šŸŽ āƒ— š’‚ āƒ— Ɨ (š’ƒ āƒ— āˆ’ š’„ āƒ—) = šŸŽ āƒ— Thus, either (š’ƒ āƒ— āˆ’ š’„ āƒ—) = šŸŽ āƒ— Or š’‚ āƒ— is parallel to (š’ƒ āƒ— āˆ’ š’„ āƒ—) Since š‘Ž āƒ— cannot both be perpendicular and parallel to (š‘ āƒ— āˆ’ š‘ āƒ—) Therefore, the only other solution is (š‘ āƒ— āˆ’ š‘ āƒ—) = 0 āƒ— š’ƒ āƒ— = š’„ āƒ— Hence proved

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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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo