If a, b, c are three vectors such that a + b + c = 0 ,Ā  then prove that a Ɨ b = b Ɨ c = c Ɨ a, and hence show that [a b c] = 0.

This is a question of CBSE Sample Paper - Class 12 - 2017/18.

You can download the question paper hereĀ  https://www.teachoo.com/cbse/sample-papers/


If a, b, c are three vectors such that a + b + c = 0, then prove a x b

Question 20 - CBSE Class 12 Sample Paper for 2018 Boards - Part 2
Question 20 - CBSE Class 12 Sample Paper for 2018 Boards - Part 3 Question 20 - CBSE Class 12 Sample Paper for 2018 Boards - Part 4

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Question 20 If š‘Ž āƒ—, š‘ āƒ—, š‘ āƒ— are three vectors such that š‘Ž āƒ— + š‘ āƒ— + š‘ āƒ— = 0 āƒ— , then prove that š‘Ž āƒ— Ɨ š‘ āƒ— = š‘ āƒ— Ɨ š‘ āƒ— = š‘ āƒ— Ɨ š‘Ž āƒ—, and hence show that [š‘Ž āƒ—" " š‘ āƒ—" " š‘ āƒ— ] = 0. Theory Here [š‘Ž āƒ—" " š‘ āƒ—" " š‘ āƒ— ] = š‘Ž āƒ—.(š‘ āƒ— Ɨ š‘ āƒ— ) Given š‘Ž āƒ— + š‘ āƒ— + š‘ āƒ— = 0 āƒ— š‘Ž āƒ—Ć—(š‘Ž āƒ—+š‘ āƒ—+š‘ āƒ— )= š‘Ž āƒ—Ć—0 āƒ— š‘Ž āƒ—Ć—š‘Ž āƒ—+š‘Ž āƒ—Ć—š‘ āƒ—+š‘Ž āƒ—Ć—š‘ āƒ—= 0 āƒ— Since š‘Ž āƒ—Ć—š‘Ž āƒ—=0 " " 0+š‘Ž āƒ—Ć—š‘ āƒ—+š‘Ž āƒ—Ć—š‘ āƒ—=" " 0 āƒ— š‘Ž āƒ—Ć—š‘ āƒ—+š‘Ž āƒ—Ć—š‘ āƒ—=" " 0 āƒ— š‘Ž āƒ—Ć—š‘ āƒ—=āˆ’š‘Ž āƒ—Ć—š‘ āƒ— Since āˆ’š‘Ž āƒ—Ć—š‘ āƒ— = š‘ āƒ—Ć—š‘Ž āƒ— š’‚ āƒ—Ć—š’ƒ āƒ—=š’„ āƒ—Ć—š’‚ āƒ— Similarly, š‘Ž āƒ— + š‘ āƒ— + š‘ āƒ— = 0 āƒ— š‘ āƒ—Ć—(š‘Ž āƒ—+š‘ āƒ—+š‘ āƒ— )= š‘ āƒ—Ć—0 āƒ— š‘ āƒ—Ć—š‘Ž āƒ—+š‘ āƒ—Ć—š‘ āƒ—+š‘ āƒ—Ć—š‘ āƒ—= 0 āƒ— Since š‘ āƒ—Ć—š‘ āƒ—=0 š‘ āƒ—Ć—š‘Ž āƒ—+0+š‘ āƒ—Ć—š‘ āƒ—= 0 āƒ— š‘ āƒ—Ć—š‘Ž āƒ—+š‘ āƒ—Ć—š‘ āƒ—=" " 0 āƒ— š‘ āƒ—Ć—š‘ āƒ—=āˆ’š‘ āƒ—Ć—š‘Ž āƒ— š‘ āƒ—Ć—š‘ āƒ—=āˆ’š‘ āƒ—Ć—š‘Ž āƒ— Since āˆ’š‘ āƒ—Ć—š‘Ž āƒ— = š‘Ž āƒ—Ć—š‘ āƒ— š‘ āƒ—Ć—š‘ āƒ—=š‘Ž āƒ—Ć—š‘ āƒ— Thus, š’‚ āƒ—Ć—š’ƒ āƒ—=š’„ āƒ—Ć—š’‚ āƒ— & š‘ āƒ—Ć—š‘ āƒ—=š‘Ž āƒ—Ć—š‘ āƒ— ∓ š’‚ āƒ—Ć—š’ƒ āƒ—=š’ƒ āƒ—Ć—š’„ āƒ—=š’„ āƒ—Ć—š’‚ āƒ— Now, we need to show that show that [š‘Ž āƒ—" " š‘ āƒ—" " š‘ āƒ— ] = 0 [š‘Ž āƒ— š‘ āƒ— š‘ āƒ— ]=š‘Ž āƒ— . (š‘ āƒ—Ć—š‘ āƒ— ) From (1): š‘ āƒ—Ć—š‘ āƒ— = š‘Ž āƒ—Ć—š‘ āƒ— =š‘Ž āƒ— . (š‘Ž āƒ—Ć—š‘ āƒ— ) Now, š‘Ž āƒ—Ć—š‘ āƒ— will be a vector perpendicular to š‘Ž āƒ— And dot product of š‘Ž āƒ— with a vector perpendicular to š‘Ž āƒ— will be 0 as angle is 90° and cos 90° = 0 ∓ [š‘Ž āƒ— š‘ āƒ— š‘ āƒ— ]=š‘Ž āƒ— . (š‘Ž āƒ—Ć—š‘ āƒ— ) = 0 Hence proved

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