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Vector Algebra

Question
CBSEENMA12034069
Wired Faculty App

Find the value of A such that the vectors 3 straight a with rightwards arrow on top plus space 4 straight b with rightwards arrow on top and 2 straight a with rightwards arrow on top space plus space straight b with rightwards arrow on top are perpendicular to each other when 

Solution

        straight a with rightwards arrow on top space equals space straight i with hat on top space plus space 2 space straight j with hat on top space minus space 3 space straight k with hat on top comma space space straight b with rightwards arrow on top space equals space straight i with hat on top space plus space straight lambda space straight j with hat on top space plus space space 5 space straight k with hat on top.
Here,  straight a with rightwards arrow on top space equals space straight i with hat on top space plus space 2 space straight j with hat on top space minus space 3 space straight k with hat on top comma space space space straight b with rightwards arrow on top space equals space straight i with hat on top space plus space straight lambda space straight j with hat on top space plus space 5 space straight k with hat on top
therefore space space space 3 space straight a with rightwards arrow on top space plus space 4 space straight b with rightwards arrow on top space equals space 3 space left parenthesis straight i with hat on top space plus space 2 space straight j with hat on top space minus space 3 space straight k with hat on top right parenthesis plus space 4 space left parenthesis straight i with hat on top space plus space straight lambda space straight j with hat on top space plus space 5 space straight k with hat on top right parenthesis
                              equals 7 straight i with hat on top plus left parenthesis 6 plus 4 straight lambda right parenthesis straight j with hat on top space plus 11 space straight k with hat on top
and 2 straight a with rightwards arrow on top plus straight b with rightwards arrow on top space equals space 2 left parenthesis straight i with hat on top plus 2 straight j with hat on top space minus 3 straight k with hat on top right parenthesis space plus space left parenthesis straight i with hat on top plus straight lambda straight j with hat on top space plus space 5 straight k with hat on top right parenthesis
                     equals space 3 straight i with hat on top plus left parenthesis 4 plus straight lambda right parenthesis straight j with hat on top space minus space straight k with hat on top
Since 3 straight a with rightwards arrow on top plus space 4 straight b with rightwards arrow on top space and space 2 straight a with rightwards arrow on top plus straight b with rightwards arrow on top are perpendicular to each other
 therefore space space space left parenthesis 3 straight a with rightwards arrow on top space plus space 4 straight b with rightwards arrow on top right parenthesis. space left parenthesis 2 straight a with rightwards arrow on top plus straight b with rightwards arrow on top right parenthesis space equals space 0 therefore space space space left curly bracket 7 straight i with hat on top plus left parenthesis 6 plus 4 straight lambda right parenthesis space straight j with hat on top space plus space 11 space straight k with hat on top right curly bracket. space open curly brackets 3 straight i with hat on top space plus left parenthesis 4 plus straight lambda right parenthesis space straight j with hat on top space minus space straight k with hat on top close curly brackets space equals space 0 therefore space left parenthesis 7 right parenthesis thin space left parenthesis 3 right parenthesis space plus space left parenthesis 6 plus 4 space straight lambda right parenthesis space left parenthesis 4 plus straight lambda right parenthesis plus left parenthesis 11 right parenthesis thin space left parenthesis negative 1 right parenthesis space equals space 0 therefore space space space 21 plus 24 plus 22 straight lambda space plus space 4 straight lambda squared minus 11 space equals space 0 therefore space space space 4 straight lambda squared space plus 22 straight lambda space plus 34 space equals space 0 rightwards double arrow space space space 2 straight lambda squared plus 11 straight lambda plus 22 space equals space 0 therefore space space space straight lambda space equals space fraction numerator negative 11 plus-or-minus square root of 121 minus 176 end root over denominator 4 end fraction equals space fraction numerator negative 11 plus-or-minus square root of negative 55 end root over denominator 2 end fraction
there exist no real value of A.

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