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Vector Algebra

Question
CBSEENMA12036276
Wired Faculty App

Let straight a with rightwards arrow on top comma space straight b with rightwards arrow on top space and space straight c with rightwards arrow on top be three non-zero vectors such that no two of these are collinear. If the vector straight a with rightwards arrow on top space plus 2 straight b with rightwards arrow on top is collinear withstraight c with rightwards arrow on top space and space straight b with rightwards arrow on top space plus 3 straight c with rightwards arrow on top is collinear with straight a with rightwards arrow on top (λ being some non-zero scalar) then straight a with rightwards arrow on top space plus 2 straight b with rightwards arrow on top space plus 6 straight c with rightwards arrow on top equals

  • λstraight a with rightwards arrow on top

  • λstraight b with rightwards arrow on top

  • λstraight c with rightwards arrow on top

  • 0

Solution

D.

0

left parenthesis straight a with rightwards arrow on top space plus 2 straight b with rightwards arrow on top space right parenthesis space equals space straight t subscript 1 straight c with rightwards arrow on top space.... space left parenthesis 1 right parenthesis and space straight b with rightwards arrow on top space plus 3 straight c with rightwards arrow on top space equals space straight t subscript 2 straight a with rightwards arrow on top space..... space left parenthesis 2 right parenthesis left parenthesis 1 right parenthesis space minus space 2 space straight x space left parenthesis 2 right parenthesis rightwards double arrow space straight a with rightwards arrow on top space left parenthesis 1 space plus 2 straight t subscript 2 right parenthesis space plus space straight c with rightwards arrow on top space left parenthesis negative straight t subscript 1 minus 6 right parenthesis space equals space 0 rightwards double arrow space 1 plus space 2 straight t subscript 2 space equals 0 rightwards double arrow space straight t subscript 2 space equals space minus space minus 1 divided by 2 space & space straight t subscript 1 space equals space minus 6 since space straight a with rightwards arrow on top space and space straight c with rightwards arrow on top space are space space non minus collinear Putting space the space value space of space straight t subscript 1 space and space straight t subscript 2 space in space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis space we space get space straight a with rightwards arrow on top space plus 2 straight b with rightwards arrow on top space plus 6 straight c with rightwards arrow on top space space equals space space 0

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