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Labeen Ramaswamy
Asked: 2022-11-05T15:00:19+05:30 In:

Let `vecr, veca, vecb` and `vecc` be four non-zero vectors such that `vecr.veca=0, |vecrxxvecb|=|vecr||vecb|,|vecrxxvecc|=|vecr||vecc|` then
`[(veca, vecb, vecc)]=`
A. `-1`
B. `0`
C. `1`
D. `2`

Let `vecr, veca, vecb` and `vecc` be four non-zero vectors such that `vecr.veca=0, |vecrxxvecb|=|vecr||vecb|,|vecrxxvecc|=|vecr||vecc|` then
`[(veca, vecb, vecc)]=`
A. `-1`
B. `0`
C. `1`
D. `2`
Parabola
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1 Answer

  1. Correct Answer – B
    We have
    `{:(vecr.veca=0impliesvecr_|_veca),(|vecrxxvecb|=|vecr||vecb|impliesvecr_|_vecb),(“and”|vecrxxvecr|=|vecr||vecc|impliesvecr_|_vecc):}}impliesveca, vecb vecc` are coplanar.
    Hence `[(veca, vecb, vecc)]=0`

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Marlo Sai Randhawa
Asked: 2022-10-29T11:34:05+05:30 In:

Let `vecr, veca, vecb` and `vecc` be four non-zero vectors such that `vecr.veca=0, |vecrxxvecb|=|vecr||vecb|,|vecrxxvecc|=|vecr||vecc|` then
`[(veca, vecb, vecc)]=`
A. |a||b||c|
B. `-|a||b||c|`
C. 0
D. none of these

Let `vecr, veca, vecb` and `vecc` be four non-zero vectors such that `vecr.veca=0, |vecrxxvecb|=|vecr||vecb|,|vecrxxvecc|=|vecr||vecc|` then
`[(veca, vecb, vecc)]=`
A. |a||b||c|
B. `-|a||b||c|`
C. 0
D. none of these
Parabola
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Leave an answer

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1 Answer

  1. Correct Answer – c
    `vecr.veca = 0, |vecr xx vecb| = |vecr||vecb| and |vecr xx vecc|`
    `|vecr||vecc|`
    ` [ veca vecb vecc] =0`

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