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Ananya Dara
Asked: 2022-11-10T18:13:03+05:30 In:

If `vecb` is not perpendicular to `vecc` . Then find the vector `vecr` satisfying the equation `vecr xx vecb = veca xx vecb and vecr. vecc=0`

If `vecb` is not perpendicular to `vecc` . Then find the vector `vecr` satisfying the equation `vecr xx vecb = veca xx vecb and vecr. vecc=0`
Parabola
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  1. Given `vecrxxvecb=vecaxxvecbRightarrow (vecr-veca)xxvecb=0`
    Hence, (`vecr – veca) and vecb` are parallel.
    `Rightarrow vecr-veca=tvecb`
    `vecr.vecc=0`
    Therefore, taking dot product of (i) by `vecc`, we get `vecr.vecc -veca.vecc=t(vecb .vecc)`
    `or 0-veca.vecc=r(vecb.vecc)ort=-((veca.vecc)/(vecb.vecc))`
    from (i) and (ii) solution of `vecr” is ” vecr =veca-((veca.vecc)/(vecb.vecc))vecb`

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Binita Rau
Asked: 2022-11-03T14:57:22+05:30 In:

If `vecb` is not perpendicular to `vecc` . Then find the vector `vecr` satisfying the equation `vecr xx vecb = veca xx vecb and vecr. vecc=0`

If `vecb` is not perpendicular to `vecc` . Then find the vector `vecr` satisfying the equation `vecr xx vecb = veca xx vecb and vecr. vecc=0`
Parabola
Leave an answer

Leave an answer

Browse

1 Answer

  1. Given `vecrxxvecb=vecaxxvecbRightarrow (vecr-veca)xxvecb=0`
    Hence, (`vecr – veca) and vecb` are parallel.
    `Rightarrow vecr-veca=tvecb`
    `vecr.vecc=0`
    Therefore, taking dot product of (i) by `vecc`, we get `vecr.vecc -veca.vecc=t(vecb .vecc)`
    `or 0-veca.vecc=r(vecb.vecc)ort=-((veca.vecc)/(vecb.vecc))`
    from (i) and (ii) solution of `vecr” is ” vecr =veca-((veca.vecc)/(vecb.vecc))vecb`

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