Given :- L1L3 + m1m3 + n1n3 = 0 ----------(i) & L2L3 + m2m3 + n2n3 = 0 ---------(ii) solve (i) & (ii), prove:-?

L3 = m1n2 - m2n1 & m3 = n1L2 - n2L1 & n3 = L1m2 - L2m1

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  • 10 years ago
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    I suppose it is not possible to proof the result as it is. However, it can proved that they are in proportion of the given quantities.

    Step 1:

    L1L3 + m1m3 = -n1n3 ==> L1L2L3 + m1m3L2 = -n1n3L2 --- (1)

    L2L3 + m2m3 = -n2n3 ==> L1L2L3 + m2m3L1 = - n2n3L1 --- (2)

    Eqn (2) - Eqn (1)

    m3(L1m2 - L2m1) = n3(n1L2 - n2L1)

    ==> m3/(n1L2 - n2L1) = n3/(L1m2 - L2m1) ----- (3)

    Similarly eliminating, (m1m3) & (m2m3), we can obtain,

    ==> L3/( m1n2 - m2n1) = n3/(L1m2 - L2m1) ---- (4)

    Thus from (3) and (4), we have,

    L3/( m1n2 - m2n1) = m3/(n1L2 - n2L1) = n3/(L1m2 - L2m1), say each of this is equal to some constant k

    So, L3 = k( m1n2 - m2n1); m3 = k(n1L2 - n2L1); and n3 = k(L1m2 - L2m1)

    Thus (L3, m3, n3) are in proportion with the given one.

    If the constant parameter k is assigned = 1,

    then, L3 = ( m1n2 - m2n1); m3 = (n1L2 - n2L1); and n3 = (L1m2 - L2m1)

    NOTE: Easy method of solving this,using cross multiplication rule of solving, However, this system of Q & A is not feasible for presenting that method. Hence, I presented with the normal method of elimination process.

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