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Let-p-q-r-be-three-mutually-perpendicular-vectors-of-the-same-magnitude-If-a-vector-x-satisfies-the-equation-p-x-q-p-q-x-r-q-r-x-p-r-0-then-x-is-given-by-




Question Number 19262 by djatmiko last updated on 08/Aug/17
Let p, q, r be three mutually perpendicular  vectors of the same magnitude. If a vector  x  satisfies the equation   p ×{x−q)×p}+q×{x−r)×q}                   +  r×{x−p)×r}=0, then x is  given by
Let\boldsymbolp,\boldsymbolq,\boldsymbolrbethreemutuallyperpendicularvectorsofthesamemagnitude.Ifavector\boldsymbolxsatisfiestheequation\boldsymbolp×{\boldsymbolx\boldsymbolq)×\boldsymbolp}+\boldsymbolq×{\boldsymbolx\boldsymbolr)×\boldsymbolq}+\boldsymbolr×{\boldsymbolx\boldsymbolp)×\boldsymbolr}=0,then\boldsymbolxisgivenby
Answered by ajfour last updated on 08/Aug/17
= (p^� .p^� )(x^� −q^� )−[p^� .(x^� −q^� )]p^� +       (q^� .q^� )(x^� −r^� )−[q^� .(x^� −r^� )]q^� +       (r^� .r^� )(x^� −p^� )−[r^� .(x^� −p^� )]r^�   = (p^2 +q^2 +r^2 )x^� −[(p^� .x^� )p^� +(q^� .x^� )q^� +(r^� .x^� )r^� ]       −(p^2 +q^2 +r^2 )(p^� +q^� +r^� )+        (p^� .q^� )p^� +(q^� .r^� )q^� +(r^� .p^� )r^� =0   p^� , q^� , and r^�  are mutually ⊥,  and of equal magnitudes ;let  their manitudes be a.  ⇒3a^2 [x^� −(p^� +q^� +r^� )]=a(x_p p^� +x_q q^� +x_r r^� )  ⇒ 3a^2 [x^� −(p^� +q^� +r^� )]=a^2 (x^� )  ⇒  x^� =(3/2)(p^� +q^� +r^� ) .
=(p¯.p¯)(x¯q¯)[p¯.(x¯q¯)]p¯+(q¯.q¯)(x¯r¯)[q¯.(x¯r¯)]q¯+(r¯.r¯)(x¯p¯)[r¯.(x¯p¯)]r¯=(p2+q2+r2)x¯[(p¯.x¯)p¯+(q¯.x¯)q¯+(r¯.x¯)r¯](p2+q2+r2)(p¯+q¯+r¯)+(p¯.q¯)p¯+(q¯.r¯)q¯+(r¯.p¯)r¯=0p¯,q¯,andr¯aremutually,andofequalmagnitudes;lettheirmanitudesbea.3a2[x¯(p¯+q¯+r¯)]=a(xpp¯+xqq¯+xrr¯)3a2[x¯(p¯+q¯+r¯)]=a2(x¯)x¯=32(p¯+q¯+r¯).

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