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Note that the coordinates of the point whose position vector is 2i+3j-k\, are (2,3,-1).

Given that \phi (x,y,z)=x^{2}y^{3}z^{4}\,

Therefore,\nabla \phi =[i{\frac  {\partial }{\partial x}}+j{\frac  {\partial }{\partial y}}+k{\frac  {\partial }{\partial z}}](x^{2}y^{3}z^{4})=2xy^{3}z^{3}i+3x^{2}y^{2}z^{4}j+4x^{2}y^{3}z^{3}k\,

=108i+108j-432k\, at the point (2,3,-1).


The rate of change of phi is maximum in the direction of gradient of phi along the vector 108(i+j-4k)\,

Therefore the magnitude of the maximim rate of change of phi at the point 2i+3j-k\, =|\nabla \phi |\, at (2,3,-1).

=|108(i+j-4k)|=108{\sqrt  {1+1+4}}=324{\sqrt  {2}}\,

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