VC3.47

From Example Problems
Jump to: navigation, search

Given \phi (x,y,z)=x^{2}+y^{2}+z^{2}=1,\psi (x,y,z)=x+y+z=1\,

(\nabla \phi )_{{0}},(\nabla \psi )_{{0}}\, denote the values at (1,0,0).

Now \nabla \phi =[i{\frac  {\partial }{\partial x}}+j{\frac  {\partial }{\partial y}}+k{\frac  {\partial }{\partial z}}](x^{2}+y^{2}+z^{2})=2xi+2yj+2zk\,

and \nabla \psi ==[i{\frac  {\partial }{\partial x}}+j{\frac  {\partial }{\partial y}}+k{\frac  {\partial }{\partial z}}](x+y+z)=i+j+k\,

Hence the required equation of the tangent line is

(r-r_{{0}})\times [(\nabla \phi )_{{0}}\times (\nabla \psi )_{{0}}]=0\,

or [(x-1)i+yj+zk]\times [2i\times (i+j+k)]=0\,

or [(x-1)i+yj+zk]\times [0i-2j+2k]=0\,

or {\begin{vmatrix}i&j&k\\x-1&y&z\\0&-2&2\end{vmatrix}}=0\,

or (2y+2z)i-2(x-1)j-2(x-1)k=0\,

y=-z,x-1=0\,

Therefore,the required equation of the tangent lines in cartesian form in (r-r_{{0}})\cdot [(\nabla \phi )_{{0}}\times (\nabla \psi )_{{0}}]=0\,

[(x-1)i+yj+zk]\cdot (2i\times (i+j+k))=0,-2y+2z=0,y-z=0\,


Main Page