VC3.47

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

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