Question

Find an equation of the tangent plane and find symmetric equations of the normal line to the surface at the given point. $$ x^{2}+y^{2}+z^{2}=9, \quad(1,2,2) $$

Short answer

The equation of the tangent plane is \(2(x-1)+4(y-2)+4(z-2) = 0\), the symmetric equations of the normal line at the given point to the surface are \(x = 1 + 2t, y = 2 + 4t, z = 2 + 4t\).

Step-by-step solution

Finding the Normal Vector

The first task is to compute the normal vector to the surface at the provided point. The gradient of the function describing the sphere is normal to the surface. The sphere's function is \(f(x,y,z) = x^{2} + y^{2} + z^{2} = 9\). Calculate the gradient of this function: \(\nabla f = (2x, 2y, 2z)\). At point (1,2,2), \(\nabla f(1,2,2) = (2*1, 2*2, 2*2) = (2,4,4)\), yielding the normal vector.

Finding the Equation of the Tangent Plane

Now we can find the equation of the tangent plane. The equation of the tangent plane is given by \(A(x-x_{0}) + B(y-y_{0}) + C(z-z_{0}) = 0\), where (A,B,C) is the normal vector and (x_{0},y_{0},z_{0}) is the provided point. Substituting in the values, the equation becomes: \(2(x-1)+4(y-2)+4(z-2) = 0\)

Finding the Symmetric Equations of the Normal Line

The symmetric equations of the normal line can be found from the point and the normal vector. The symmetric equations are \(x=x_{0}+At, y=y_{0}+Bt, z=z_{0}+Ct\), where t is the parameter. Inserting the given values, the symmetric equations of the normal line would be \(x = 1 + 2t, y = 2 + 4t, z = 2 + 4t\). The parameter t can be any real number.