Question

Find an equation of the tangent plane and find symmetric equations of the normal line to the surface at the given point. $$ x^{2}-y^{2}+z^{2}=0, \quad(5,13,-12) $$

Short answer

The equation of the tangent plane is \(10x - 26y - 24z = 384\). The symmetric equations of the normal line are \(x = 5 + 10t\), \(y = 13 - 26t\), and \(z = -12 - 24t\).

Step-by-step solution

Calculate the Partial Derivatives

First, we need to compute the partial derivatives of \( f(x,y,z) = x^{2} - y^{2} + z^{2} \). The partial derivatives are as follows: \( f_{x} = 2x \), \( f_{y} = -2y \), \( f_{z} = 2z \).Using the point (5, 13, -12), \( f_{x}(5, 13, -12) = 2(5) = 10 \), \( f_{y}(5, 13, -12) = -2(13) = -26 \), \( f_{z}(5, 13, -12) = 2(-12) = -24 \).

Find the Gradient Vector

The gradient vector is the vector that consists of the partial derivatives, i.e., \( \nabla f = (f_{x}, f_{y}, f_{z}) = (10, -26, -24) \). This vector is normal (perpendicular) to the tangent plane at the point (5, 13, -12).

Determine the Equation of the Tangent Plane

The equation of the tangent plane to a surface at a given point is given by \( F_{x}(x-a) + F_{y}(y-b) + F_{z}(z-c) = 0 \), where (a, b, c) is the point on the surface and \( F_{x} \), \( F_{y} \), and \( F_{z} \) are the values of the partial derivatives at that point. Plugging in the known values we get \( 10*(x-5) - 26*(y-13) - 24*(z+12) = 0 \). This simplifies to \( 10x - 26y - 24z = 384 \).

Find the Symmetric Equations of the Normal Line

The symmetric equations of the normal line are given by \( x = a + at \), \( y = b + bt \), and \( z = c + ct \), where (a, b, c) is the point of tangency, and (a, b, c) is the directional vector. Substituting the given values, we have \( x = 5 + 10t \), \( y = 13 - 26t \), and \( z = -12 - 24t \).