Question

Find an equation of the tangent plane and find symmetric equations of the normal line to the surface at the given point. $$ x y-z=0, \quad(-2,-3,6) $$

Short answer

The equation of the tangent plane to the surface \(xy - z = 0\) at the point (-2,-3,6) is \( -3x - 2y + 12 = z \), and the symmetric equations of the normal line is \( -(x + 2)/3 = -(y + 3)/2 = z - 6 \).

Step-by-step solution

Calculate gradient of the given surface equation

The gradient vector \(\nabla f(x, y, z)\) of a function \(f(x, y, z)\) is \(\nabla f(x, y, z) = (f_x, f_y, f_z)\). Based on the given surface equation, \(x*y - z = 0\), it can be rewritten as \(f(x, y, z) = xy - z\). The gradient would hence be \(\nabla f(x, y, z) = (f_x, f_y, f_z) = (y, x, -1)\)

Find the normal vector

Substitute the point (-2,-3,6) into the gradient equation to get the normal vector. The normal vector is hence \(N = (f_x(-2, -3, 6), f_y(-2, -3, 6), f_z(-2, -3, 6)) = (-3, -2, -1)\)

Calculate the equation of the tangent plane

The equation of the tangent plane at point P(x_0, y_0, z_0) of the surface z = f(x, y) is given by: (f_x(x_0, y_0) * (x - x_0) + f_y(x_0, y_0) * (y - y_0) = z - z_0). Substituting the normal vector and the given point, we get \[ -3(x - (-2)) - 2(y -(-3)) = z - 6 \], which simplifies to \[-3x - 6 - 2y -(-6) = z - 6\]. Hence, the equation of the tangent plane is \( -3x - 2y + 12 = z \]

Calculate the symmetric equations of the normal line

The symmetric equations of the line is given by \[(x - x_0)/a = (y - y_0)/b = (z - z_0)/c\], where (a, b, c) is the normal vector and (x_0, y_0, z_0) is the given point. Substituting respectively, we get \[(x - (-2))/(-3) = (y - (-3))/(-2) = (z - 6)/(-1)\], which simplifies to \[-(x + 2)/3 = -(y + 3)/2 = z - 6\]