Question

Find an equation of the tangent plane and find symmetric equations of the normal line to the surface at the given point. $$ x y z=10, \quad(1,2,5) $$

Short answer

The equation of the tangent plane is \(10(x - 1) + 5(y - 2) + 2(z - 5) = 0\). The symmetric equations of the normal line are \( x = 10t + 1 \), \( y = 5t + 2 \) , \( z = 2t + 5 \).

Step-by-step solution

Calculate the Gradient Vector

Find the partial derivatives of \(f(x, y, z) = xyz -10\) with respect to \(x\), \(y\), and \(z\). \(f_x = yz\), \(f_y = xz\), \(f_z = xy\).Plug in the given point \((1, 2, 5)\) to get the gradient vector, which is also the normal vector to the surface at the point. So \(f_x(1,2,5) = 2*5 = 10\), \(f_y(1,2,5) = 1*5 = 5\), \(f_z(1,2,5) = 1*2 = 2\). Therefore, the normal vector is \(<10, 5, 2>\).

Find the Equation of the Tangent Plane

The equation of a plane is given by \(A(x−x_0) + B(y−y_0) + C(z−z_0) = 0\) where \(A,B, C \) are the components of the normal vector and \( x_0, y_0, z_0 \) are the coordinates of the given point. For our problem, A=10, B=5, C=2 and \(x_0 = 1\), \(y_0 = 2\), \(z_0 = 5\). So the equation of the tangent plane becomes: \(10(x - 1) + 5(y - 2) + 2(z - 5) = 0\).

Determine the Symmetric Equations of the Normal Line

The symmetric equations of the line passing through a point \((x_0, y_0, z_0) \) and parallel to a vector \(A, B, C\) are \(x = A + x_0t, y = B + y_0t, z = C + z_0t \). For our problem, A=10, B=5, C=2 and \(x_0 =1\), \(y_0 = 2\), \(z_0 = 5\). Therefore, the equations of the normal line are: \( x = 10t + 1 \), \( y = 5t + 2 \), \( z = 2t + 5 \).