Question

Find equations of (a) the tangent plane and (b) the normal line to the given surface at the specified point.

\(xy + yz + zx = 5,\;\;\;(1,2,1)\)

Short answer

1. The equation of the tangent plane to the surface \(xy + yz + zx = 5\) at the point \((1,2,1)\) is \(3x + 2y + 3z = 10\).
2. The equation of normal line to the surface \(xy + yz + zx = 5\) at the point \((1,2,1)\) is \(\frac{{(x - 1)}}{3} = \frac{{(y - 2)}}{2} = \frac{{(z - 1)}}{3}\).

Step-by-step solution

Step 1:

Result used:

"The tangent plane to the level surface at the point\(P\left( {{x_0},{y_0},{z_0}} \right)\)is defined as\({F_x}\left( {{x_0},{y_0},{z_0}} \right)\left( {x - {x_0}} \right) + {F_y}\left( {{x_0},{y_0},{z_0}} \right)\left( {y - {y_0}} \right) + {F_z}\left( {{x_0},{y_0},{z_0}} \right)\left( {z - {z_0}} \right) = 0{\rm{ }}\)

equation of the tangent plane

The given surface is, \(xy + yz + zx = 5\).

Let the surface function be, \(F(x,y,z) = xy + yz + zx - 5 \cdot (1)\)

The equation of the tangent plane to the given surface at the point \((1,2,1)\) is defined by,

\({F_x}(1,2,1)(x - 1) + {F_y}(1,2,1)(y - 2) + {F_z}(1,2,1)(z - 1) = 0.(2)\)

Take partial derivative with respect to \(x\)

Take partial derivative with respect to \(x\) at the point \((1,2,1)\) in the equation (1),

\(\begin{aligned}{l}{F_x}(x,y,z) = \frac{\partial }{{\partial x}}(xy + yz + zx - 5)\\ = (1)y + 0 + z(1) - 0\\ = y + z\end{aligned}\)

The value of \({F_x}(x,y,z)\) at the point \((1,2,1)\) is,

\(\begin{aligned}{l}{F_x}(1,2,1) = 2 + 1\\{F_x}(1,2,1) = 3\end{aligned}\)

Thus, the value of \({F_x}(1,2,1) = 3\).

Take partial derivative with respect to \(y\)

Take partial derivative with respect to \(y\) at the point \((1,2,1)\) in the equation (1),

\(\begin{aligned}{l}{F_y}(x,y,z) = \frac{\partial }{{\partial y}}(xy + yz + zx - 5)\\ = x(1) + (1)z + 0 - 0\\{F_y}(x,y,z) = x + z\end{aligned}\)

Take partial derivative with respect to \(z\)

Take partial derivative with respect to at the point \((1,2,1)\) in the equation (1),

\(\begin{aligned}{l}{F_z}(x,y,z) = \frac{\partial }{{\partial z}}(xy + yz + zx - 5)\\ = 0 + y(1) + (1)x\\ = x + y\end{aligned}\)

The value of \({F_z}(x,y,z)\) at the point \((1,2,1)\) is,

\(\begin{aligned}{l}{F_z}(1,2,1) = 1 + 2\\{F_z}(1,2,1) = 3\end{aligned}\)

Thus, the value of \({F_z}(1,2,1) = 3\).

 Step 6: substitute respective values in the equation (2)

Substitute the respective values in the equation (2) and obtain,

\(\begin{aligned}{l}3(x - 1) + 2(y - 2) + 3(z - 1) = 0\\3x - 3 + 2y - 4 + 3z - 3 = 0\\3x + 2y + 3z - 10 = 0\\3x + 2y + 3z = 10\end{aligned}\)

Thus, the equation of the tangent plane to the surface \(xy + yz + zx = 5\) at the point \((1,2,1)\) is \(3x + 2y + 3z = 10\).

(b) Step 7: normal to the level surface

Result used:

"The normal to the level surface at the point \(P\left( {{x_0},{y_0},{z_0}} \right)\) is defined as \(\frac{{\left( {x - {x_0}} \right)}}{{{F_x}\left( {{x_0},{y_0},{z_0}} \right)}} = \frac{{\left( {y - {y_0}} \right)}}{{{F_y}\left( {{x_0},{y_0},{z_0}} \right)}} = \frac{{\left( {z - {z_0}} \right)}}{{{F_z}\left( {{x_0},{y_0},{z_0}} \right)}}\) ".

calculate equation of normal line to the surface

The given surface function is, \(xy + yz + zx = 5\).

Let the surface function be, \(F(x,y,z) = xy + yz + zx - 5\).

The normal line is defined as, \(\frac{{(x - 1)}}{{{F_x}(1,2,1)}} = \frac{{(y - 2)}}{{{F_y}(1,2,1)}} = \frac{{(z - 1)}}{{{F_z}(1,2,1)}}............(3)\)

From part (a), the values of \({F_x}(1,2,1) = 3,{F_y}(1,2,1) = 2\) and \({F_z}(1,2,1) = 3\).

Substitute the respective values in the equation (3) and obtain the equation of normal line.

\(\frac{{(x - 1)}}{3} = \frac{{(y - 2)}}{2} = \frac{{(z - 1)}}{3}\).

Thus, the equation of normal line to the surface \(xy + yz + zx = 5\) at the point \((1,2,1)\) is \(\frac{{(x - 1)}}{3} = \frac{{(y - 2)}}{2} = \frac{{(z - 1)}}{3}\)..