Find an equation of the tangent plane to the surface represented by the vector-valued function at the given point. $$\overrightarrow { \mathbf { r } } ( u , v ) = ( 2 u + v ) \hat { \mathbf { i } } + ( u - v ) \hat { \mathbf { j } } + v \hat { \mathbf { k } } , ( 4 , - 4,4 )$$
A) $( x - 4 ) - 2 ( y + 4 ) - 3 ( z - 4 ) = 0$
B) $2 ( x - 4 ) + 2 ( y + 4 ) - 3 ( z - 4 ) = 0$
C) $z = 0$
D) $( x - 4 ) - 2 ( y + 4 ) - 3 ( z - 4 ) = 0$.
E) $2 ( x - 4 ) - 2 ( y + 4 ) + 1 ( z - 4 ) = 0$.

Question

Quizplus · Accepted Answer

The tangent plane to a surface at a point is perpendicular to the gradient of the surface at that point. The gradient of the surface is given by the vector$$\overrightarrow { \mathbf { r } }_u 	imes \overrightarrow { \mathbf { r } }_v = \left| \begin{array} {ccc} \hat { \mathbf { i } } & \hat { \mathbf { j } } & \hat { \mathbf { k } } \\ 2 & 1 & 0 \\ 1 & -1 & 1 \end{array} ight| = -2 \hat { \mathbf { i } } - 2 \hat { \mathbf { j } } + 3 \hat { \mathbf { k } }$$At the point (4, -4, 4), the gradient is$$\overrightarrow { \mathbf { r } }_u 	imes \overrightarrow { \mathbf { r } }_v = -2 \hat { \mathbf { i } } - 2 \hat { \mathbf { j } } + 3 \hat { \mathbf { k } }$$So the equation of the tangent plane is$$-2 ( x - 4 ) - 2 ( y + 4 ) + 3 ( z - 4 ) = 0$$

Find an Equation of the Tangent Plane to the Surface