Consider the plane that is parallel to the \( x z \)-plane and goes through the point \( \left(x_{0}, y_{0}, z_{0}\right)=(1,8,4) \). Which position vector is normal to the plane? \( \square[7,0,0] \) Use the normal position vector above and the point \( \left(x_{0}, y_{0}, z_{0}\right) \) to find the equation of the plane. Equation of plane: \( y= \) Ex: 3

The normal position vector for a plane parallel to the \( xz \)-plane would indeed point directly along the \( y \)-axis. Thus, the normal vector can be represented as \( \mathbf{n} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \), reflecting that the orientation of the plane does not change along the \( y \)-axis, while \( x \) and \( z \) can vary freely. Now, using the point \( (1, 8, 4) \) and the normal vector, the equation of the plane can be constructed. Since our normal vector indicates that the plane is horizontal and \( y \) is constant, the equation becomes \( y = 8 \). This means the plane runs parallel to the \( xz \)-plane at the height of \( 8 \) units along the \( y \)-direction!