1. **State the problem:** Find the normal vector and the equation of the tangent plane to the surface defined by the plane $z = x + 3$ inside the cylinder $x^2 + y^2 = 1$ at the point $P = (1, 0, 4)$. 2. **Understand the surface:** The surface is given by $z = x + 3$. This can be rewritten as $F(x,y,z) = z - x - 3 = 0$. 3. **Find the gradient (normal vector):** The normal vector to the surface at any point is given by the gradient of $F$, i.e., $$\nabla F = \left(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}\right)$$ Calculate each partial derivative: $$\frac{\partial F}{\partial x} = -1, \quad \frac{\partial F}{\partial y} = 0, \quad \frac{\partial F}{\partial z} = 1$$ So, $$\nabla F = (-1, 0, 1)$$ 4. **Evaluate the normal vector at point $P$:** Since the gradient is constant, the normal vector at $P$ is $$\mathbf{n} = (-1, 0, 1)$$ 5. **Equation of the tangent plane:** The tangent plane at point $P = (x_0, y_0, z_0)$ with normal vector $\mathbf{n} = (a, b, c)$ is given by $$a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$$ Substitute $a = -1$, $b = 0$, $c = 1$, and $P = (1, 0, 4)$: $$-1(x - 1) + 0(y - 0) + 1(z - 4) = 0$$ Simplify: $$-x + 1 + z - 4 = 0 \implies -x + z - 3 = 0$$ Or equivalently, $$z - x - 3 = 0$$ 6. **Check if $P$ lies inside the cylinder:** The cylinder is defined by $$x^2 + y^2 = 1$$ At $P = (1, 0, 4)$, $$1^2 + 0^2 = 1$$ So $P$ lies on the boundary of the cylinder. **Final answers:** - Normal vector at $P$ is $\boxed{(-1, 0, 1)}$. - Equation of the tangent plane at $P$ is $\boxed{z - x - 3 = 0}$.