91Ó°ÊÓ

To find a point on the plane, we can evaluate \(\mathbf{r}(t)\) for an arbitrary value of \(t\). Let's use \(t=0\). Then, \(\mathbf{r}(0) = 0 \mathbf{i} + 0^{2} \mathbf{k} = \mathbf{0}\), so the point \((0,0,0)\) lies on the plane.

In order to find a normal vector to the plane, we can take the cross product of two vectors that are tangent to the curve. One such tangent vector is the derivative of \(\mathbf{r}(t)\) with respect to \(t\), which is denoted as \(\mathbf{r'}(t)\). Let's calculate \(\mathbf{r'}(t)\): \[\mathbf{r'}(t) = \frac{d}{dt}(t \mathbf{i} + t^{2} \mathbf{k}) = \mathbf{i} + 2t\mathbf{k}.\] Now, we need another vector tangent to the curve in order to compute the cross product. We can obtain this by taking the derivative of \(\mathbf{r'}(t)\) with respect to \(t\), denoted as \(\mathbf{r''}(t)\). Let's calculate \(\mathbf{r''}(t)\): \[\mathbf{r''}(t) = \frac{d}{dt}(\mathbf{i} + 2t\mathbf{k}) = 2\mathbf{k}.\] Now we will find the cross product of \(\mathbf{r'}(t)\) and \(\mathbf{r''}(t)\): \[\mathbf{n} = \mathbf{r'}(t) \times \mathbf{r''}(t) = (\mathbf{i} + 2t\mathbf{k}) \times 2\mathbf{k} = 2\mathbf{i} \times 2\mathbf{k} = 4\mathbf{j}.\] Since the cross product is independent of \(t\), the normal vector is a constant vector and equal to \(4\mathbf{j}\).

Now that we have a point on the plane, \((0,0,0)\), and a normal vector, \(4\mathbf{j}\), we can write the equation of the plane using the point-normal form: The point-normal form of the plane is given by: \[\mathbf{n} \cdot (\mathbf{r} - \mathbf{r_0}) = 0,\] where \(\mathbf{n}\) is the normal vector, \(\mathbf{r}\) is a point on the plane, and \(\mathbf{r_0}\) is a fixed point on the plane. In this case, \(\mathbf{n} = 4\mathbf{j}\), \(\mathbf{r} = (x,y,z)\), and \(\mathbf{r_0} = (0,0,0)\). Plugging these values into the equation, we obtain: \(4\mathbf{j} \cdot (x\mathbf{i} + y\mathbf{j} + z\mathbf{k} - 0\mathbf{i} - 0\mathbf{j} - 0\mathbf{k}) = 0\). This simplifies to: \(4y=0\), which is equivalent to \(y=0\). So, the curve lies in the plane \(y = 0\).