To find the equation of a plane through a given point $P(2, -3, 2)$ that also contains the line given by the parametric equation $\vec{r}(t) = \langle 2 + 3t, t, -5 - 3t \rangle$, follow these steps:

Step 1: Determine two points on the line.

The parametric equation $\vec{r}(t) = \langle 2 + 3t, t, -5 - 3t \rangle$ gives points on the line for different values of $t$. Let's calculate two points by choosing different values of $t$.

• For $t = 0$: $\vec{r}(0) = \langle 2 + 3(0), 0, -5 - 3(0) \rangle = \langle 2, 0, -5 \rangle$ So, one point is $Q(2, 0, -5)$.

• For $t = 1$: $\vec{r}(1) = \langle 2 + 3(1), 1, -5 - 3(1) \rangle = \langle 5, 1, -8 \rangle$ So, another point is $R(5, 1, -8)$.

Step 2: Find two vectors in the plane.

Now we have three points: $P(2, -3, 2)$, $Q(2, 0, -5)$, and $R(5, 1, -8)$. These points define two vectors that lie in the plane.

• Vector $\overrightarrow{PQ} = Q - P = \langle 2 - 2, 0 - (-3), -5 - 2 \rangle = \langle 0, 3, -7 \rangle$.
• Vector $\overrightarrow{PR} = R - P = \langle 5 - 2, 1 - (-3), -8 - 2 \rangle = \langle 3, 4, -10 \rangle$.

Step 3: Compute the normal vector to the plane.

The normal vector to the plane can be found by taking the cross product of the two vectors $\overrightarrow{PQ}$ and $\overrightarrow{PR}$.

\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 3 & -7 \\ 3 & 4 & -10 \end{vmatrix}$$ The cross product is calculated as: $$\overrightarrow{n} = \hat{i} \left( (3)(-10) - (4)(-7) \right) - \hat{j} \left( (0)(-10) - (3)(-7) \right) + \hat{k} \left( (0)(4) - (3)(3) \right)$$ $$\overrightarrow{n} = \hat{i}(-30 + 28) - \hat{j}(0 + 21) + \hat{k}(0 - 9)$$ $$\overrightarrow{n} = \langle -2, -21, -9 \rangle$$ Thus, the normal vector is $$\langle -2, -21, -9 \rangle$$. ### Step 4: Write the equation of the plane. The equation of a plane is given by: $$a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$$ where $$\langle a, b, c \rangle$$ is the normal vector and $$(x_0, y_0, z_0)$$ is a point on the plane (we'll use $$P(2, -3, 2)$$). Substituting the normal vector $$\langle -2, -21, -9 \rangle$$ and point $$P(2, -3, 2)$$ into the equation: $$-2(x - 2) - 21(y + 3) - 9(z - 2) = 0$$ Simplifying: $$-2(x - 2) - 21(y + 3) - 9(z - 2) = -2x + 4 - 21y - 63 - 9z + 18 = 0$$ $$-2x - 21y - 9z - 41 = 0$$ Thus, the equation of the plane is: $$2x + 21y + 9z = -41$$ Let me know if you need further clarification on any step or more details! --- Here are 5 related questions: 1. How do you compute a normal vector using the cross product of two vectors? 2. What is the geometric interpretation of the normal vector in the context of planes? 3. How do you find parametric equations for lines in 3D? 4. How does the cross product differ from the dot product in vector algebra? 5. How do you verify that a point lies on a plane? **Tip**: When taking the cross product, always ensure that you set up the determinant correctly to avoid sign errors.