To find the equation of a plane that contains the line and is perpendicular to another plane, we need to follow a systematic process.

Given:

1. The parametric form of the line is: $\frac{x + 2}{3} = t, \quad y = 4 + 2t, \quad z = 3 - t$ or equivalently: $x = -2 + 3t, \quad y = 4 + 2t, \quad z = 3 - t$

2. The plane $\pi$ has the equation: $x - 2y + z = 5$

We need to find the equation of the plane that contains the given line and is perpendicular to $\pi$.

Step 1: Find the direction vector of the line

From the parametric equations of the line: $x = -2 + 3t, \quad y = 4 + 2t, \quad z = 3 - t$ The direction vector $\vec{d}$ of the line is: $\vec{d} = \langle 3, 2, -1 \rangle$

Step 2: Find the normal vector of the given plane $\pi$

The equation of the plane is $x - 2y + z = 5$, so its normal vector is: $\vec{n}_{\pi} = \langle 1, -2, 1 \rangle$

Step 3: The plane we are finding is perpendicular to $\pi$

Since the desired plane is perpendicular to $\pi$, its normal vector $\vec{n}_{\text{plane}}$ must lie in the direction of both the normal vector $\vec{n}_{\pi}$ of $\pi$ and the direction vector $\vec{d}$ of the line.

The normal vector of the required plane can be found by taking the cross product of $\vec{d}$ and $\vec{n}_{\pi}$: $\vec{n}_{\text{plane}} = \vec{d} \times \vec{n}_{\pi}$ $\vec{n}_{\text{plane}} = \langle 3, 2, -1 \rangle \times \langle 1, -2, 1 \rangle$

Step 4: Compute the cross product

We use the determinant formula for the cross product:

\hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & -1 \\ 1 & -2 & 1 \end{vmatrix}$$ $$\vec{n}_{\text{plane}} = \hat{i}((2)(1) - (-1)(-2)) - \hat{j}((3)(1) - (-1)(1)) + \hat{k}((3)(-2) - (2)(1))$$ $$\vec{n}_{\text{plane}} = \hat{i}(2 - 2) - \hat{j}(3 + 1) + \hat{k}(-6 - 2)$$ $$\vec{n}_{\text{plane}} = \hat{i}(0) - \hat{j}(4) + \hat{k}(-8)$$ $$\vec{n}_{\text{plane}} = \langle 0, -4, -8 \rangle$$ Thus, the normal vector of the required plane is $$\langle 0, -4, -8 \rangle$$, or equivalently $$\langle 0, 1, 2 \rangle$$ after dividing by $$-4$$. ### Step 5: Find a point on the plane We can use a point from the line. Let $$t = 0$$, then the point on the line is: $$P( x = -2, y = 4, z = 3 )$$ ### Step 6: Write the equation of the plane The equation of a plane with normal vector $$\langle A, B, C \rangle$$ passing through a point $$P(x_0, y_0, z_0)$$ is: $$A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$$ Substituting $$A = 0$$, $$B = 1$$, $$C = 2$$, and the point $$P(-2, 4, 3)$$: $$0(x + 2) + 1(y - 4) + 2(z - 3) = 0$$ $$y - 4 + 2(z - 3) = 0$$ $$y - 4 + 2z - 6 = 0$$ $$y + 2z = 10$$ ### Final Answer: The equation of the plane is: $$y + 2z = 10$$ Would you like more details or have any questions? ### Related questions: 1. How do you find the equation of a plane using three points? 2. What is the significance of the cross product in geometry? 3. How do you verify if a line lies in a plane? 4. What are other ways to define a plane besides normal vectors? 5. Can two planes be both perpendicular and parallel to the same line? ### Tip: The cross product of two vectors gives a vector perpendicular to both, which is very useful in geometry, especially for finding normals to planes.