To find the parametric equations of the line through the point $(0, 1, 2)$ that is:

1. Parallel to the plane $x + y + z = 5$
2. Perpendicular to the line $x = 1 + t, y = 1 - t, z = 2t$,

we need to determine the direction vector of the line.

Step 1: Find the normal vector to the plane.

The equation of the plane is $x + y + z = 5$, so the normal vector to the plane is given by the coefficients of $x$, $y$, and $z$ in the equation. Therefore, the normal vector to the plane is:

$\vec{n}_\text{plane} = (1, 1, 1)$

Step 2: Find the direction vector of the given line.

The parametric equations of the given line are:

$x = 1 + t, \quad y = 1 - t, \quad z = 2t$

From these equations, we can extract the direction vector of the given line:

$\vec{d}_\text{line} = (1, -1, 2)$

Step 3: Find the direction vector of the new line.

We are asked to find a line that is parallel to the plane and perpendicular to the given line. Therefore, the direction vector of the new line must be perpendicular to both the normal vector of the plane and the direction vector of the given line.

To find a vector that is perpendicular to both $\vec{n}_\text{plane} = (1, 1, 1)$ and $\vec{d}_\text{line} = (1, -1, 2)$, we compute the cross product:

$\vec{v} = \vec{n}_\text{plane} \times \vec{d}_\text{line}$

Calculating the cross product:

\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -1 & 2 \end{vmatrix}$$ $$\vec{v} = \hat{i} \left( \begin{vmatrix} 1 & 1 \\ -1 & 2 \end{vmatrix} \right) - \hat{j} \left( \begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} \right) + \hat{k} \left( \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} \right)$$ $$= \hat{i} \left( (1)(2) - (1)(-1) \right) - \hat{j} \left( (1)(2) - (1)(1) \right) + \hat{k} \left( (1)(-1) - (1)(1) \right)$$ $$= \hat{i} (2 + 1) - \hat{j} (2 - 1) + \hat{k} (-1 - 1)$$ $$= \hat{i} (3) - \hat{j} (1) + \hat{k} (-2)$$ $$\vec{v} = (3, -1, -2)$$ ### Step 4: Parametric equations of the new line. Now that we have the direction vector $$\vec{v} = (3, -1, -2)$$, and the point $$(0, 1, 2)$$, the parametric equations of the line can be written as: $$x(t) = 0 + 3t = 3t$$ $$y(t) = 1 - t$$ $$z(t) = 2 - 2t$$ Thus, the parametric equations of the line are: $$x(t) = 3t, \quad y(t) = 1 - t, \quad z(t) = 2 - 2t$$ --- Would you like a more detailed explanation of any step, or do you have any further questions? ### Related Questions: 1. How do we calculate the cross product of two vectors? 2. What does it mean for a vector to be perpendicular to two other vectors? 3. How do parametric equations relate to the geometry of a line in 3D space? 4. How would the solution change if the line was required to be parallel to another vector? 5. What are the conditions for a line to be parallel to a plane? ### Tip: When working with planes and lines in 3D, remember that the cross product of two vectors gives you a new vector that is perpendicular to both, which is very useful for finding direction vectors that meet specific geometric conditions.