1. Key Concepts and Formulas

   • Equation of a Line and General Point: A line passing through $(x_0, y_0, z_0)$ with direction vector $\vec{d} = (a, b, c)$ can be represented as $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c} = \lambda$. A general point on this line is $(x_0 + a\lambda, y_0 + b\lambda, z_0 + c\lambda)$.
   • Foot of Perpendicular: If N is the foot of the perpendicular from point P to a line L, then the vector $\vec{PN}$ is perpendicular to the direction vector of line L. Their dot product is zero: $\vec{PN} \cdot \vec{d}_L = 0$.
   • Line Parallel to a Plane: If a line is parallel to a plane, its direction vector is perpendicular to the normal vector of the plane. Their dot product is zero.
   • Angle between Two Vectors: The cosine of the angle $\theta$ between two vectors $\vec{a}$ and $\vec{b}$ is given by $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{||\vec{a}|| \cdot ||\vec{b}||}$. For the acute angle, we take the absolute value of the dot product in the numerator.

2. Step-by-Step Solution

   Step 1: Determine the coordinates of N, the foot of the perpendicular from P to L.

   • What we are doing: We are finding the specific point N on line L such that the line segment PN is perpendicular to L.
   • Why: N is a crucial point for defining the vector $\vec{PN}$, which is one of the lines for which we need to find the angle.
   • Math: The given line L is $\frac{x}{1} = \frac{y}{0} = \frac{z}{-1}$. This line passes through the origin (0, 0, 0) and has a direction vector $\vec{d}_L = (1, 0, -1)$. Any general point N on L can be represented as $(t, 0, -t)$ for some scalar $t$. The coordinates of point P are (1, 2, -1). The vector $\vec{PN}$ is given by $N - P = (t-1, 0-2, -t - (-1)) = (t-1, -2, -t+1)$. Since PN is perpendicular to L, their direction vectors must be orthogonal. Therefore, their dot product is zero: $\vec{PN} \cdot \vec{d}_L = 0$ $(t-1)(1) + (-2)(0) + (-t+1)(-1) = 0$ $t-1 + 0 + t-1 = 0$ $2t - 2 = 0$ $2t = 2$ $t = 1$ Substitute $t=1$ back into the general point N to find its coordinates: $N = (1, 0, -1)$ Now, we can find the vector $\vec{PN}$: $\vec{PN} = N - P = (1-1, 0-2, -1-(-1)) = (0, -2, 0)$ And its magnitude: $||\vec{PN}|| = \sqrt{0^2 + (-2)^2 + 0^2} = \sqrt{4} = 2$

   Step 2: Determine the coordinates of Q, the intersection point of line PQ with L.

   • What we are doing: We are finding the point Q on line L such that the line segment PQ is parallel to the given plane.
   • Why: Q is the other crucial point for defining the vector $\vec{PQ}$, the second line for which we need to find the angle.
   • Math: Line PQ passes through P(1, 2, -1). Point Q lies on line L, so its coordinates can be written as $(k, 0, -k)$ for some scalar $k$. The vector $\vec{PQ}$ is given by $Q - P = (k-1, 0-2, -k - (-1)) = (k-1, -2, -k+1)$. The line PQ is parallel to the plane $x + y + 2z = 0$. The normal vector to this plane is $\vec{n} = (1, 1, 2)$. Since line PQ is parallel to the plane, its direction vector $\vec{PQ}$ must be perpendicular to the normal vector of the plane. Therefore, their dot product is zero: $\vec{PQ} \cdot \vec{n} = 0$ $(k-1)(1) + (-2)(1) + (-k+1)(2) = 0$ $k-1 - 2 - 2k + 2 = 0$ $-k - 1 = 0$ $-k = 1$ $k = -1$ Substitute $k=-1$ back into the general point Q to find its coordinates: $Q = (-1, 0, -(-1)) = (-1, 0, 1)$ Now, we can find the vector $\vec{PQ}$: $\vec{PQ} = Q - P = (-1-1, 0-2, 1-(-1)) = (-2, -2, 2)$ And its magnitude: $||\vec{PQ}|| = \sqrt{(-2)^2 + (-2)^2 + 2^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3}$

   Step 3: Calculate the acute angle $\alpha$ between $\vec{PN}$ and $\vec{PQ}$.

   • What we are doing: We are applying the dot product formula to find the cosine of the angle.
   • Why: This is the final step to answer the question.
   • Math: We have $\vec{PN} = (0, -2, 0)$ and $\vec{PQ} = (-2, -2, 2)$. First, calculate the dot product $\vec{PN} \cdot \vec{PQ}$: $\vec{PN} \cdot \vec{PQ} = (0)(-2) + (-2)(-2) + (0)(2) = 0 + 4 + 0 = 4$ Next, use the formula for $\cos\alpha$: $\cos\alpha = \frac{|\vec{PN} \cdot \vec{PQ}|}{||\vec{PN}|| \cdot ||\vec{PQ}||}$ Since $\vec{PN} \cdot \vec{PQ} = 4 > 0$, the angle is already acute, so the absolute value doesn't change the result. $\cos\alpha = \frac{4}{2 \cdot 2\sqrt{3}} = \frac{4}{4\sqrt{3}} = \frac{1}{\sqrt{3}}$

3. Common Mistakes & Tips

   • Incorrect Direction Vector: Ensure the direction vector of the line L is correctly identified, especially with $y/0$ terms (which means the y-component is 0).
   • Dot Product for Perpendicularity/Parallelism: Remember that a line perpendicular to another line (or a plane) means their direction vectors (or direction vector and normal vector) have a dot product of zero. A line parallel to a plane means its direction vector is perpendicular to the plane's normal vector.
   • Magnitude Calculation: Double-check the calculation of vector magnitudes to avoid arithmetic errors that propagate to the final angle.

4. Summary

   We first determined the coordinates of N by finding the foot of the perpendicular from P to line L, utilizing the orthogonality of $\vec{PN}$ and the direction vector of L. Next, we found the coordinates of Q by using the condition that the line PQ is parallel to the given plane, implying that $\vec{PQ}$ is orthogonal to the plane's normal vector. Finally, with vectors $\vec{PN}$ and $\vec{PQ}$ established, we used the dot product formula to calculate the cosine of the acute angle between them. Our calculations yielded $\cos\alpha = \frac{1}{\sqrt{3}}$.

5. Final Answer

The final answer is $\boxed{\frac{1}{\sqrt{3}}}$.