Key Concepts and Formulas

This problem requires us to find two points of intersection of a line with two different planes and then calculate the distance between these two points. The essential concepts and formulas are:

1. Parametric Form of a Line: A line passing through a point $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ can be expressed in its parametric form as: $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = \lambda$ where $\lambda$ is a scalar parameter. Any point on the line can then be represented by its coordinates $(x_1 + a\lambda, y_1 + b\lambda, z_1 + c\lambda)$. This form is crucial for representing a general point on the line.

2. Intersection of a Line and a Plane: A point where a line intersects a plane must lie on both the line and the plane. To find this point, we substitute the parametric coordinates of a general point on the line into the equation of the plane. Solving the resulting linear equation for $\lambda$ gives the specific parameter value for the intersection point. Substituting this $\lambda$ back into the parametric form yields the coordinates of the intersection.

3. Distance Formula in 3D: The distance between two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ in three-dimensional space is given by: $PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

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Step-by-Step Solution

Step 1: Representing a General Point on the Line

Our first step is to express the coordinates of any arbitrary point on the given line using a single parameter. This is done by setting the given symmetric form of the line equal to $\lambda$.

The given line is: $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 1}{-1}$

Let's set each fraction equal to a parameter $\lambda$: $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 1}{-1} = \lambda$

Now, we can express $x, y, z$ in terms of $\lambda$:

• From $\frac{x - 2}{3} = \lambda \Rightarrow x - 2 = 3\lambda \Rightarrow x = 3\lambda + 2$
• From $\frac{y + 1}{2} = \lambda \Rightarrow y + 1 = 2\lambda \Rightarrow y = 2\lambda - 1$
• From $\frac{z - 1}{-1} = \lambda \Rightarrow z - 1 = -\lambda \Rightarrow z = -\lambda + 1$

So, any point on the line can be represented as $A(3\lambda + 2, 2\lambda - 1, -\lambda + 1)$. This parametric form is fundamental for finding the intersection points.

Step 2: Finding Point P (Intersection with the First Plane)

Point P is the intersection of the line with the plane $2x + 3y - z + 13 = 0$. Since P lies on both the line and the plane, its coordinates must satisfy both equations. We substitute the parametric coordinates of point A (from Step 1) into the equation of the first plane.

Substitute $(3\lambda + 2, 2\lambda - 1, -\lambda + 1)$ into $2x + 3y - z + 13 = 0$: $2(3\lambda + 2) + 3(2\lambda - 1) - (-\lambda + 1) + 13 = 0$

Now, we solve this linear equation for $\lambda$: $6\lambda + 4 + 6\lambda - 3 + \lambda - 1 + 13 = 0$

Combine the terms involving $\lambda$: $(6\lambda + 6\lambda + \lambda) = 13\lambda$ Combine the constant terms: $(4 - 3 - 1 + 13) = 13$

So, the equation simplifies to: $13\lambda + 13 = 0$ $13\lambda = -13$ $\lambda = -1$

Now, substitute $\lambda = -1$ back into the parametric form of the point to find the coordinates of P:

• $x_P = 3(-1) + 2 = -3 + 2 = -1$
• $y_P = 2(-1) - 1 = -2 - 1 = -3$
• $z_P = -(-1) + 1 = 1 + 1 = 2$

Thus, the coordinates of point P are $(-1, -3, 2)$.

Step 3: Finding Point Q (Intersection with the Second Plane)

Point Q is the intersection of the line with the plane $3x + y + 4z = 16$. We follow the same procedure as for point P: substitute the parametric coordinates of point A into the equation of the second plane.

Substitute $(3\lambda + 2, 2\lambda - 1, -\lambda + 1)$ into $3x + y + 4z = 16$: $3(3\lambda + 2) + (2\lambda - 1) + 4(-\lambda + 1) = 16$

Now, solve this equation for $\lambda$: $9\lambda + 6 + 2\lambda - 1 - 4\lambda + 4 = 16$

Combine the terms involving $\lambda$: $(9\lambda + 2\lambda - 4\lambda) = 7\lambda$ Combine the constant terms: $(6 - 1 + 4) = 9$

So, the equation simplifies to: $7\lambda + 9 = 16$ $7\lambda = 16 - 9$ $7\lambda = 7$ $\lambda = 1$

Substitute $\lambda = 1$ back into the parametric form of the point to find the coordinates of Q:

• $x_Q = 3(1) + 2 = 3 + 2 = 5$
• $y_Q = 2(1) - 1 = 2 - 1 = 1$
• $z_Q = -(1) + 1 = -1 + 1 = 0$

Thus, the coordinates of point Q are $(5, 1, 0)$.

Step 4: Calculating the Distance PQ

Now that we have the coordinates of both intersection points, $P(-1, -3, 2)$ and $Q(5, 1, 0)$, we can use the 3D distance formula to find the length of the segment PQ.

$PQ = \sqrt{(x_Q - x_P)^2 + (y_Q - y_P)^2 + (z_Q - z_P)^2}$ $PQ = \sqrt{(5 - (-1))^2 + (1 - (-3))^2 + (0 - 2)^2}$ $PQ = \sqrt{(5 + 1)^2 + (1 + 3)^2 + (-2)^2}$ $PQ = \sqrt{(6)^2 + (4)^2 + (-2)^2}$ $PQ = \sqrt{36 + 16 + 4}$ $PQ = \sqrt{56}$

To simplify the radical $\sqrt{56}$, we look for perfect square factors. We know that $56 = 4 \times 14$. $PQ = \sqrt{4 \times 14} = \sqrt{4} \times \sqrt{14} = 2\sqrt{14}$

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Common Mistakes & Tips

• Sign Errors are Critical: Be extremely cautious with negative signs, especially when distributing terms, combining like terms, and applying the distance formula. A simple sign error can lead to a completely different result. For example, $-(-\lambda+1)$ must be $\lambda-1$, not $-\lambda+1$.
• Algebraic Precision: Double-check your arithmetic when solving for $\lambda$ and calculating coordinates. Even minor calculation errors will propagate and lead to incorrect final answers.
• Correct Variable Substitution: Ensure that the parametric form of the line is correctly substituted into each plane equation. Keep track of which $\lambda$ value corresponds to which point.
• Radical Simplification: Always simplify square roots completely. Look for the largest perfect square factor to express the radical in its simplest form, matching the format of the options provided.

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Summary

This problem is a classic application of 3D geometry concepts. We systematically found the distance between two points, P and Q, which are defined as the intersection points of a given line with two different planes. The method involved first representing any point on the line using a parametric form. Then, by substituting these parametric coordinates into each plane's equation, we determined the specific parameter values for points P and Q, thus finding their exact coordinates. Finally, the standard 3D distance formula was applied to calculate the length of the segment PQ. This approach is highly effective for solving various problems involving lines and planes in 3D space.

The final answer is $\boxed{2\sqrt{14}}$, which corresponds to option (C).