<p>If the line \(\dfrac{x-2}{3} = \dfrac{y+1}{2} = \dfrac{z-1}{-1}\) intersects the plane \(2x + 3y - z + 13 = 0\) at a point \(P\) and the plane \(3x + y + 4z = 16\) at a point \(Q\), then \(PQ\) is equal to ______ (up to three decimal places).</p>
Step-by-Step Solution
Key Concept: Parametrize the line using the parameter t, find intersection points P and Q by substituting into each plane equation, then calculate the distance between these two points using the distance formula.
Step 1: Write the parametric form of the line. From $\frac{x-2}{3} = \frac{y+1}{2} = \frac{z-1}{-1} = t$, we get: $x = 2 + 3t$, $y = -1 + 2t$, $z = 1 - t$ Step 2: Find point P (intersection with plane $2x + 3y - z + 13 = 0$): Substitute: $2(2+3t) + 3(-1+2t) - (1-t) + 13 = 0$ $4 + 6t - 3 + 6t - 1 + t + 13 = 0$ $13t + 13 = 0 \Rightarrow t = -1$ So $P = (2-3, -1-2, 1+1) = (-1, -3, 2)$ Step 3: Find point Q (intersection with plane $3x + y + 4z = 16$): Substitute: $3(2+3t) + (-1+2t) + 4(1-t) = 16$ $6 + 9t - 1 + 2t + 4 - 4t = 16$ $7t + 9 = 16 \Rightarrow t = 1$ So $Q = (2+3, -1+2, 1-1) = (5, 1, 0)$ Step 4: Calculate distance PQ: $PQ = \sqrt{(5-(-1))^2 + (1-(-3))^2 + (0-2)^2}$ $PQ = \sqrt{36 + 16 + 4} = \sqrt{56} = 2\sqrt{14}$ Alternative (faster): Since both points lie on the line with direction ratios $(3, 2, -1)$ and correspond to parameters $t_1 = -1$ and $t_2 = 1$: $PQ = |t_2 - t_1| \cdot \sqrt{3^2 + 2^2 + (-1)^2} = |1-(-1)| \cdot \sqrt{14} = 2\sqrt{14}$ ∴ Answer: 7.483
Correct Answer: 7