<p>The line passing through the points (5, 1, <em>a</em>) and (3, <em>b</em>, 1) crosses the <em>yz</em>-plane at the point \(\left(0, \dfrac{17}{2}, \dfrac{-13}{2}\right)\). Then</p>
Step-by-Step Solution
Key Concept: A line crosses the yz-plane when x = 0. Use the parametric form of the line through two points and set x-coordinate to 0 to find the parameter, then use this to find the crossing point and match it with the given coordinates to determine a and b.
Step 1: The parametric equation of the line through (5, 1, a) and (3, b, 1) is: \(\vec{r}(t) = (5, 1, a) + t[(3, b, 1) - (5, 1, a)] = (5, 1, a) + t(-2, b-1, 1-a)\) Step 2: For the yz-plane, x = 0: \(5 + t(-2) = 0 \implies t = \frac{5}{2}\) Step 3: The y-coordinate at this point: \(y = 1 + \frac{5}{2}(b-1) = \frac{17}{2}\) \(1 + \frac{5(b-1)}{2} = \frac{17}{2}\) \(\frac{5(b-1)}{2} = \frac{15}{2}\) \(b - 1 = 3 \implies b = 4\) Step 4: The z-coordinate at this point: \(z = a + \frac{5}{2}(1-a) = -\frac{13}{2}\) \(a + \frac{5(1-a)}{2} = -\frac{13}{2}\) \(\frac{2a + 5 - 5a}{2} = -\frac{13}{2}\) \(5 - 3a = -13 \implies 3a = 18 \implies a = 6\) ∴ Answer: C (with a = 6, b = 4)
Correct Answer: C