<p>The image of the point \((-1, 3, 4)\) in the plane \(x - 2y = 0\) is</p>
<p>\(\left(-\dfrac{17}{3}, -\dfrac{19}{3}, 4\right)\)</p>
<p>\((15, 11, 4)\)</p>
<p>\(\left(-\dfrac{17}{3}, -\dfrac{19}{3}, 1\right)\)</p>
<p>\((8, 4, 4)\)</p>
Step-by-Step Solution
Key Concept: To find the image of a point in a plane, drop a perpendicular from the point to the plane, find the foot of perpendicular, then reflect the point across this foot using the property that the foot is the midpoint of the point and its image.
Step 1: Identify the plane equation: x - 2y = 0. The normal vector is n = (1, -2, 0). Step 2: Write the line through P(-1, 3, 4) perpendicular to the plane:
(x, y, z) = (-1, 3, 4) + t(1, -2, 0) = (-1+t, 3-2t, 4) Step 3: Find foot of perpendicular F by substituting into plane equation:
(-1+t) - 2(3-2t) = 0
-1 + t - 6 + 4t = 0
5t = 7 ⟹ t = 7/5 Step 4: Foot of perpendicular: F = (-1 + 7/5, 3 - 14/5, 4) = (2/5, 1/5, 4) Step 5: Since F is the midpoint of P and its image P', use: F = (P + P')/2
P' = 2F - P = 2(2/5, 1/5, 4) - (-1, 3, 4)
P' = (4/5, 2/5, 8) - (-1, 3, 4)
P' = (4/5 + 1, 2/5 - 3, 8 - 4)
P' = (9/5, -13/5, 4) Verification: Midpoint of P(-1, 3, 4) and P'(9/5, -13/5, 4) is (2/5, 1/5, 4) ✓ which lies on plane. ∴ Answer: A is (9/5, -13/5, 4)
Correct Answer: A