--- title: "Image of a point in a plane" subject: "Mathematics" chapter: 3 key: "semester_2/mathematics:image-point-plane" source: https://www.bctengineer.com/q/semester_2/mathematics/image-point-plane site: BCT Engineering --- # Image of a point in a plane Mathematics · Chapter 3: Three Dimensional Solid Geometry - Asked in: 2069 Bhadra, 5 marks - Video: [How to find image of a point in a plane | Kamaldheeriya](https://www.youtube.com/watch?v=a2vwF5wynRQ) — Kamaldheeriya Maths easy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/mathematics/image-point-plane - Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-3 ## Question Find the image of the point (2,−1,3) in the plane 3x−2y−z−9 = 0. ## Short answer Substituting a point on the normal line into the plane gives t=2/7 for the foot midpoint; doubling t gives the image (26/7, -15/7, 17/7). ## Answer The image of a point in a plane is the reflection of the point in that plane. It lies on the normal through the point, at the same distance on the other side, and the foot of the perpendicular is the mid-point of the point and its image. Let P(2, −1, 3) and the plane be 3x − 2y − z − 9 = 0. Its normal direction is (3, −2, −1). #### Step 1: Normal line through P Any point of the normal through P is (2 + 3t, −1 − 2t, 3 − t). #### Step 2: Foot of the perpendicular Put it in the plane: ```text 3(2+3t) − 2(−1−2t) − (3−t) − 9 = 0 6 + 9t + 2 + 4t − 3 + t − 9 = 0 14t − 4 = 0 → t = 2/7 ``` Foot F = (2 + 6/7, −1 − 4/7, 3 − 2/7) = (20/7, −11/7, 19/7). Check: 3(20/7) − 2(−11/7) − 19/7 = (60 + 22 − 19)/7 = 9, so F is on the plane. #### Step 3: Image F is the mid-point of P and its image Q, so the image is given by t = 4/7 on the normal: ```text Q = (2 + 12/7, −1 − 8/7, 3 − 4/7) = (26/7, −15/7, 17/7) ``` Check: the mid-point of P and Q is ((2+26/7)/2, (−1−15/7)/2, (3+17/7)/2) = (20/7, −11/7, 19/7) = F. **Answer: the image is (26/7, −15/7, 17/7).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Where does the image Q of P in a plane lie? A. On the line from P to the origin B. On the normal through P, with the foot of the perpendicular as the midpoint of PQ C. On the plane itself D. On the line through P parallel to the plane Answer: B. Reflection keeps PQ perpendicular to the plane and bisected by it. (Key point: Geometry of image) 2. With the normal point (2+3t, -1-2t, 3-t) and plane 3x-2y-z-9 = 0, what is t at the foot? A. t = 2/7 B. t = 4/7 C. t = 1 D. t = 2 Answer: A. 14t - 4 = 0 gives t = 2/7. (Key point: Foot parameter) 3. Given the foot at t = 2/7, which parameter gives the image? A. t = 2/7 B. t = -2/7 C. t = 4/7 D. t = 1/7 Answer: C. The foot is the midpoint, so the image is at twice the parameter. (Key point: From foot to image) 4. What is the image? A. (20/7, -11/7, 19/7) B. (26/7, -15/7, 17/7) C. (26/7, 15/7, 17/7) D. (2/7, -1/7, 3/7) Answer: B. Substituting t = 4/7 gives (26/7, -15/7, 17/7). 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