--- title: "Coplanar lines, intersection point and plane" subject: "Mathematics" chapter: 3 key: "semester_2/mathematics:coplanar-lines-intersection-point-plane" source: https://www.bctengineer.com/q/semester_2/mathematics/coplanar-lines-intersection-point-plane site: BCT Engineering --- # Coplanar lines, intersection point and plane Mathematics · Chapter 3: Three Dimensional Solid Geometry - Asked in: 2070 Magh, 5 marks - Video: [Coplanar lines/Point of intersection and the plane containing the lines](https://www.youtube.com/watch?v=67E9CE3byQM) — PHYSICSLEARNING - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/mathematics/coplanar-lines-intersection-point-plane - Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-3 ## Question Show that the lines (x−5)/4 = (y−7)/4 = (z−3)/(−5) and (x−8)/7 = (y−4)/1 = (z−5)/3 are coplanar. Find their point of intersection and equation of plane in which they lie. ## Short answer Using (z+3)/(-5) for first line, lines through (5,7,-3) and (8,4,5) are coplanar since determinant is zero. Intersection point: (1,3,2). Plane: 17x-47y-24z+172=0. ## Answer **How I read the question.** As printed, the first line is (x − 5)/4 = (y − 7)/4 = (z − 3)/(−5). With this line the two lines are not coplanar (the determinant below is 144, not 0). The standard textbook version has (z + 3)/(−5), and the coplanarity then works exactly, so I solve that intended version: - L₁: (x − 5)/4 = (y − 7)/4 = (z + 3)/(−5), through A(5, 7, −3), direction d₁ = (4, 4, −5). - L₂: (x − 8)/7 = (y − 4)/1 = (z − 5)/3, through B(8, 4, 5), direction d₂ = (7, 1, 3). #### Coplanarity AB = (3, −3, 8). ```text | 3 −3 8 | | 4 4 −5 | = 3(12 + 5) + 3(12 + 35) + 8(4 − 28) | 7 1 3 | = 51 + 141 − 192 = 0 ``` The determinant is zero, so the lines are coplanar. (They are not parallel, since (4, 4, −5) is not proportional to (7, 1, 3).) #### Point of intersection L₁: (5 + 4t, 7 + 4t, −3 − 5t); L₂: (8 + 7s, 4 + s, 5 + 3s). ```text 5 + 4t = 8 + 7s → 4t − 7s = 3 7 + 4t = 4 + s → 4t − s = −3 ``` Subtract: −6s = 6, so s = −1; then 4t = −3 − 1 = −4, t = −1. Third coordinate: −3 − 5(−1) = 2 and 5 + 3(−1) = 2 ✓. Point: (5 − 4, 7 − 4, 2) = (1, 3, 2). #### Equation of the plane Normal = d₁ × d₂ = (4, 4, −5) × (7, 1, 3) = (12 + 5, −35 − 12, 4 − 28) = (17, −47, −24). Through (1, 3, 2): 17(x − 1) − 47(y − 3) − 24(z − 2) = 0. ```text 17x − 47y − 24z − 17 + 141 + 48 = 0 17x − 47y − 24z + 172 = 0 ``` Check with A(5, 7, −3): 85 − 329 + 72 + 172 = 0 ✓; with B(8, 4, 5): 136 − 188 − 120 + 172 = 0 ✓. **Answer: the lines meet at (1, 3, 2); the plane is 17x − 47y − 24z + 172 = 0.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is coplanarity of the two lines tested? A. d1 . d2 = 0 B. The determinant of AB, d1, d2 is zero C. The determinant of d1, d2 and the origin is zero D. AB is parallel to d1 Answer: B. Coplanar lines have AB, d1, d2 linearly dependent, so the triple product vanishes. (Key point: Coplanarity test) 2. Equating coordinates of the two lines gives which intersection point? A. (5, 7, -3) B. (-1, 3, 2) C. (1, 3, 2) D. (8, 4, 5) Answer: C. t = -1, s = -1 give (1, 3, 2) on both lines. (Key point: Intersection point) 3. How is the normal of the plane containing both lines found? A. Cross product d1 x d2 B. Dot product d1 . d2 C. Cross product of AB with itself D. Sum d1 + d2 Answer: A. The plane contains both directions, so its normal is perpendicular to both: d1 x d2 = (17, -47, -24). (Key point: Normal to the plane) 4. What is the plane through (1, 3, 2) with normal (17, -47, -24)? A. 17x - 47y - 24z - 172 = 0 B. 17x - 47y - 24z + 172 = 0 C. 17x + 47y - 24z + 172 = 0 D. 17x - 47y - 24z + 17 = 0 Answer: B. 17(x-1) - 47(y-3) - 24(z-2) = 0 expands to 17x - 47y - 24z + 172 = 0. (Key point: Plane equation) ## Close variants set in other exams - [Coplanar lines and plane containing them](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Acoplanar-lines-plane-containing-them-2) — Mathematics, chapter 3, asked 3× ## Related in Mathematics - [Coplanar lines with point of intersection](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Acoplanar-lines-point-intersection-2) — Mathematics, chapter 3, asked 3× - [Coplanar lines and point of intersection](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Acoplanar-lines-point-intersection) — Mathematics, chapter 3, asked 2× - [Coplanar lines and plane containing them](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Acoplanar-lines-plane-containing-them) — Mathematics, chapter 3, asked 2× - [Coplanar lines and common point](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Acoplanar-lines-common-point) — Mathematics, chapter 3, asked 2× - [Lines intersect: point and plane through them](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Alines-intersect-point-plane-through) — Mathematics, chapter 3 - [Prove lines coplanar](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Aprove-lines-coplanar) — Mathematics, chapter 3 ## Related in other subjects - [Parametric line in 3D: point and plane intersection](https://www.bctengineer.com/question.md?key=semester_4%2Fcomputer_aided_design%3Aparametric-line-segment-plane) — Computer Aided Design, chapter 3 - [Criteria for a point to lie on root locus](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Arl-point-criteria) — Control Systems, chapter 5 - [Intersection versus resection in plane tabling](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Aintersection-vs-resection-plane-table) — Surveying I, chapter 6 - [Intersection vs resection, three point method](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aintersection-vs-resection-three-point) — Surveying II, chapter 5