--- title: "Plane through intersection, perpendicular to plane" subject: "Mathematics" chapter: 3 key: "semester_2/mathematics:plane-through-intersection-perpendicular-plane" source: https://www.bctengineer.com/q/semester_2/mathematics/plane-through-intersection-perpendicular-plane site: BCT Engineering --- # Plane through intersection, perpendicular to plane Mathematics · Chapter 3: Three Dimensional Solid Geometry - Asked in: 2074 Bhadra, 5 marks - Video: [10b) Find equation of plane through intersection of x+y+z=6 2x+3y+4z+5=0 perpendicular to 4x+5y–3z=8](https://www.youtube.com/watch?v=Z17Pi0qB8II) — Simply Australia Things - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/mathematics/plane-through-intersection-perpendicular-plane - Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-3 ## Question Obtain the equation of the plane passing through the line of intersection of two planes 7x−4y+7z+16 = 0 and 4x−3y−2z+13 = 0 and perpendicular to plane x−y−2z+5 = 0. ## Short answer Plane through intersection: (7x-4y+7z+16)+λ(4x-3y-2z+13)=0. Normal (7+4λ, -4-3λ, 7-2λ) dot (1,-1,-2) = 0 gives λ=3/11. Substituting yields 89x - 53y + 71z + 215 = 0. ## Answer **Given planes.** P₁: 7x − 4y + 7z + 16 = 0 and P₂: 4x − 3y − 2z + 13 = 0. #### Plane through their line of intersection (7x − 4y + 7z + 16) + λ(4x − 3y − 2z + 13) = 0 ```text (7 + 4λ)x − (4 + 3λ)y + (7 − 2λ)z + (16 + 13λ) = 0 ... (1) ``` Its normal is (7 + 4λ, −(4 + 3λ), 7 − 2λ). #### Perpendicular to x − y − 2z + 5 = 0 Normals must be perpendicular: their dot product is zero. The given plane has normal (1, −1, −2). ```text (7 + 4λ)(1) + (−(4 + 3λ))(−1) + (7 − 2λ)(−2) = 0 7 + 4λ + 4 + 3λ − 14 + 4λ = 0 11λ − 3 = 0 → λ = 3/11 ``` #### Substitute in (1) Multiply (1) by 11 and use λ = 3/11: ```text (77 + 12)x − (44 + 9)y + (77 − 6)z + (176 + 39) = 0 89x − 53y + 71z + 215 = 0 ``` Check: the normal (89, −53, 71) with (1, −1, −2): 89 + 53 − 142 = 0 ✓. **Answer: 89x − 53y + 71z + 215 = 0.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is a plane through the line of intersection written? A. (7x - 4y + 7z + 16) + lambda(4x - 3y - 2z + 13) = 0 B. (7x - 4y + 7z + 16) x lambda = 4x - 3y - 2z + 13 C. 7x - 4y + 7z + 16 = lambda D. (7x - 4y + 7z) + lambda(4x - 3y - 2z) = 0 Answer: A. The two planes are combined with the multiplier lambda. (Key point: Family of planes) 2. How is perpendicularity to x - y - 2z + 5 = 0 imposed? A. The normals are parallel B. The dot product of the normals is zero C. The constant terms are equal D. The cross product of the normals is zero Answer: B. Two planes are perpendicular if and only if their normals are. (Key point: Perpendicularity condition) 3. What is lambda? A. -3/11 B. 11/3 C. 3/11 D. 3 Answer: C. 11 lambda - 3 = 0. (Key point: Value of lambda) 4. What is the final plane? A. 89x - 53y + 71z + 215 = 0 B. 89x + 53y + 71z + 215 = 0 C. 89x - 53y + 71z - 215 = 0 D. 77x - 44y + 71z + 176 = 0 Answer: A. Multiply by 11 and substitute lambda = 3/11. 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