Determine the equation of the plane that passes through (P(2,,-1,,3)) and contains the line of intersection of (3x-2y+z=4) and (x+y+z=1).
Answer
2x-13y-4z=5
The correct answer is 2x-13y-4z=5 — Simplify: (2x-13y-4z=5).
Step-by-step solution
- Find normals of given planes: (mathbf{n}_1=(3,,-2,,1)) and (mathbf{n}_2=(1,,1,,1)).
- Compute direction vector of their intersection line: (mathbf{d}=mathbf{n}_1\times\mathbf{n}_2=(-3,,-2,,5)).
- Find a point on the intersection line by solving the two equations: choose (y=0), giving (x=\frac32), (z=-\frac12). So (Q(\frac32,0,-\frac12)).
- Form vector (mathbf{PQ}=Q-P=(-\frac12,,1,,-\frac72)).
- Compute normal of required plane as (mathbf{n}=mathbf{d}\times\mathbf{PQ}=(2,,-13,,-4)).
- Write plane equation using point (P): (2(x-2)-13(y+1)-4(z-3)=0).
- Simplify: (2x-13y-4z=5).
Common mistake: using wrong normals or failing to cross‑product for direction vector
Concept tested: plane through point containing line of intersection
Published · How this solution was prepared and checked
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