A line and two geometric conditions
A coordinate exercise places point A at(1,0,2). One plane passes through A and has normal vector n=(1,2,-1). A separate line starts at A with direction vector d=(2,1,0). The line is not assumed to lie in the first plane. A second plane, representing an abstract checkpoint, has equation x+y+z=9. A student confuses the normal vector with a direction along the first plane and uses it to construct the route. Another student finds the line parameter at the checkpoint but reports the parameter itself as the travel distance. Treat coordinates as consistent length units, use exact arithmetic, and distinguish points, vectors, parameters and distances. No speed, travel time, collision or physical safety property is being asserted by this abstract model.
1. Write the equation of the plane with normal n through A.
Read the worked answer
Use n·(r-A)=0: (x-1)+2y-(z-2)=0, hence x+2y-z=-1. Substituting A gives1+0-2=-1. The normal is perpendicular to directions lying in this plane.
2. Find where the specified line meets x+y+z=9.
Read the worked answer
The line is(x,y,z)=(1+2t,t,2). Substitution gives1+2t+t+2=9, so3t=6 and t=2. The intersection is(5,2,2), whose coordinates sum to9.
3. Calculate the distance from A to that intersection and decide whether d lies parallel to the first plane.
Read the worked answer
The difference vector is(4,2,0), so the distance is√(16+4)=2√5, not the parameter2. Also n·d=2+2=4≠0, so d is not parallel to the first plane.