Question 421 of 521MediumMCQ
The values of a for which the two points (1,a,1) and (−3,0,a) lie on the opposite sides of the plane 3x+4y−12z+13=0, satisfy: Solutions · 1
Official solution
Plane values: at (1,a,1): 3+4a−12+13=4+4a; at (−3,0,a): −9+0−12a+13=4−12a. Opposite sides ⇔ product <0:
\[(4+4a)(4-12a)<0\implies 3a^2+2a-1>0\implies (3a-1)(a+1)>0\implies a<-1\text{ or }a>\tfrac13.\] Answer the question to unlock