Question #cca71

1 Answer
Mar 20, 2017

The Line crosses the Plane at the pt. (1,-2,7).

Explanation:

In the Usual Notation, the eqn. of the Line L thro. the pts.

(3,-4,-5) and (2,-3,1) is given, by,

L : (x-3)/(3-2)=(y-(-4))/(-4-(-3))=(z-(-5))/(-5-1), i.e.,

L : (x-3)/1=(y+4)/-1=(z+5)/-6=k, k in RR, or,

L : x=k+3, y=-k-4, z=-6k-5, k in RR............(1).

The eqn. of the Plane pi thro. the pts. (0,4,3),(1,2,3),(4,2,-3) is

pi : |(x-0,y-4,z-3),(1-0,2-4,3-3),(4-0,2-4,-3-3)|=0.

pi : |(x,y-4,z-3),(1,-2,0),(4,-2,-6)|=0.

pi :12x-(-6)(y-4)+6(z-3)=0.

pi : 2x+(y-4)+(z-3)=0, or, 2x+y+z=7..........(2).

To determine, pi nn L, we solve (1) and (2).

Subst.ing x,y,z" from "(1)" into "(2), we get,

2(k+3)+(-k-4)+(-6k-5)=7," for some k" in RR.

rArr -5k=10 :. k=-2.

:. x=k+3=-2+3=1, y=-k-4=2-4=-2, z=-6k-5=12-5=7.

:. pi nn L ={(1,-2,7)}.

Enjoy Maths.!