Question

Question #cca71

Answer 1

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.!