Question #cca71

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Answer 1 · 2017-03-20T12:45:25

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

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)}.#

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