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Question
Find the coordinate of the point P where the line through A(3, –4, –5) and B(2, –3, 1) crosses the plane passing through three points L(2, 2, 1), M(3, 0, 1) and N(4, –1, 0).
Also, find the ratio in which P divides the line segment AB.
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Solution
The equation of the plane passing through three given points can be given by
`|(x-2,y-2,z-1),(x-3,y-0,z-1),(x-4,y+1,z-0)|=0`
Performing elementary row operations R2 →R1−R2 and R3 →R1−R3, we get
`=>|(x-2,y-2,z-1),(3-2,0-2,0),(4-2,-1-2,-1)|=0`
`=>|(x-2,y-2,z-1),(1,-2,0),(2,-3,-1)|=0`
Solving the above determinant, we get
⇒(x−2)(2−0)−(y−2)(−1−0)+(z−1)(−3+4)=0
⇒(2x−4)+(y−2)+(z−1)=0
⇒2x+y+z−7=0
Therefore, the equation of the plane is 2x+y+z−7=0
Now, the equation of the line passing through two given points is
`(x-3)/(2-3)=(y+4)/(-3+4)=(z+5)/(1+5)=lambda`
`=>(x-3)/(-1)=(y+4)/1=(z+5)/6=lambda`
⇒x=(−λ+3), y=(λ−4), z=(6λ−5)
At the point of intersection, these points satisfy the equation of the plane 2x+y+z−7=0.
Putting the values of x, y and z in the equation of the plane, we get the value of λ.
2(−λ+3)+(λ−4)+(6λ−5)−7=0
⇒−2λ+6+λ−4+6λ−5−7=0
⇒5λ=10
⇒λ=2
Thus, the point of intersection is P(1, −2, 7).
Now, let P divide the line AB in the ratio m : n.
By the section formula, we have
`1=(2m+3n)/(m+n)`
⇒m+2n=0
⇒m=−2n
`=>m/n=(-2)/1`
Hence, P externally divides the line segment AB in the ratio 2 : 1
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