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Question

The image of line x−13=y−31=z−4−5 in the plane 2x−y+z+3=0

A
x33=y+51=z25
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B
x33=y+51=z25
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C
x+33=y51=z25
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D
x+33=y51=z+25
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Solution

The correct option is C x+33=y51=z25
Given,
Image of the line x13=y31=z45
in the plane 2xy+z+3=0
firstly.
let us find out the given line is whether parallel or perpendicular.
if the given line is parallel to the given plane them
we get,
3×x coefficient +1×y co-efficient +(5)×z co-efficient
3×2+1(1)+(5)×1=615
=66=0 equation (1)
equation (1) show that given lime is parallel to the given plane
Now, let is represent it through a figure.
Representation of given data diagramatically for the lar line, equation will be,
x12=y31=z41=k [where k is coefficient]
Now in finding k value we get
x12=k ; y31=k ; z41=k
x1=2k y3=k z4=k
x2k+1 ; y=k+3 ; z=k+4
Let us assume N as a point of the lar
as we have plane equation as 2xy+z+3=0
Now, let us substitude x, y, z Values thus obtained in the plane equation to find k we get.
2(2k+1)1[k+3]+1[k+4]+3=0
4k+23+k+k+4+3=0
6k+6 0 6k=6 k=66=1
if k=1 then, let us find out x, y, z values,
x=2(1)+1=1, y=(1)+3=4, z=(1)+4=3
N(x, y, z)=N(1, 4, 3)
As Q is the image of line P then N will be the point bisecting both Q and N will be the mid point.
As N is the midpoint for P and Q we get,
1=x1+12, 4y1+32, 3=z1+42
x1=21 ; y1=83 ; z1=64
x1=3 y1=5 z1=2
So, now the line equation Q will be.
xx13=yy11=zz15
x(3)3=y(5)1=z(2)5 x+33=y51=z25
the line equation of the image Q is
x+33=y51=z25 so, correct option is [C]

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