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If the lines `(x-1)/2=(y+1)/3=(z-1)/4 ` and `(x-3)/1=(y-k)/2=z/1` intersect each other then find value of k

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#### Solution

Let `(x-1)/2=(y+1)/3=(z-1)/4 =u ` where is any constant.

So for any point on this line has co-ordinates in the form (2u+1,3u-1,4u+1)

`(x-3)/1=(y-k)/2=z/1=v`

So for any point on this line has co-ordinates in the form (v+3,2v+k,v).

Point of intersection of these two lines will have co-ordinates of the form

(2u +1, 3u −1,4u +1) and (v +3, 2v + k,v) .

Equating the x, y and z co-ordinates for both the forms we get three equations

2u+1=v+3

2u-v=2.............(1)

3u-1=2v+k

3u-2v=k+1.......(2)

4u+1=v

4u-v=-1...........(3)

Subtracting equation (1)from equation(3) we get,

2u = -3

u=-3/2

Substitute value of u in equation (1) we get,

2(-3/2) - v=2

v=-5

Substitute value of v and in equation (2) we get,

3(-3/2) - 2(-5)=k+1

k=9/2

the value of k is 9/2

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