If the lines `(x-1)/2=(y+1)/3=(z-1)/4a n d(x-3)/1=(y-k)/2=z/1` intersect, then find the value of `kdot`

Options

A. `(9)/(2)`
B. `(1)/(2)`
C. `(5)/(2)`
D. `(7)/(2)`

Answer

A

Explanation

The given lines are intersecting, since directing ratios of lines are not proportional.
Let `(x-1)/(2)=(y+1)/(3)=(z-1)/(4)= lambda`
`:.` The coordinates of a point on the line are
`(2 lambda + 1, 3 lambda-1, 4lambda+1)`
Similarly, let `(s-3)/(1)=(y-k)/(2)=(z-0)/(1)= mu`
`:.` The coordinates of a point on the line are
`( mu + 3 , 2 mu + k, mu)`
since, two lines intersect for some values of `lambda` and `mu`.
`:. ( 2 lambda+1, 3lambda-1,4 lambda+1)=(mu+3, 2 mu+k, mu)`
On comparing corresponding coordinates ,we get
`2lambda+1=mu+3, 3 lambda-1=2mu + k and 4 lambda + 1 = mu`
`2lambda-mu=2, 3 lambda-2mu=k+1 and 4 lambda - mu = - 1`
Now, `4 lambda-mu - 2 lambda+mu=-1-2`
`rArr 2lambda=-3`
`rArr lambda=(-3)/(2)`
On putting `lambda=(-3)/(2)` in `2lambda-mu=2`, we get
`2xx(-3/2)-mu=2`
`rArr -3-mu=2`
`rArr mu=-3-2=-5`
On putting `lambda=(-3)/(2)` and `mu=-5` in `3lambda-2mu=k+1`,
we get
`3xx(-3/2)-2xx-5=k+1 rArr (-9)/(2) +10-1=k`
`rArr k=(-9+18)/(2)=(9)/(2)`
Hence, option (a) is correct.