If lines `(x-1)/(2)=(y+1)/(3)=(z-1)/(4) and (x-3)/(1)=(y-lamda)/(2)=(z)/(1)` intersect each other, then `lamda=` . . .

Options

A. `(7)/(2)`
B. `(3)/(2)`
C. `(9)/(2)`
D. `(5)/(2)`

Answer

C

Explanation

Given lines are
`(x-1)/(2)=(y+1)/(3)=(z-1)/(4)=lamda_(1)` (let) . . . (i)
and `(x-3)/(1)=(y-lamda)/(2)=(z)/(1)=lamda_(2)` (let) . . (ii)
Then any point on line (i) is `(2lamda_(1)+1,3lamda_(1)-1,4lamda_(1)+1)` and any point on line (ii) is `(lamda_(2)+3,2lamda_(2)+lamda,lamda_(2))`
Clearly, then lines (i) and (ii) will intersect if
`(2lamda_(1)+1,3lamda_(1)-1,4lamda_(1)+1)=(lamda_(2)+3,2lamda_(2)+lamda,lamda_(2))`
for some particular value of `lamda_(1) and lamda_(2)`
`implies2lamda_(1)+1=lamda_(2)+3,3lamda_(1)-1=2lamda_(2)+lamda`
and `4lamda_(1)+1=lamda_(2)`
`implies2lamda_(1)-lamda_(2)=2,3lamda_(1)-2lamda_(2)=lamda+1 and 4lamda_(1)-lamda_(2)=-1`
on solving `2lamda_(1)-lamda_(2)=2 and 4lamda_(1)-lamda_(2)=-1`
we get `lamda_(1)=-(3)/(2) and lamda_(2)=-5`
Now, putting the values of `lamda_(1) and lamda_(2) ` in
`3lamda_(1)-2lamda_(2)=lamda+1`
`implies3((-3)/(2))-2(-5)=lamda+1implies(-9)/(2)+10=lamda+1`
`implies lamda+1=(11)/(2)implieslamda=(9)/(2)`