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Two lines L1x=5, y/(3-alpha)=z/(-2)a n d...

Two lines `L_1x=5, y/(3-alpha)=z/(-2)a n dL_2: x=alphay/(-1)=z/(2-alpha)` are coplanar. Then `alpha` can take value (s) a. `1` b. `2` c. `3` d. `4`

A

1

B

2

C

3

D

4

Text Solution

Verified by Experts

The correct Answer is:
A, D
If two straight lines are coplanar,
i.e.`" "(x-x_(1))/(a_(1))=(y-y_(1))/(b_(1))=(z-z_(1))/(c_(1))`
and`" "(x-x_(2))/(a_(2))=(y-y_(2))/(b_(2))=(z-z_(2))/(c_(2))` are coplanar

Then, `(x_(2)-x_(1),y_(2)-y_(1),z_(2)-z_(1)),(a_(1),b_(1),c_(1))` and `(a_(2),b_(2),c_(2))` are coplanar.
i.e.`" "|{:(x_(2)-x_(1),y_(2)-y_(1),z_(2)-z_(1)),(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2)):}|=0`
Here, `" "x=5,(y)/(3-alpha)=(z)/(-2)`
`implies" "(x-5)/(0)=(y-0)/(-(alpha-3))=(z-0)/(-2)" "...(i)`
and`" "x-alpha,(y)/(-1)=(z)/(2-alpha)`
`implies" "(x-alpha)/(0)=(y-0)/(-1)=(z-0)/(2-alpha)" "...(ii)`
`implies" "|{:(5-alpha," "0," "0),(" "0,3-alpha," "-2),(" "0,-1,2-alpha):}|=0`
`implies" "(5-alpha)[(3-alpha)(2-alpha)-2]=0`
`implies" "(5-alpha)[a^(2)-5alpha+4]=0`
`implies" "(5-alpha)(alpha-1)(alpha-4)=0`
`:." "alpha=1,4,5`
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