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If lines x=y=za n dx=y/2=z/3 and third l...

If lines `x=y=za n dx=y/2=z/3` and third line passing through `(1,1,1)` form a triangle of area `sqrt(6)` units, then the point of intersection of third line with the second line will be a. `(1,2,3)` b. `2,4,6` c. `4/3,6/3,(12)/3` d. none of these

A

`(4/3,8/3,12/3)`

B

(1,2,3)

C

(2,4,6)

D

(3,6,9)

Text Solution

Verified by Experts

The correct Answer is:
C
A(1,1,1) lies on x=y=z. Any point on second line is `(lambda,2lambda,3lambda)`.

`costheta=6/sqrt(42), therefore sintheta=sqrt(6)/sqrt(42)`
Area of `triangleOAB=1/2(OA)(OB)sintheta=sqrt(6)`
`=rArr 1/2(sqrt(3))(sqrt(14)lambda)sqrt(6)/sqrt(42) = sqrt(6) rArr lambda=2`
`therefore B=(2,4,6)`.
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