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Let the position vectors of the points `Pa n dQ` be `4 hat i+ hat j+lambda hat ka n d2 hat i- hat j+lambda hat k ,` respectively. Vector ` hat i- hat j+6 hat k` is perpendicular to the plane containing the origin and the points `Pa n dQ` . Then `lambda` equals `1//2` b. `1//2` c. `1` d. none of these

A

`-1//2`

B

`1//2`

C

1

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
a
A vector perpendicular to the plane of O , P and Q is
` vec(OP) xx vec(OQ) `
Now, `vec(OP) xx vec(OQ) = |{:(hati,hatj,hatk),(4,1,lambda),(2,-1,lambda):}|`
` = 2lambda hati - 2hatj - 6hatk`
therefore, `hatim -hatj + 6hatk` is parallel to
`2lambda hati - 2lambdahatj - 6hatk`
Hence, `1/(2lambda) = (-1)/(-2lambda) = 6/(-6)`
`lambda = -1/2`
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