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Let A( vec a)a n dB( vec b) be points on...

Let `A( vec a)a n dB( vec b)` be points on two skew lines ` vec r= vec a+lambda vec pa n d vec r= vec b+u vec q` and the shortest distance between the skew lines is `1, w h e r e vec pa n d vec q` are unit vectors forming adjacent sides of a parallelogram enclosing an area of 1/2 units. If angle between`A B` and the line of shortest distance is `60^0,` then `A B=` a. `1/2` b. `2` c. `1` d. `lambda R={10}`

A

`(1)/(2)`

B

2

C

1

D

`lamdaepsiR-{0}`

Text Solution

Verified by Experts

The correct Answer is:
b
`1=|(vecb-veca).(vecpxxvecq)/(|vecpxxq|)|`
or `|vecb-veca|cos60^(@)=1impliesAB=2`
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