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
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}`
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1 Answers
Correct Answer - b
`1=|(vecb-veca).(vecpxxvecq)/(|vecpxxq|)|`
or `|vecb-veca|cos60^(@)=1impliesAB=2`
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Answered