दो रेखाओं जिनके सदिश समीकरण निम्नलिखित हैं, के बीच कि न्यूनतम दुरी ज्ञात करें|
(i) `vecr = (3-t)hati + (4+2t)hatj + (t-2)hatk` तथा (and)
`vecr = (1+s)hati + (3s-7)hatj + (2s-2)hatk`

Options

A. `-sqrt(15)`
B. `-sqrt(35)`
C. `-sqrt(37)`
D. `-sqrt(45)`

Answer

B

Explanation

(i) दी हुई रेखाएँ हैं, `vecr = (3-t)hati + (4+2t)hatj + (t-2)hatk`.............(1)
(1)से, `vecr =(3hati + 4hatj -2hatk)+t(-hati + 2hatj+hatk)`
(1) से, `vecr =(3hati +4hatj -2hatk) + t(-hati + 2hatj +hatk)`
`rArr vecr = veca_(1) + tvecb_(1)`
जहाँ `veca_(1) = 3hati + 4hatj - 2hatk` तथा `vecb_(1) = -hati + 2hatj + hatk`
(2) से, `vecr =(hati - 7hatj -2hatk)+s (hati + 3hatj + 2hatk)`
`rArr vecr = veca_(2) + svecb_(2)`
जहाँ `veca_(2) = hati-7hatj - 2hatk` or `vecb_(2) = hati + 3hatj + 2hatk`
रेखाओं (1) तथा (2) के बीच कि न्यूनतम दुरी
`d=|((veca_(2)-veca_(1))(vecb_(1)xx vecb_(2)))/(|vecb_(1) xx vecb_(2)|)|`..........(1)
अब `veca_(2)-veca_(1) = (hati - 7hatj - 2hatk) -(3hati + 4hatj -2hatk) = -2hati-11hatj`........(2)
`vecb_(1) xx vecb_(2) = |:(hati, hatj, hatk),(-1,2,1),(1,3,2):|=hati + 3hatj - 5hatk`..........(3)
`|vecb_(1) xx vecb_(2)| = sqrt(1^(2) + 3^(2) + (-5)^(2))= sqrt(35)`..........(4)
तथा `(veca_(2)-veca_(1)).(vecb_(1) xx vecb_(2)) = (-2hati-11hatj).(hati + 3hatj - 5hatk)`
`=(-2) xx 1 +(-11) xx 3+0 =-35`............(5)
`therefore d=|-35/sqrt(35)|=35/sqrt(35)=sqrt(35)` इकाई