यदि बिंदुO के सापेक्ष A और B के स्थिति सदिश क्रमश : `2hati - hatj + hatk" और" 5hati + 3hatj + hat k" हों तो " vec(AB)`तथा उसका मापांक ज्ञात कीजिए ।

यदि A (-hati + 3hatj + 2hatk) , B (-4hati + 2hatj - 2hatk) तथा C(5hati + (lambda +1)hatj + mu hatk) संरेखीय है तब :

if a=2hati+hatj+2hattk and b=5hati-3hatj+hatk , then the projection of 'b' and 'a' is

Angle between the planes: (i) vec(r). (hati - 2 hatj - hatk) = 1 and vec(r). (3 hati - 6 hatj + 2 hatk) = 0 (ii) vec(r). (2 hati + 2 hatj - 3 hatk ) = 5 and vec(r) . ( 3 hati - 3 hatj + 5 hatk ) = 3

If vec A=hati +7hatj +hatk and vec B =2 hati+3hatj +4hatk then the component of vec A along vec B is

If vec(a) = hati - 2hatj - 3hatk, vec(b) = 2 hati - hatj - hatk and vec(c) = hati +3 hatj - 2hatk find (vec(a) xx vec(b)) xx vec(c)

If the position vectors of A and B respectively hati+3hatj-7hatk and 5 hati-2hatj+4hatk , then find vec(AB)

If vec(A) = hati + 3hatj + 2hatk and vec(B) = 3hati + hatj + 2hatk , then find the vector product vec(A) xx vec(B) .

Consider the equations of the straight lines given by : L_(1) : vec(r) = (hati + 2 hatj + hatk ) + lambda ( hati - hatj + hatk) L_(2) : vec(r) = (2 hati - hatj - hatk) + mu ( 2 hati + hatj + 2 hatk) . If vec(a_(1))= hati + 2 hatj + hatk, " " vec(b_(1)) = hati - hatj + hatk , vec(a_(2)) = 2 hat(i) - hatj - hatk, vec(b_(2)) = 2 hati + hatj + 2 hatk , then find : (i) vec(a_(2)) - vec(a_(1)) " " (ii) vec(b_(2)) - vec(b_(1)) (iii) vec(b_(1))xx vec(b_(2)) " " (iv) vec(a_(1)) xx vec(a_(2)) (v) (vec(b_(1)) xx vec(b_(2))).(vec(a_(1)) xxvec(a_(2))) (vi) the shortest distance between L_(1) and L_(2) .