lf `OA = 3i + j-k,|AB|=2/sqrt6` and AB has the direction ratios 1,-1,2 then `|OB|

If vec OA=3hat i+hat j-hat k|vec AB|=2sqrt(6) and AB has the direction ratios 1,-1, and 2 then |OB| is

Straight lines OA and OB are drawn from the origin O, if the direction ratios of the lines OA and OB are 1, -1, -1 and 2, -1, 1 respectively then find the direction cosines of the normal to the plane AOB.

If vec(OA) = hati - 2hatk and vec(OB) = 3 hati - 2hatj then the direction cosines of the vectore vec(AB) are

If vec OA = i-j and vec OB = j-k , the unit vector perpendicular to the plane AOB is

vec OA = i+xj+2k, vec OB = 2i+k and vec OC =-i+j+k and vec AB is perpendicular to vec BC ,then x =

If the position vectors of the points A and B respectively are i+2j-3k and j-k find the direction cosines of AB

If vec AB=2hat i+hat j3hat k and the coordinates of A are (1,2,-1), then the coordinates of B are

if a=(i+j+k),a.b=1 and ab=j-k then b is equal to

If vec(OA)=hat(i)-2hat(k) " and " vec(OB)=3hat(i)-2hat(j) then the direction cosines of the vector vec(AB) are -

A : If the positon vectors of A, B are (2, 3, 5), (1, 1, 7) then the unit vector in the direction of vec(AB) is (-i-2j+2k)/(3) R : Unit vector in the direction of vec(AB) is (vec(AB))/|vec(AB)|