एक रेखा l जो मूलबिन्दु से गुजरती है, रेखाओंl_(1) : (3+t)hati + (-1 + 2t)hatj + (4+2t)hatk, -inft...

A line l passing through the origin is perpendicular to the lines 1: (3 + t ) hati + (-1 +2 t ) hatj + (4 + 2t) hatk -oolt t lt ooand 1__(2) : (3 + 2s )hati + (3+2s) hati + (3+ 2s) hatj + ( 2+s) hatk , -oo lt s lt oo Then the coordinate(s) of the point(s) on 1 _(2) at a distance of sqrt17 from the point of intersection of 1 and 1_(1) is (are)

Find the shortest distance between the following (6 -7) lines whose vector equations are : (i) vec(r) = (1 -t) hati + (t - 2) hatj + (3 - 2t) hatk and vec(r) = (s + 1) hat(i) + (2s - 1) hatj - (2s + 1) hatk (ii) vec(r) = (3 -t) hati + (4 + 2t) hatj + (t - 2) hatk and vec(r) = (1 + s) hati + (3s - 7 ) hatj + (2s -2) hatk. where t and s are scalars.

Find the Cartesian equation of the following planes : a. vec(r). (hati + hatj - hatk ) = 2 b. vec(r). (2 hati + 3 hatj - 4 hatk ) = 1 (c ) vec(r). [ (s - 2t) hati + (3 - t ) hatj + (2 s + t ) hatk] = 15

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) .

The cartesian form of the equation of plane barr=(s+t)hati+(2+t)hatj+(3s+2t)hatk is

Find the Cartesian from the equation of the plane vecr=(s-2t)hati+(3-t)hatj+(2s+t)hatk .

Find the cartesian form of the equation of the plane. barr = (hati + hatj) + s(hati - hatj + 2hatk) + t(hati + 2hatj +hatk) .

The position vector of a particle is given by vecr=(2 sin 2t)hati+(3+ cos 2t)hatj+(8t)hatk . Determine its velocity and acceleration at t=pi//3 .