If `A=(1,0,1), B=(0,-1,0), C=(-1,0,1), D=(0,1,-1)` then angle between `vec(AB)` and `vec(CD)` is (A) `pi/6` (B) `pi/4` (C) `pi/3` (D) `pi/2`

If =(0,1)B=(1,0),C=(1,2),D=(2,1) ,prove that vec AB=vec CD

If |vecc|=2 , |veca|=|vecb|=1 and vecaxx(vecaxxvecc)+vecb=vec0 then the acute angle between veca and vecc is (A) pi/6 (B) pi/4 (C) pi/3 (D) (2pi)/3

If A(8,2,0),B(4,6,-7),C(-3,1,2)and D(-9,-2,4) are four given point then find the angle between vec(AB)and vec(CD) .

sin^(-1)((-1)/2) (a) pi/3 (b) -pi/3 (c) pi/6 (d) -pi/6

sin^(-1)(0) is equal to : (a) 0 (b) pi/6 (c) pi/2 (d) pi/3

int_-1^1 sin^-1(x/(1+x^2))dx= (A) pi/4 (B) pi/2 (C) pi (D) 0

sin^(-1)x+cos^(-1)x,x in[-1,1] = (A) 0 (B) pi/2 (C) pi (D) 2pi

If vec a,vec b, and vec c are such that [vec avec bvec c]=1,vec c=lambdavec a xxvec b, angle, between vec a and vec b is (2 pi)/(3),|vec a|=sqrt(2),|vec b|=sqrt(3) and |vec c|=(1)/(sqrt(3)) then the angel between vec a and vec b is (pi)/(6) b.(pi)/(4) c.(pi)/(3) d.(pi)/(2)

sin^(-1)x+cos^(-1)x,x in[-1,1]= (A) 0 (B) (pi)/(2) (C) pi (D) 2 pi

vec a+vec b+vec c=vec 0,|vec a|=3,|vec b|=5,|vec c|=7 then the angle between vec a and vec b is (pi)/(6) b.(2 pi)/(3) c.(5 pi)/(3)d*(pi)/(3)