The position verctors of the points A and B with respect of O are `2hati+2hatj+hatk and 2hati+4hatj+4hatk`, the length of the internal bisector of `angleBOA` of `DeltaAOB` is

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

If A(veca)=2hati+2hatj+2hatk & B(vecb)=2hati+4hatj+4hatk find the length of angle bisector of angleAOB

The position vectors of the points A, B, C are 2 hati + hatj - hatk , 3 hati - 2 hatj + hatk and hati + 4hatj - 3 hatk respectively . These points

If a=2hati+2hatj-8hatk and b=hati+3hatj-4hatk , then the magnitude of a+b is equal to

If the sides of an angle are given by vectors veca=hati-2hatj+2hatk and vecb=2hati+hatj+2hatk , then find the internal bisector of the angle.

The position vectors of A and B are 2hati-9hatj-4hatk and 6hati-3hatj+8hatk respectively, then the magnitude of AB is

The position vectors of the points A, B, and C are hati + hatj + hatk.hati+ 5hatj-hatk and 2hati + 3hatj + 5hatk , respectively . The greatest angle of triangle ABC is -

The position vectors of the points A,B and C are (2hati + hatj - hatk), (3hati - 2hatj + hatk) and (hati + 4hatj - 3hatk) respectively. Show that the points A,B and C are collinear.