`(3vec i+4vec j-5vec k)*vec i=`

If vec r = 3vec i + 2vec j-5vec kvec a = 2vec i-vec j + vec k, vec b = vec i + 3vec j-2vec k and vec c = -2vec i + vec j-3vec k such that vec r = lambdavec a + muvec b + deltavec c then mu, (lambda) / (2), delta are in? (A) AP (B) GP (C) HP (D) AGP

Let ABCD be the parallelogram whose sides AB and AD are represented by the vectors 2vec i+4vec j-5vec k and vec i+2vec j+3vec k respectively. Then if vec a is a unit vector parallel to vec AC ,then vec a is equal to

The position vectors of the points A,B,C , D are respectively 2 vec i + vec j - vec k, vec i + vec j + vec k, vec i- 2 vec j+ 3 vec k and 3 vec i- vec j + 2 vec k . Evaluate [vec (AB), vec (AC), vec (AD)] .

If three vector 2vec i-vec j-vec k,vec i+2vec j-3vec k,3vec i+lambdavec j+5vec k are coplanar then the value of lambda is :

(vec i + xvec j + 3vec k) xx (vec i-vec j + vec k) = 5vec i + xvec j-3vec k then x =

The values of a for which the points A,B,C with position vectors 2vec i-vec j+vec k+vec i-3vec j-5vec k and avec i-3vec j+vec k,vec i-3vec j-5vec k and avec i-3vec j+vec k are the vertices of a right angled triangle with /_C=(pi)/(2) are

The line of intersection of the planes vec r*(3vec i-vec j+vec k)=1 and vec r*(vec i+4vec j-2vec k)=2, is

consider the points A,B,C and D with position vector 7vec i-4vec j+7vec k,vec i-6vec j+10vec k,-vec i-3vec j+4vec k and 5vec i-vec j+vec k respectively then ABCD is

If the position vectors of three consecutive vertices of any parallelogram are respectively vec i+vec j+vec k+vec i+3vec j+5vec k,7vec i+9vec j+11vec k then the position vector of its fourth vertex is:

Show that the four points A,B,C and D whose position vectors are (4vec i+5vec j+vec k),(-vec j-vec k),(3vec i+9vec j+4vec k) and (-4vec i+4vec j+4vec k) respectively are coplanar