Solve the vector equation `rxxb=axxb`, `r*c=0` provided that c is not perpendicular to b.

Solve the vector equation vecr xx vecb = veca xx vecb, vecr.vecc = 0 provided that vecc is not perpendicular to vecb

The equation ax+by +c=0 represents a plane perpendicular to the

From the matrix equation AB=AC, we conclude B=C provided.

If a vector vec a is perpendicular to two non collinear vectors vec b\ a n d\ vec c ,\ t h e n\ \ vec a\ is perpendicular to every vector in the plane of \ vec b\ a n d\ vec c

vec a , vec b and vec c are the position vectors of points A ,B and C respectively, prove that : vec a× vec b+ vec b× vec c+ vec c× vec a is vector perpendicular to the plane of triangle A B Cdot

Statement I: If a is perpendicular to b and c, then axx(bxxc)=0 Statement II: if a is perpendicular to b and c, then bxxc=0

The vector equation of a plane is vecr.(6hati-3hatj-2hatk)+2 = 0 . Convert it into normal form. Also find the length of perpendicular from origin and the d.c.'s of this perpendicular vector.

If a, b and c are any three non-zero vectors, then the component of atimes(btimesc) perpendicular to b is

Solve for vectors A and B, where A+B=a,AxxB=b,A*a=1 .

Find a unit vector , C=AxxB is a vector perpendicular to both A and B Henc , a unit vector hatn perpendicular to A and B can be Written as hatn=(C)/(C)=(AxxB)/(|AxxB|) Here, AxxB=|underset(1,-1,1)underset(2,3,1)hati,hatj,hatk| =hati(3+1)+hatj(1-2)+hatk(-2-3) =4hati-hatj-5hatk Further, |AxxB|=sqrt((4)^(2)+(-1)^(2)+(-5)^(2))=sqrt(42) therefore"the desired unit vector is "hatn=(AxxB)/(|AxxB|) or hatn=(1)/(sqrt(42))(4hati -hatj+5hatk)