Vector normal to the plane; passing through `veca` and `vecb`

The vector equation of the plane passing through the origin and the line of intersection of the planes vecr.veca=lamdaandvecr.vecb=mu is

p (2,3,-4) ,vecb =2 hati - hatj + 2 hatk Vector equation of a plane passing through the point P perpendicular to the vector vecb

Find a unit vectro perpendicular to the plane of two vectors veca and vecb where veca=hati-hatj and vecb=hatj+hatk

Length of the perpendicular from origin to the plane passing through the three non-collinear points veca, vecb, vecc is

Show that the equation of as line perpendicular to the two vectors vecb and vecc and passing through point veca is vecr=veca+t(vecbxxvecc) where t is a scalar.

Find the vector and cartesian equations of the planes (a) that passes through the point (1,0,-2) and the normal to the plane is (b) that passes through the point (1,4,6) and the normal vector to the plane is the normal vector to the plane is the normal vector to the plane is hat i-2hat j-hat k

If the angles between the vectors veca and vecb,vecb and vecc,vecc an veca are respectively pi/6,pi/4 and pi/3 , then the angle the vector veca makes with the plane containing vecb and vecc , is

If veca =hati + hatj-hatk, vecb = - hati + 2hatj + 2hatk and vecc = - hati +2hatj -hatk , then a unit vector normal to the vectors veca + vecb and vecb -vecc , is

If veca,vecb and vecc are three non-coplanar vectors then the length of projection of vector veca in the plane of the vectors vecb and vecc may be given by