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.

The length of the perpendicular from the origin to the plane passing through the points veca and containing the line vecr = vecb + lambda vecc is

If vecb is not perpendicular to vecc . Then find the vector vecr satisfying the equation vecr xx vecb = veca xx vecb and vecr. vecc=0

If vecb is not perpendicular to vecc . Then find the vector vecr satisfying the equation vecr xx vecb = veca xx vecb and vecr. vecc=0

If veca is perpendicular to vecb and vecr is a non-zero vector such that pvecr + (vecr.vecb)veca=vecc , then vecr is:

If veca and vecb are mutually perpendicular unit vectors, vecr is a vector satisfying vecr.veca =0, vecr.vecb=1 and (vecr veca vecb)=1 , then vecr is:

Find vector vecr if vecr.veca=m and vecrxxvecb=vecc, where veca.vecb!=0

Let veca and vecb be two non- zero perpendicular vectors. A vector vecr satisfying the equation vecr xx vecb = veca can be

Prove that the perpendiculasr distanceof as point with position vector veca from the plane thorugh three points with position vectors vecb,vecc, vecd is ([veca vecc vecd]+[veca vecd vecb]+[veca vecb vecc]-[vecb vecc vecd])/(|vecbxxvecc+veccxxvecd+vecdxvecb|)

Show that the plane through the points veca,vecb,vecc has the equation [vecr vecb vecc]+[vecr vecc veca]+[vecr veca vecb]=[veca vecb vecc]

Let vecn be a unit vector perpendicular to the plane containing the point whose position vectors are veca, vecb and vecc and, if abs([(vecr -veca)(vecb-veca)vecn])=lambda abs(vecrxx(vecb-veca)-vecaxxvecb) then lambda is equal to