A straighat line `vecr=veca+lamdavecb` meets the plane `vecr.vecn=p` in the point whose position vector is (A) `veca+((veca.hatn)/(vecb.hatn))vecb` (B) `veca+((p-veca.hatn)/(vecb.hatn))vecb` (C) `veca-((veca.hatn)/(vecb.hatn))vecb` (D) none of these

The angle theta the line vecr=vecr+lamdavecb and the plane vecr.hatn=d is given by (A) sin^-1((vecb.hatn)/(|vecn||vecb|)) (B) cos^-1((vecb.hatn)/(|vecb|)) (C) sin^-1((veca.hatn)/(|veca|)) (D) cos^-1((veca.hatn)/(|veca|))

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

For two vectors veca and vecb,veca,vecb=|veca||vecb| then (A) veca||vecb (B) veca_|_vecb (C) veca=vecb (D) none of these

If veca,vecb and vecc are three mutually perpendicular unit vectors then (vecr.veca)veca+(vecr.vecb)vecb+(vecr.vecc)vecc= (A) ([veca vecb vecc]vecr)/2 (B) vecr (C) 2[veca vecb vecc] (D) none of these

The resolved part of the vector veca along the vector vecb is veclamda and that perpendicular to vecb is vecmu . Then (A) veclamda=((veca.vecb).veca)/veca^2 (B) veclamda=((veca.vecb).vecb)/vecb^2 (C) vecmu=((vecb.vecb)veca-(veca.vecb)vecb)/vecb^2 (D) vecmu=(vecbxx(vecaxxvecb))/vecb^2

If vecasxxvecb=0 and veca.vecb=0 then (A) veca_|_vecb (B) veca||vecb (C) veca=0 and vecb=0 (D) veca=0 or vecb=0

Distance of the point P(vecp) from the line vecr=veca+lamdavecb is (a) |(veca-vecp)+(((vecp-veca).vecb)vecb)/(|vecb|^(2))| (b) |(vecb-vecp)+(((vecp-veca).vecb)vecb)/(|vecb|^(2))| (c) |(veca-vecp)+(((vecp-vecb).vecb)vecb)/(|vecb|^(2))| (d)none of these

If veca is parallel to vecb xx vecc, then (veca xx vecb) .(veca xx vecc) is equal to (a) |veca|^(2)(vecb.vecc) (b) |vecb|^(2)(veca .vecc) (c) |vecc|^(2)(veca.vecb) (d) none of these

If hatn is a unit vector in the direction of the vector vecA , them :

If hatn is a unit vector in the direction of the vector vecA , them :