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)/(|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

Show that (veca xx vecb)^(2) = |veca| ^(2) |vecb|^(2) - (veca.vecb)^(2) = |(veca.veca)/(veca. vecb)(veca.vecb)/(vecb.vecb)|

The points with position vectors veca + vecb, veca-vecb and veca +k vecb are collinear for all real values of k.

If vecr.veca=vecr.vecb=vecr.vecc=1/2 for some non zero vector vecr and veca,vecb,vecc are non coplanar, then the area of the triangle whose vertices are A(veca),B(vecb) and C(vecc0 is (A) |[veca vecb vecc]| (B) |vecr| (C) |[veca vecb vecr]vecr| (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

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

The vector vecA and vecB are such that |vecA+vecB|=|vecA-vecB| . The angle between vectors vecA and vecB is -

If hatn is the unit vector in the direction of vecA then hatn is equal to