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

(vecA+2vecB).(2vecA-3vecB) :-

Find |veca| and |vecb| if (veca+vecb).(veca-vecb)=8 and |veca|=8|vecb|.

Find |veca| and |vecb| if (veca+vecb).(veca-vecb)=8 and |veca|=8|vecb| .

Prove that (veca+3vecb)xx(veca+vecb)+(3veca-5vecb)xx(veca-vecb)=0

Prove that (vecaxxvecb)^(2)= |(veca.veca" "veca.vecb),(veca.vecb" "vecb.vecb)|

Find |veca|and |vecb|, if (veca+vecb).(veca-vecb) =8 and |veca|=8|vecb|

For two vectors veca and vecb,veca,vecb=|veca||vecb| then (A) veca||vecb (B) veca_|_vecb (C) veca=vecb (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 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 the vectors veca, vecb, and vecc are coplanar show that |(veca,vecb,vecc),(veca.veca, veca.vecb,veca.vecc),(vecb.veca,vecb.vecb,vecb.vecc)|=0