If →a,→b,→c are three non- coplanar vectors for which [→a→b→c]≠→a′,→b′,→c′ constitute the reciprocal system of vectors, then any vector →r can be expressed as
A
→r=(→r×→a).→a′+(→r×→b).→b′+(→r×→c).→c′
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
→r=(→r×→a′).→a+(→r×→b′).→b+(→r×→c′).→c
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
→r=(→r.→a).→a′+(→r.→b).→b′+(→r.→c).→c′
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
→r=(→r.→a′)→a+(→r.→b′)→b+(→r.→c′)→c
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is C→r=(→r.→a′)→a+(→r.→b′)→b+(→r.→c′)→c Let r be expressed as a linear combination of the non-coplanar vector a,b,c in the form r=xa+yb+zc ...(1) where x,y,z are some scalar Multiplying both sides of (1) scalarly with b×c, we get r⋅(b×c)=xa⋅(b×c)+yb⋅(b×c)+zc⋅(b×c)=x[abc]+y[bbc]+z[cbc]=x[abc] Since [bbc]=0=[cbc] ∴x=r⋅(b×c)[abc]=r⋅(b×c)[abc]=r⋅a′, since a′=b×c[abc] Similarly multiplying both sides of (1) scalarly with c×a and a×b, we can show that y=r⋅b′ and z=r⋅c′ Substitute the values of x,y and z in (1) , we get r=(r⋅a′)a+(r⋅b′)b+(r⋅c′)c