Q.
Let a,b and c be non-coplanar unit vectors, equally inclined to one another at an angle θ. If a×b+b×c=pa+qb+rc, then find scalars pq and r in terms of θ.
Since, a,b,c are non-coplanar vectors. ⇒ [abc]=0
Also, a×b+b×c=pa+qb+rc
Taking dot product with a,b and c respectively both sides, we get p+qcosθ+rcosθ=[abc].....(i) pcosθ+q+rcosθ=0.....(ii)
and pcosθ+qcosθ+r=[abc].....(iii)
On adding above equations, p+q+r=2cosθ+12[abc]....(iv)
On multiplying Eq. (iv) by cosθ and subtracting Eq. (i), we get p(cosθ−1)=2cosθ+12[abc]cosθ−[abc] ⇒p=(1−cosθ)(2cosθ+1)[abc]
Similarly, q=(1+2cosθ)(1−cosθ)−2[abc]cosθ
and r=(1+2cosθ)(1−cosθ)[abc]
Now, [abc]2=∣∣a⋅ab⋅ac⋅aa⋅bb⋅bc⋅ba⋅cb⋅cc⋅c∣∣=∣∣1cosθcosθcosθ1cosθcosθcosθ1∣∣
Applying R1→R1+R2+R3 =(1+2cosθ)∣∣1cosθcosθ11cosθ1cosθ1∣∣
Applying C2→C2−C1,C3→C3−C1 =(1+2cosθ)⋅∣∣1cosθcosθ01−cosθ0001−cosθ∣∣ =(1+2cosθ)⋅(1−cosθ)2 ⇒[abc]=(1+2cosθ)⋅(1−cosθ) ∴p=1+2cosθ1 q=1+2cosθ−2cosθ
and r=1+2cosθ1