Let a→=i^+2j^+3k^, b→=3i^+j^–k^ and c→ be three vectors such that c→ is coplanar with a→ and b→. If the vector c→ is perpendicular to b→ and a→·c→=5, then |c→| is equal to [2025]
(1)
|b→|=9+1+1=11, a→·b→=3+2–3=2
As c→ is perpendicular to b→ and coplanar with a→ and b→.
∴ c→=λ(b→×(a→×b→))
=λ((b→·b→)a→–(a→·b→)b→)=λ(11a→–2b→)
⇒ c→=λ(11i^+22j^+33k^–6i^–2j^+2k^)=5λ(i^+4j^+7k^)
As a→·c→=5 ⇒ 5λ(1+8+21)=5
⇒ λ=130 ∴ c→=16(i^+4j^+7k^)
⇒ |c→|=161+16+49=116