Let a→=3i^–j^+2k^, b→=a→×(i^–2k^) and c→=b→×k^. Then the projection of c→–2j^ on a→ is : [2025]
(3)
Given: a→=3i^–j^+2k^
and b→=a→×(i^–2k^)=|i^j^k^3–1210–2|=2i^+8j^+k^
and c→=b^×k^=|i^j^k^281001|=8i^–2j^
⇒ c→–2j^=8i^–4j^
Now, Projection of c→–2j^ on a→ is given by
=(c→–2j^)·a^=(8i–4j)·(3i–j+2k)32+12+22
=24+4+014=2814=214.