Q.

Let a unit vector which makes an angle of 60° with 2i^+2j^–k^ and an angle of 45° with i^–k^ be C→. Then C→+(–12i^+132j^–23k^) is          [2024]

1 23i^+132j^–12k^  
2 –23i^+23j^+(12+223)k^  
3 23i^–12k^  
4 (13+12)i^+(13–132)j^+(13+23)k^  

Ans.

(3)

Let C→=ai^+bj^+ck^, then we have |C→|=1

⇒ a2+b2+c2=1          ... (i)

Angle between C→ and  2i^+2j^–k^ is 60°.

∴  (ai^+bj^+ck^)·(2i^+2j^–k^)=(4+4+1) cos 60°          (∵  a2+b2+c2=1)

⇒ 2a+2b–c=32          ... (ii)

Angle between C→ and i^–k^ is 45°.

∴  (ai^+bj^+ck^)·(i^–k^)=1+0+1 cos 45°

⇒ a–c=1          ... (iii)

Solving (i), (ii) and (iii), we get a+2b=12 and a2+b2+(a–1)2=1

⇒ 2a2–2a+b2=0 ⇒ 2a2–2a+(2a–14)2=0

⇒ 32a2–32a+4a2–4a+1=0 ⇒ 36a2–36a+1=0

⇒ a=36±(36)2–4(36)2×36=12±23

∴   b=1–2a4 ⇒ b=1–1∓2234=∓132

∴   c=–12±23

∴   C→+(–12i^+132j^–23k^)=23i^–12k^