Q.

If a→ is a non-zero vector such its projections on the vectors 2i^–j^+2k^, i^+2j^–2k^ and k^ are equal, then a unit vector along a→ is:          [2025]

1 1155(7i^+9j^+5k^)  
2 1155(7i^+9j^–5k^)  
3 1155(–7i^+9j^–5k^)  
4 1155(–7i^+9j^+5k^)  

Ans.

(1)

Let a→=xi^+yj^+zk^,u→=2i^–j^+2k^, v→=i^+2j^–2k^ and w→=k^

When, a→·u→|u→|=a→·v→|v→|=a→·w→|w→|

Now, a→·u→|u→|=a→·w→|w→|

⇒ (xi^+yj^+zk^)·(2i^–j^+2k^)3=(xi^+yj^+zk^)·(k^)1

⇒ 2x–y+2z=3z

⇒ 2x–y–z=0          ... (i)

Also, a→·v→|v→|=a→·w→|w→|

⇒ x+2y–2z3=z

⇒ x+2y–5z=0          ... (ii)

and a→·u→|u→|=a→·v→|v→|

⇒ 2x–y+2z=x+2y–2z

⇒ x–3y+4z=0          ... (iii)

From (i), (ii) and (iii), we get x = 7, y = 9 and z = 5

a→=7i^+9j^+5k^(7)2+(9)2+(5)2=7i^+9j^+5k^155