Q.

Let a unit vector u^=xi^+yj^+zk^ make angles π2,π3 and 2π3 with the vectors 12i^+12k^, 12j^+12k^ and 12i^+12j^ respectively. If v=12i^+12j^+12k^, then |u^v|2 is equal to          [2024]

1 9  
2 112  
3 7  
4 52  

Ans.

(4)

Let a=12i^+12k^b=12j^+12k^ and c=12i^+12j^

Now, u^·a=|u^|·|a|cosθ,

 x2+z2=1·1·cosπ2  x+z=0          ... (i)

Again u^·b=y2+z2=1·1·cosπ3  y+z=12          ... (ii)

and u^·c=x2+y2=1·1·cos2π3  x+y=12          ... (iii)

Solving (i), (ii) and (iii), we get

x=12, y=0, z=12       u^=12i^+12k^

So, |u^v|2=|2i^12j^|2=2+12=52.