The vector $\hat i + x\hat j + 3\hat k$ is rotated through an angle A Λ and doubled in magnitude, then it becomes $4\hat i + (4x-2)\hat j + 2\hat k$. The value of x are

-2/3, 2

We have,

$2|\hat i +x\hat j+3\hat k|=|4\hat i + (4x-2)\hat j+2\hat k|$

$⇒2\sqrt{x^2+10}=\sqrt{20+ (4x-2)^2}$

$⇒4(x^2 + 10)=20+ (4x-2)^2$

$⇒3x^2-4x-4=0⇒x=2,-2/3$