Let $\vec b = 4\hat i+ 3\hat j$ and $\vec c$ be a vector perpendicular to $\vec b$ band lying in the xy-plane. A vector in the xy-plane having projections 1 and 2 along $\vec b$ and $\vec c$ is

$2\hat i-\hat j$

Let the required vector be $\vec a = x\hat i+y\hat j$. It is given that the vector $\vec c$ is in xy-plane and is perpendicular to $\vec b=4\hat i + 3\hat j$. Therefore,

$\vec c=λ(3\hat i-4\hat j)⇒\hat c= (3\hat i-4\hat j)$

Now, $\vec a.\hat b=1$ and $\vec a .\hat c = 2$

$⇒4x+3y=5$ and $3x-4y=10$

Solving these two equations, we get $x = 2, y = -1$.

Hence, $\vec b = 2\hat i-\hat j$