Two particles are projected simultaneously in the same vertical plane, from the same point, but with different speeds and at different angles to the horizontal. The path followed by one, as seen by the other, till both are in air is

The correct option is B A straight line making a constant angle $(\ne {90}^{\circ })$ with the horizontal
The velocities of the two particles after time $t$ are
$\overrightarrow{v_1}=\left(u_1\cos {\theta }_1\right)\hat{i}+(u_1\sin {\theta }_1-gt)\hat{j}$ and $\overrightarrow{v_2}=\left(u_2\cos {\theta }_2\right)\hat{i}+(u_2\sin {\theta }_2-gt)\hat{j}$
Also, $\overrightarrow{a_1}=-g\hat{k}$ and $\overrightarrow{a_2}=-g\hat{k}$

$\because$ $\overrightarrow{a_{12}}=0$, there is no relative acceleration between the particles.
Therefore, relative velocity $\overrightarrow{v_{12}}=(u_1\cos {\theta }_1-u_2\cos {\theta }_2)\hat{i}+(u_1\sin {\theta }_1-u_2\sin {\theta }_2)\hat{j}$
$\therefore \overrightarrow{v_{12}}=\text{constant}$
So, path of one projectile w.r.t another projectile is always a straight line.
The straight line path can be in any direction depending on initial direction and velocity of both projectiles.