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**Hint:**The x,y and z are the three directions which are mutually perpendicular to each other, the particle A is in plane x-z and the particle B is in the plane y-z. The relative velocity is the difference between the velocities of the two bodies.

**Formula used:**

The relative velocity of the two particles is given by,

$ \Rightarrow {v_B} - {v_A}$

Where the velocity of particle B is ${v_B}$ and the velocity of particle A is${v_A}$.

**Complete step by step solution:**

In this problem it is given that a particle ‘A’ is projected with velocity $4\sqrt 2 m{s^{ - 1}}$ making an angle $45^\circ $ to the horizontal. Particle B is projected at $5m{s^{ - 1}}$ an angle $\theta = {\tan ^{ - 1}}\left( {\dfrac{1}{3}} \right)$ to y axis in y-z plane then we need to find the velocity of B wrt A.

According to the condition the figure will be.

The velocity of particle A is ${v_A} = 4\hat i + 4\hat k$ and the velocity of particle B is equal to ${v_B} = 3\hat j + 4\hat k$.

The relative velocity of the particle B with respect to particle A is equal to,

$ \Rightarrow {v_B} - {v_A} = \left( {3\hat j + 4\hat k} \right) - \left( {4\hat i + 4\hat k} \right)$

$ \Rightarrow {v_B} - {v_A} = 3\hat j + 4\hat k - 4\hat i - 4\hat k$

$ \Rightarrow {v_B} - {v_A} = 3\hat j + 4\hat k - 4\hat i - 4\hat k$

$ \Rightarrow {v_B} - {v_A} = 3\hat j - 4\hat i$

The magnitude of the velocity is equal to,

$ \Rightarrow {v_B} - {v_A} = \sqrt {{3^2} + {4^2}} $

$ \Rightarrow {v_B} - {v_A} = \sqrt {9 + 16} $

$ \Rightarrow {v_B} - {v_A} = \sqrt {25} $

$ \Rightarrow {v_B} - {v_A} = 5m{s^{ - 1}}$

Which means option A is correct. The magnitude of the acceleration will not change with time and therefore the option B is wrong.

The relative velocity is ${v_B} - {v_A} = 3\hat j - 4\hat i$ which means it is in the x-y plane, the option C is also correct. The angle of the initially equal to,

$ \Rightarrow \theta = {\tan ^{ - 1}}\left( { - \dfrac{4}{3}} \right)$

As the angle is negative and therefore we add $\dfrac{\pi }{2}$ so the angle becomes $\theta + \dfrac{\pi }{2}$, so the option D is correct.

**The wrong option is option B, so the answer for this problem is option B.**

**Note:**The students are advised to remember the formula of the relative velocity and also the diagram of the velocity of the particle A and particle B should be drawn very carefully as the answer is very dependent on the diagram.

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