vectors

At one instant ,ships X and y are at a distance of d from each other.THe velocities of ship x and ship y are u and v respectively. Angle a and b are acute. Find the tangent of angle betweeen the direction of relative velocity and $\displaystyle \vec{xy}$

My work:

the relative velocity vector is

xVy=Vx-Vy=(u+vsin(a+b))i+(-v cosb)j

relative displacement vector, xRy=(-d)i

Before i proceed with the dot product formulas, can i confirm whether the above is crrect?

Quote:

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Originally Posted by hooke

At one instant ,ships X and y are at a distance of d from each other.THe velocities of ship x and ship y are u and v respectively. Angle a and b are acute. Find the tangent of angle betweeen the direction of relative velocity and $\displaystyle \vec{xy}$

My work:

the relative velocity vector is

xVy=Vx-Vy=(u+vsin(a+b))i+(-v cosb)j

relative displacement vector, xRy=(-d)i

Before i proceed with the dot product formulas, can i confirm whether the above is crrect?

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Dear hooke,

Your expressions are incorrect.

According to your figure,

Velocity of x relative to y=velocity of x relative to earth-velocity of y relative to earth= $\displaystyle u\underline{i}-(-v\cos{(a+b)}\underline{i}+v\sin{(a+b)}\underline{j })$

Displacement of x relative to y = $\displaystyle (-d\cos{a})\underline{i}+(d\sin{a})\underline{j}$

Hope this helps.