Finding the Tangent of an Angle Between Vectors Using the Dot Product

[thereddevils]

Homework Statement

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 \vec{XY}

Homework Equations

The Attempt at a Solution

Isn't the \vec{XY} the same as the relative velocity vector? I need more hints on this problem. Thanks.

 

[Rasalhague]

No, that's a displacement vector. It points in the direction from X to Y, and its length is the distance between X and Y.

Your first problem is to find relative velocity in terms of u and v. Imagine you're on board one ship and think of what the velocity of the other ship would be relative to you. What happens to the velocity of the other ship if you switch to a frame of reference in which your ship's velocity is zero?

For the angle, are you familiar with the dot product and cross product of vectors and how they can be defined using trigonometric functions?

 

[thereddevils]

No, that's a displacement vector. It points in the direction from X to Y, and its length is the distance between X and Y.

Your first problem is to find relative velocity in terms of u and v. Imagine you're on board one ship and think of what the velocity of the other ship would be relative to you. What happens to the velocity of the other ship if you switch to a frame of reference in which your ship's velocity is zero?

For the angle, are you familiar with the dot product and cross product of vectors and how they can be defined using trigonometric functions?

I am confused, for the relative velocity i found it to be

\sqrt{v^2+u^2+2uv\cos (a+b)}

How can the relative velocity vector be different from the displacement vector XY. I know they are theoritically different, but how do you represent it on the diagram? Perhaps you can post a diagram here. Thanks

 

[Rasalhague]

Okay, here's a diagram. You could take either ship as being at rest, but suppose we take X, then the relative velocity will be the velocity of ship Y in a reference frame where X has zero velocity. To convert velocities from the reference frame you start with to that one, just subtract u. Does that make sense?

Some formulas you could use in the next step. The dot product:

\textbf{p}\cdot \textbf{q}=\left | \textbf{p} \right | \; \left | \textbf{q} \right | \; \cos \theta

where p and q are vectors, |p| and |q| their magnitudes,

\left | \textbf{p} \right |=\sqrt{\textbf{p}\cdot \textbf{p}}

and theta the angle between them. The cross product:

\textbf{p}\times \textbf{q}=\left | \textbf{p} \right | \; \left | \textbf{q} \right | \; \sin \theta \; \hat{\textbf{n}}

where \hat{\textbf{n}} is a unit vector at right angles to p and q, so that

\hat{\textbf{n}} \cdot \hat{\textbf{n}} = 1

By the way, one problem with your calculation of the relative velocity is that the value you got is a number rather than a 2-dimensional vector: it doesn't specify a direction in the plane, so we couldn't say what angle it would make with the vector XY.

 

[thereddevils]

Okay, here's a diagram. You could take either ship as being at rest, but suppose we take X, then the relative velocity will be the velocity of ship Y in a reference frame where X has zero velocity. To convert velocities from the reference frame you start with to that one, just subtract u. Does that make sense?

Some formulas you could use in the next step. The dot product:

\textbf{p}\cdot \textbf{q}=\left | \textbf{p} \right | \; \left | \textbf{q} \right | \; \cos \theta

where p and q are vectors, |p| and |q| their magnitudes,

\left | \textbf{p} \right |=\sqrt{\textbf{p}\cdot \textbf{p}}

and theta the angle between them. The cross product:

\textbf{p}\times \textbf{q}=\left | \textbf{p} \right | \; \left | \textbf{q} \right | \; \sin \theta \; \hat{\textbf{n}}

where \hat{\textbf{n}} is a unit vector at right angles to p and q, so that

\hat{\textbf{n}} \cdot \hat{\textbf{n}} = 1

By the way, one problem with your calculation of the relative velocity is that the value you got is a number rather than a 2-dimensional vector: it doesn't specify a direction in the plane, so we couldn't say what angle it would make with the vector XY.

ok

Is my relative velocity correct?

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

 

[Rasalhague]

ok

Is my relative velocity correct?

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

I don't understand the notation in your answer. v-u is the velocity of ship Y relative to ship X (that's the unlabelled vector in my diagram). Alternatively, you could use the velocity of ship X relative to ship Y, which is u-v.

 

[thereddevils]

I don't understand the notation in your answer. v-u is the velocity of ship Y relative to ship X (that's the unlabelled vector in my diagram). Alternatively, you could use the velocity of ship X relative to ship Y, which is u-v.

the vector is not in the direction of i and j instead its placed at angle, don't we need to resolve first?

Also, if the relative velocity is simply u-v ,then the answer wouldn't be interms of a and b too. Btw the answer given is in terms of u,v,a and b.

 

[Rasalhague]

the vector is not in the direction of i and j instead its placed at angle, don't we need to resolve first?

Also, if the relative velocity is simply u-v ,then the answer wouldn't be interms of a and b too. Btw the answer given is in terms of u,v,a and b.

Whether a vector is in the direction of i or j depends on which directions i and j point in. They aren't mentioned in the problem statement, so I guess that's up to you. A convenient choice might be to make i parallel to the displacement vector from ship X to ship Y. Then

\textbf{i}=\frac{\vec{XY}}{|\vec{XY}|} \enspace\enspace\enspace \textbf{j}=\frac{\textbf{u}-(\textbf{u}\cdot\textbf{i})\textbf{i}}{|\textbf{u}-(\textbf{u}\cdot\textbf{i})\textbf{i}|}

and

\textbf{u}=|\textbf{u}| \, \cos(a) \, \textbf{i}+|\textbf{u}|\,\sin(a) \,\textbf{j}

\textbf{v}=-|\textbf{v}| \, \cos(b) \, \textbf{i}+|\textbf{v}|\,\sin(b) \,\textbf{j}

Then you could define a relative velocity r, with respect to ship X, as

\textbf{r}=\textbf{v}-\textbf{u}=-(|\textbf{v}|\cos(b) + |\textbf{u}| \cos(a) ) \textbf{i}+(|\textbf{v}|\sin(b)-|\textbf{u}|\sin(a)) \,\textbf{j}

and, having defined basis vectors i and j, you can get the tangent of the angle you're after using just the dot product and the definition of the tangent of an angle as its sine divided by its cosine.