# At a given instant the position of a plane

At a given instant the position of a plane at A and a train at B are measured relative to a radar antenna at O. Determine the distance d between A and B at this instant. To solve the problem, formulate a position vector, directed from A to B, and then determine its magnitude. Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

#### Solution:

We will first find out the locations of point A and point B and write them in Cartesian vector form. To do so, we will use trigonometry. Notice how we can create a right angle triangle as highlighted in the image. Using sine and cosine functions, we can figure out the z-component of point A (the height) and F’. F’ is the force that lies on the xy plane.

z-component = $5\sin(60^0)\,=\,4.33$ km

$F'\,=\,5\cos(60^0)\,=\,2.5$ km

(Remember, sin is opposite over hypotenuse, and cosine is adjacent over hypotenuse. In this triangle, the hypotenuse is 5)

Next, we will figure out the x and y components of point A. In this new triangle we highlighted, notice how F’ is now the hypotenuse. We will use F’ and trigonometry to figure out the x and y components.

x-component = $2.5\cos(35^0)\,=\,2.05$ km

y-component = $2.5\sin(35^0)\,=\,1.43$ km

(remember we found that F’=2.5)

We can now write point A in Cartesian vector form.

$A:(-2.05i-1.43j+4.33k)$ km

(Note the negative signs in front of the i(x-term) and the j(y-term). This is because from the diagram, we see that point A lies on the negative quadrant of the x and y axes)

We will now focus on point B. Again, we will use trigonometry to figure out the new F’ in this triangle, and the z-component (height).

z-component = $2\sin(25^0)\,=\,0.84$ km

$F'\,=\,2\cos(25^0)\,=\,1.81$ km

We will now figure out the x and y-components. As before, note how F’ is now the hypotenuse of the triangle we just formed.

x-component = $1.81\sin(40^0)\,=\,1.16$ km

y-component = $1.81\cos(40^0)\,=\,1.39$ km

(remember we found that F’=1.81)

Let us write point B in Cartesian vector form.

$B:(1.16i+1.39j-0.84k)$ km

(Note the negative signs in front of the k(z-term). Again, this is because from the diagram, we see that point B lies below the xy plane. In other words, it is in the negative z-axis.)

We can now figure out the position vector, $r_{AB}$.

$r_{AB}\,=\,\left\{(1.16-(-2.05))i+(1.39-(-1.43))j+(-0.84-4.33)k\right\}$

$r_{AB}\,=\,\left\{3.21i+2.82j-5.17k\right\}$ km

A position vector, denoted $\mathbf{r}$ is a vector beginning from one point and extending to another point. It is calculated by subtracting the corresponding vector coordinates of one point from the other. If the coordinates of point A was $(x_A,y_A,z_A)$ and the coordinates of point B was$(x_B,y_B,z_B)$, then $r_{AB}\,=\,(x_B-x_A)i+(y_B-y_A)j+(z_B-z_A)k$

The distance between the two points is equal to the magnitude of the position vector $r_{AB}$.

magnitude of $r_{AB}\,=\,\sqrt{(3.21)^2+(2.82)^2+(-5.17)^2}$

magnitude of $r_{AB}\,=\,6.71$ km

The magnitude is equal to the square root of the sum of the squares of the vector. If the position vector was $r\,=\,ai+bj+ck$, then the magnitude would be, $r_{magnitude}\,=\,\sqrt{(a^2)+(b^2)+(c^2)}$. In the simplest sense, you take each term of a vector, square it, add it together, and then take the square root of that value.