# Determine the angle ϴ between the pipe segments 2

Determine the angle ϴ between the pipe segments BA and BC. Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

#### Solution:

We will write two position vectors, from points B to A and B to C. To do so, we must first write the locations of points A, B and C in Cartesian vector form. Using the diagram, the locations of the points are:

$A:(0i+0j+0k)$ m

$B:(1.5i+2j+0k)$ m

$C:(3.5i+3j-2k)$ m

We can now write our position vectors.

$r_{BA}\,=\,\left\{(0-1.5)i+(0-2)j+(0-0)k\right\}$

$r_{BA}\,=\,\left\{-1.5i-2j+0k\right\}$ m

$r_{BC}\,=\,\left\{(3.5-1.5)i+(3-2)j+(-2-0)k\right\}$

$r_{BC}\,=\,\left\{2i+1j-2k\right\}$ m

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 $r_{AB}\,=\,(x_B-x_A)i+(y_B-y_A)j+(z_B-z_A)k$

Let us now find the magnitude of our position vectors.

magnitude of $r_{BA}\,=\,\sqrt{(-1.5)^2+(-2)^2+(0)^2}\,=\,2.5$ m

magnitude of $r_{BC}\,=\,\sqrt{(2)^2+(1)^2+(-2)^2}\,=\,3$ m

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.

Next, we will find the dot product between the two position vectors.

$r_{BA}\cdot r_{BC}\,=\,\left[(-1.5)(2)+(-2)(1)+(0)(-2)\right]$

$r_{BA}\cdot r_{BC}\,=\,-5$

The dot product is the product of the magnitudes of two vectors and the cosine angle $\theta$ between them. For example, if we had two vectors, $\mathbf{A}$ and $\mathbf{B}$, the dot product of the two is: $A\cdot B\,=\,AB\cos \theta$. In cartesian vector form, $A\cdot B\,=\,A_xB_x+A_yB_y+A_zB_z$. The angle formed between two vectors can be found by: $\theta\,=\,\cos^{-1}\left(\dfrac{A_xB_x+A_yB_y+A_zB_z}{(\text{magnitude of}\,A)(\text{magnitude of}\,B)}\right)$

Now that we know the magnitudes of each position vector, and the dot product value of our position vectors, we can find the angle between the two pipe segments.

$\theta\,=\,\cos^{-1}\left(\dfrac{-5}{(2.5)(3)}\right)$

$\theta\,=\,131.8^0$

$\theta\,=\,131.8^0$

## 2 thoughts on “Determine the angle ϴ between the pipe segments”

• mike

With great appreciation

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