# Determine the angle ϴ between the sides 1

Determine the angle ϴ between the sides of the triangular plate.

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

#### Solution:

Let us first determine the locations of points A, B and C and write them in Cartesian vector form.

Using the diagram, we see that the locations of the points are:

$A:(0i+1j+1k)$ m

$B:(0i+3j+4k)$ m

$C:(3i+5j+0k)$ m

Next, we will find the position vectors for points from A to B and A to C.

$r_{AB}\,=\,\left\{(0-0)i+(3-1)j+(4-1)k\right\}$

$r_{AB}\,=\,\left\{0i+2j+3k\right\}$ m

$r_{AC}\,=\,\left\{(3-0)i+(5-1)j+(0-1)k\right\}$

$r_{AC}\,=\,\left\{3i+4j-1k\right\}$ 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 each position vector.

magnitude of $r_{AB}\,=\,\sqrt{(0)^2+(2)^2+(3)^2}\,=\,3.6$ m

magnitude of $r_{AC}\,=\,\sqrt{(3)^2+(4)^2+(-1)^2}\,=\,5.1$ 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.

We will now find the dot product between the two position vectors.

$r_{AB}\cdot r_{AC}\,=\,\left[(0)(3)+(2)(4)+(3)(-1)\right]$

$r_{AB}\cdot r_{AC}\,=\,5$

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, we can solve for the angle, again, using the dot product.

$\theta\,=\,\cos^{-1}\left(\dfrac{5}{(3.6)(5.1)}\right)$

$\theta\,=\,74.2^0$

$\theta\,=\,74.2^0$