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

#### Solution:

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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

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

#### Final Answer:

\theta\,=\,74.2^0

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