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.
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.
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