Question

trying to find |A⃗ +B⃗ +C⃗ | = from a group of arbitary vectors

Answer 1

If the new vector is defined as:
\[\vec{V} = \vec{A} + \vec{B} + \vec{C} \]
\[\vec{V} = \langle a_x+b_x+c_x, a_y+by+c_y, ...\rangle \] Or generally, each orthogonal componnt of V can be written as:
\[ v_i = a_i + b_i + v_i\]

then \[|\vec{V}| = |\vec{A}+\vec{B}+\vec{C}| = \left ( \sum_{i=1}^n v_i^2 \right ) ^\frac{1}{2} =\left ( \sum_{i=1}^n a_i+b_i+c_i \right ) ^\frac{1}{2} \]

If its in 3D euclidean, n = 3.

Answer 2

typo in the last bit:
\[\left (\sum_{i=1}^n (a_i+b_i+c_i)^2 \right ) ^{\frac{1}{2}}\]

Answer 3

can that also be used just to find the magnitude of the vectors?

Answer 4

Yes, for euclidean vectors that will return the magnitude of a vector in general. But since v was the sum of three vectors, we can just add their components and then take the distance function shown.

Answer 5

because i tried to simply add the magnetudes and it threw me a wrong answer

Answer 6

That is correct! The sum of the magnitudes is the the same as the magnitude of the sum. In three dimensions, the equation I wrote would produce:

\[\sqrt{(a_x+b_x+c_x)^2 + (a_y+b_y + c_y)^2 + (a_z+b_z+c_z)^2 } \]

Note that this is not equal to:

\[ \sqrt{(a_x+b_x+c_x)^2 + (a_y+b_y + c_y)^2 + (a_z+b_z+c_z)^2 } \neq \\ \sqrt{a_x^2 + a_y^2 + a_z^2} + \sqrt{b_x^2 + b_y^2 + b_z^2} +\sqrt{x_x^2 + x_y^2 + x_z^2}\]