Question

if C is the vector sum of A and B C=A+B what must be true about directions and magnitude of A and B if C=A+B? what must be true about the directions and magnitude of A and B if C=0​

Answer 1

a) Magnitudes: $\| \vec A\| \ge 0$, $\|\vec B\| \ge 0$, $\|\vec C\| \ge 0$; Directions: $\theta_{A} \in (-\infty, +\infty)$ for $\|\vec A\|\ne 0$. Undefined for $\|\vec A\| = 0$, $\theta_{B} \in (-\infty, +\infty)$ for $\|\vec B\|\ne 0$. Undefined for $\|\vec B\| = 0$, $\theta_{C} \in (-\infty, +\infty)$ for $\|\vec C\|\ne 0$. Undefined for $\|\vec C\| = 0$.

b) Magnitudes: $\|\vec A\| \ge 0$, $\|\vec B\| \ge 0$, $\|\vec C\| = 0$; Directions: $|\theta_{A}-\theta_{B}| = 180^{\circ}$, $\theta_{C}$ is undefined.

a) Let suppose that $\vec A \ne \vec O$, $\vec B \ne \vec O$ and $\vec C \ne \vec O$, where $\vec O$ is known as Vector Zero. By definitions of Dot Product and Inverse Trigonometric Functions we derive expression for the magnitude and directions of $\vec A$, $\vec B$ and $\vec C$:

Magnitude ($\vec A$)

$\|\vec A\| = \sqrt{\vec A\,\bullet\,\vec A}$

$\| \vec A\| \ge 0$

Magnitude ($\vec B$)

$\|\vec B\| = \sqrt{\vec B\,\bullet\,\vec B}$

$\|\vec B\| \ge 0$

Magnitude ($\vec C$)

$\|\vec C\| = \sqrt{\vec C\,\bullet \,\vec C}$

$\|\vec C\| \ge 0$

Direction ($\vec A$)

$\vec A \,\bullet \,\vec u = \|\vec A\|\cdot \|u\|\cdot \cos \theta_{A}$

$\theta_{A} = \cos^{-1} \frac{\vec A\,\bullet\,\vec u}{\|\vec A\|\cdot \|u\|}$

$\theta_{A} = \cos^{-1} \frac{\vec A\,\bullet\,\vec u}{\|\vec A\|}$

$\theta_{A} \in (-\infty, +\infty)$ for $\|\vec A\|\ne 0$. Undefined for $\|\vec A\| = 0$.

Direction ($\vec B$)

$\vec B\,\bullet \, \vec u = \|\vec B\|\cdot \|\vec u\| \cdot \cos \theta_{B}$

$\theta_{B} = \cos^{-1} \frac{\vec B\,\bullet\,\vec u}{\|\vec B\|\cdot \|\vec u\|}$

$\theta_{B} = \cos^{-1} \frac{\vec B\,\bullet\,\vec u}{\|\vec B\|}$

$\theta_{B} \in (-\infty, +\infty)$ for $\|\vec B\|\ne 0$. Undefined for $\|\vec B\| = 0$.

Direction ($\vec C$)

$\vec C \,\bullet\,\vec u = \|\vec C\|\cdot\|\vec u\|\cdot \cos \theta_{C}$

$\theta_{C} = \cos^{-1}\frac{\vec C\,\bullet\,\vec u}{\|\vec C\|\cdot\|\vec u\|}$

$\theta_{C} = \cos^{-1} \frac{\vec C\,\bullet\,\vec u}{\|\vec C\|}$

$\theta_{C} \in (-\infty, +\infty)$ for $\|\vec C\|\ne 0$. Undefined for $\|\vec C\| = 0$.

Please notice that $\vec u$ is the Vector Unit.

b) Let suppose that $\vec A \ne \vec O$ and $\vec B \ne \vec O$ and $\vec C = \vec O$. Hence, $\vec A = -\vec B$. In other words, we find that both vectors are antiparallel to each other, that is, that angle between $\vec A$ and $\vec B$ is 180°. From a) we understand that $\|\vec A\| \ge 0$, $\|\vec B\| \ge 0$, but $\|\vec C\| = 0$.

Then, we have the following conclusions:

Magnitude ($\vec A$)

$\|\vec A\| \ge 0$

Magnitude ($\vec B$)

$\|\vec B\| \ge 0$

Magnitude ($\vec C$)

$\|\vec C\| = 0$

Directions ($\vec A$, $\vec B$):

$|\theta_{A}-\theta_{B}| = 180^{\circ}$

Direction ($\vec C$):

Undefined

Answer 2

The vector sum is the algebraic sum if the two vectors have the same direction.

The sum vector is zero if the two vectors have the same magnitude and opposite direction

Vector addition is a process that can be performed graphically using the parallelogram method, see  attached, where the second vector is placed at the tip of the first and the vector sum goes from the origin of the first vector to the tip of the second.

There are two special cases where the vector sum can be reduced to the algebraic sum if the vectors are parallel

case 1. if the two vectors are parallel, the sum vector has the magnitude of the sum of the magnitudes of each vector

case 2. If the two vectors are antiparallel and the magnitude of the two vectors is the same, the sum gives zero.

In summary in the sum of vectors If the vectors are parallel it is reduced to the algebraic sum, also in the case of equal magnitude and opposite direction the sum is the null vector