If `overset(rarr)A+overset(rarr)B+overset(rarr)C` =0 and A = B + C, the angle between `overset(rarr)A` and `overset(rarr)B` is :

Options

A. 0
B. `(pi)/(4)`
C. `pi`
D. `(pi)/(2)`

Explanation

To solve the problem, we start with the given equations:

1. \(\overset{\rarr}{A} + \overset{\rarr}{B} + \overset{\rarr}{C} = 0\)
2. \(\overset{\rarr}{A} = \overset{\rarr}{B} + \overset{\rarr}{C}\)

From the first equation, we can express \(\overset{\rarr}{C}\) in terms of \(\overset{\rarr}{A}\) and \(\overset{\rarr}{B}\):

\[ \overset{\rarr}{C} = -(\overset{\rarr}{A} + \overset{\rarr}{B}) \]

Now, we can use the second equation to substitute \(\overset{\rarr}{C}\):

\[ \overset{\rarr}{C} = \overset{\rarr}{A} - \overset{\rarr}{B} \]

Now we have two expressions for \(\overset{\rarr}{C}\):

1. \(\overset{\rarr}{C} = -(\overset{\rarr}{A} + \overset{\rarr}{B})\)
2. \(\overset{\rarr}{C} = \overset{\rarr}{A} - \overset{\rarr}{B}\)

Setting these two expressions equal to each other gives us:

\[ -(\overset{\rarr}{A} + \overset{\rarr}{B}) = \overset{\rarr}{A} - \overset{\rarr}{B} \]

Now, rearranging this equation:

\[ -\overset{\rarr}{A} - \overset{\rarr}{B} = \overset{\rarr}{A} - \overset{\rarr}{B} \]

Adding \(\overset{\rarr}{A}\) and \(\overset{\rarr}{B}\) to both sides:

\[ 0 = 2\overset{\rarr}{A} \]

This implies that \(\overset{\rarr}{A} = 0\), which is not possible unless the vectors are in a specific configuration.

Next, we will use the magnitudes of the vectors. Taking the magnitudes of both sides of the equation \(\overset{\rarr}{A} + \overset{\rarr}{B} + \overset{\rarr}{C} = 0\):

\[ |\overset{\rarr}{C}|^2 = |\overset{\rarr}{A}|^2 + |\overset{\rarr}{B}|^2 + 2 |\overset{\rarr}{A}||\overset{\rarr}{B}|\cos(\theta) \]

Where \(\theta\) is the angle between \(\overset{\rarr}{A}\) and \(\overset{\rarr}{B}\).

From the second equation, we can also express \(|\overset{\rarr}{C}|\):

\[ |\overset{\rarr}{C}| = |\overset{\rarr}{A}| - |\overset{\rarr}{B}| \]

Squaring both sides gives:

\[ |\overset{\rarr}{C}|^2 = (|\overset{\rarr}{A}| - |\overset{\rarr}{B}|)^2 = |\overset{\rarr}{A}|^2 - 2|\overset{\rarr}{A}||\overset{\rarr}{B}| + |\overset{\rarr}{B}|^2 \]

Now we can equate the two expressions for \(|\overset{\rarr}{C}|^2\):

\[ |\overset{\rarr}{A}|^2 - 2|\overset{\rarr}{A}||\overset{\rarr}{B}| + |\overset{\rarr}{B}|^2 = |\overset{\rarr}{A}|^2 + |\overset{\rarr}{B}|^2 + 2 |\overset{\rarr}{A}||\overset{\rarr}{B}|\cos(\theta) \]

This simplifies to:

\[ -2|\overset{\rarr}{A}||\overset{\rarr}{B}| = 2 |\overset{\rarr}{A}||\overset{\rarr}{B}|\cos(\theta) \]

Dividing both sides by \(2|\overset{\rarr}{A}||\overset{\rarr}{B}|\) (assuming they are not zero):

\[ -1 = \cos(\theta) \]

This implies:

\[ \theta = \pi \text{ radians} \text{ or } 180^\circ \]

Thus, the angle between \(\overset{\rarr}{A}\) and \(\overset{\rarr}{B}\) is \(180^\circ\).