# In triangle ABC which of the following is not true:

Question:

In triangle ABC which of the following is not true:

A. $\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}$

B. $\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}-\overrightarrow{\mathrm{AC}}=\overrightarrow{0}$

C. $\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}-\overrightarrow{\mathrm{CA}}=\overrightarrow{0}$

D. $\overrightarrow{\mathrm{AB}}-\overrightarrow{\mathrm{CB}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}$

Solution:

On applying the triangle law of addition in the given triangle, we have:

$\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{AC}}$              ...(1)

$\Rightarrow \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=-\overrightarrow{\mathrm{CA}}$

$\Rightarrow \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}$         ...(2)

$\therefore$ The equation given in alternative $\mathrm{A}$ is true.

$\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{AC}}$

$\Rightarrow \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}-\overrightarrow{\mathrm{AC}}=\overrightarrow{0}$

$\therefore$ The equation given in alternative $B$ is true.

From equation $(2)$, we have:

$\overrightarrow{\mathrm{AB}}-\overrightarrow{\mathrm{CB}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}$

$\therefore$ The equation given in alternative $\mathrm{D}$ is true.

Now, consider the equation given in alternative $\mathrm{C}$ :

$\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}-\overrightarrow{\mathrm{CA}}=\overrightarrow{0}$

$\Rightarrow \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{CA}}$             $\ldots(3)$

From equations (1) and (3), we have:

$\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{CA}}$

$\Rightarrow \overrightarrow{\mathrm{AC}}=-\overrightarrow{\mathrm{AC}}$

$\Rightarrow \overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{AC}}=\overrightarrow{0}$

$\Rightarrow 2 \overrightarrow{\mathrm{AC}}=\overrightarrow{0}$

$\Rightarrow \overrightarrow{\mathrm{AC}}=\overrightarrow{0}$, which is not true.

Hence, the equation given in alternative $C$ is incorrect.

The correct answer is $\mathbf{C}$.