In Δ ABC , Which of the following are True?(A) AB+BC+CA=0 (B) AB+BC-A…

In $ \Delta ABC $, Which of the following are True?

$Figure of triangle ABC$

(A) $ \overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\vec{0} $

(B) $ \overrightarrow{AB}+\overrightarrow{BC}-\overrightarrow{AC}=\vec{0} $

(D) $ \overrightarrow{AB}-\overrightarrow{CB}+\overrightarrow{CA}=\vec{0} $

Choose the correct answer from the options given below:

Answer & explanation

Correct answer: option 1

The correct answer is Option (1) → (A), (B) and (D) only

By the triangle law of vector addition,

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

Reversing a vector changes its sign:

$ \displaystyle \overrightarrow{CA}=-\overrightarrow{AC} $

$ \displaystyle \overrightarrow{CB}=-\overrightarrow{BC} $

So $ \overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC} $, $ \overrightarrow{CA}=-\overrightarrow{AC} $ and $ -\overrightarrow{CB}=\overrightarrow{BC} $.

Checking the statements

(A) $ \overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\vec{0} $ — Correct. The sum is $ \overrightarrow{AC}+\overrightarrow{CA}=\overrightarrow{AC}-\overrightarrow{AC}=\vec{0} $.

(B) $ \overrightarrow{AB}+\overrightarrow{BC}-\overrightarrow{AC}=\vec{0} $ — Correct. The sum is $ \overrightarrow{AC}-\overrightarrow{AC}=\vec{0} $.

(C) $ \overrightarrow{AB}+\overrightarrow{BC}-\overrightarrow{CA}=\vec{0} $ — Incorrect. The sum is $ \overrightarrow{AC}+\overrightarrow{AC}=2\overrightarrow{AC}\ne\vec{0} $.

(D) $ \overrightarrow{AB}-\overrightarrow{CB}+\overrightarrow{CA}=\vec{0} $ — Correct. With $ -\overrightarrow{CB}=\overrightarrow{BC} $ it is $ \overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\vec{0} $.