Vector operation of addition

Question

According to Law of Chasles, we know that:

$\vec{AB}+\vec{BC} = \vec{AC}$

However, for this question, it makes me confused about this law:

The question is to calculate $\vec{AB}-\vec{CD}+\vec{AD}+4\vec{BA}-\vec{BC}$

So if we calculate this according to the law, the procedure would be following:

• $\vec{AB}+\vec{DC}+\vec{AD}+4\vec{BA}+\vec{CB}$
• $\vec{AB}+\vec{AC}+4\vec{BA}+\vec{CB}$
• $\vec{AB}+4\vec{BA}+\vec{AB}$
• $\vec{AB}$ (because $4\vec{BA}+\vec{AB} = 4\vec{BB} = 0$ )

However, the right answer is $2\vec{BA}$, which I obtained using another way of calculation:

• $\vec{AB}+\vec{DC}+\vec{AD}+4\vec{BA}+\vec{CB}$
• $\vec{AB}+\vec{AC}+4\vec{BA}+\vec{CB}$
• $\vec{AB}+4\vec{BA}+\vec{AB}$
• $2\vec{AB}+4\vec{BA}$
• $2\vec{BA}$ (because $2\vec{AB}+4\vec{BA} = 4\vec{BA} - 2\vec{BA}$ )

So the question here is why the first method is not right?

Answer 1

If im not mistaken your conclusion is wrong in the last bullet is where it fails. Notice that: $$\vec{BA}=-\vec{AB}$$

So: $$4\vec{BA}+\vec{AB} \neq 4\vec{BB}$$

It is nothing but $-3\vec{AB}$. Instead we van calculate and get: $$\vec{AB}+4\vec{BA}+\vec{AB}=2\vec{AB}-4\vec{AB}=-2\vec{AB}=2\vec{BA}$$

As required.

Answer 2

$$\vec{AB}-\vec{CD}+\vec{AD}+4\vec{BA}-\vec{BC}=$$

$$ AB+DC+AD+4BA+CB=$$

$$ AB + ( AD+DC+CB)+4BA=$$

$$AB+AB+4BA=2BA$$