If $\vec a,\vec b,\vec c$ are vectors such that $\vec a+\vec b+\vec c=0$ and $|\vec a| = 1,|\vec b|=2,|\vec c|= 5$, then the expression $\vec a.\vec b+\vec b.\vec c+\vec c.\vec a$ equals
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → -15
Given: $\vec a+\vec b+\vec c=0 \;\Rightarrow\; \vec c=-(\vec a+\vec b)$
Substitute $\vec c=-(\vec a+\vec b)$:
$\vec a\cdot\vec b+\vec b\cdot[-(\vec a+\vec b)] +[-(\vec a+\vec b)]\cdot\vec a$
$= \vec a\cdot\vec b - \vec b\cdot\vec a - \vec b\cdot\vec b - \vec a\cdot\vec a - \vec b\cdot\vec a$
Since $\vec a\cdot\vec b = \vec b\cdot\vec a$:
$= -(|\vec a|^{2} + |\vec b|^{2} + \vec a\cdot\vec b)$
Now use the given magnitudes:
$|\vec a|=1,\; |\vec b|=2,\; |\vec c|=5$
Also, $|\vec c|^{2} = |\vec a+\vec b|^{2}$
$25 = 1^{2}+2^{2} + 2(\vec a\cdot\vec b)$
$25 = 5 + 2(\vec a\cdot\vec b)$
$\vec a\cdot\vec b = 10$
So the required expression is:
$-(1^{2} + 2^{2} + 10) = -(1+4+10)= -15$
The expression equals $-15$.