Question

Three vector abc satisfy the condition a + b + c is equal to zero evaluate the quantity musical to a into b

Answer 1

Vector Analysis

Complete question: Three vectors $\vec{a}$, $\vec{b}$ and $\vec{c}$ satisfy the condition $\vec{a}+\vec{b}+\vec{c}=\vec{0}$. Evaluate the quantity $\mu=\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}$, if $|\vec{a}|=1$, $|\vec{b}|=4$ and $|\vec{c}|=2$.

To remember:

• We must remember that vector dot product is commutative,
• $\quad\quad$i.e. $\vec{a}.\vec{b}=\vec{b}.\vec{a}$
• And: $\vec{a}.\vec{a}=|\vec{a}|^{2}$

Solution:

Given, $\vec{a}+\vec{b}+\vec{c}=\vec{0}\quad...(1)$

Taking dot product of $\vec{a}$ to both sides of $(1)$, we get

$\quad \vec{a}.\vec{a}+\vec{a}.\vec{b}+\vec{c}.\vec{a}=0$

$\Rightarrow \vec{a}.\vec{b}+\vec{c}.\vec{a}=-\vec{a}.\vec{a}$

$\Rightarrow \vec{a}.\vec{b}+\vec{c}.\vec{a}=-|\vec{a}|^{2}$

$\Rightarrow \vec{a}.\vec{b}+\vec{c}.\vec{a}=-1^{2}$

$\Rightarrow \vec{a}.\vec{b}+\vec{c}.\vec{a}=-1\quad...(2)$

Taking dot product of $\vec{b}$ to both sides of $(1)$, we get

$\quad \vec{a}.\vec{b}+\vec{b}.\vec{b}+\vec{b}.\vec{c}=0$

$\Rightarrow \vec{a}.\vec{b}+\vec{b}.\vec{c}=-\vec{b}.\vec{b}$

$\Rightarrow \vec{a}.\vec{b}+\vec{b}.\vec{c}=-|\vec{b}|^{2}$

$\Rightarrow \vec{a}.\vec{b}+\vec{b}.\vec{c}=-4^{2}$

$\Rightarrow \vec{a}.\vec{b}+\vec{b}.\vec{c}=-16\quad...(3)$

Taking dot product of $\vec{c}$ to both sides of $(1)$, we get

$\quad \vec{c}.\vec{a}+\vec{b}.\vec{c}+\vec{c}.\vec{c}=0$

$\Rightarrow \vec{c}.\vec{a}+\vec{b}.\vec{c}=-\vec{c}.\vec{c}$

$\Rightarrow \vec{c}.\vec{a}+\vec{b}.\vec{c}=-|\vec{c}|^{2}$

$\Rightarrow \vec{c}.\vec{a}+\vec{b}.\vec{c}=-2^{2}$

$\Rightarrow \vec{c}.\vec{a}+\vec{b}.\vec{c}=-4\quad...(4)$

Now, $(2)+(3)+(4)\implies$

$\quad 2\:(\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a})=-1-16-4$

$\Rightarrow 2\:(\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a})=-21$

$\Rightarrow \boxed{\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}=-\frac{21}{2}}$

Answer: $\displaystyle \bold{\mu=-\frac{21}{2}}$