Physics - Vectors Question with Solution | TestHub

Given $\vec{a}+\vec{b}+\vec{c}=0$

Squaring both sides, we get:

$|\vec{a}{|}^2+|\vec{b}{|}^2+|\vec{c}{|}^2+2(\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a})=0$

Substituting the given magnitudes $|\vec{a}|=1$, $|\vec{b}|=2$, and $|\vec{c}|=4$:

$1^2+2^2+4^2+2(\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a})=0$

$21+2(\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a})=0$

$\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a}=-\frac{21}{2}$

From the given equation, we can also write:

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

Squaring both sides:

$|\vec{a}{|}^2+|\vec{b}{|}^2+2\vec{a}\cdot \vec{b}=|\vec{c}{|}^2$

$1^2+2^2+2\vec{a}\cdot \vec{b}=4^2$

$5+2\vec{a}\cdot \vec{b}=16$

$2\vec{a}\cdot \vec{b}=11\Rightarrow \vec{a}\cdot \vec{b}=\frac{11}{2}$

The expression to be evaluated is $4\vec{a}\cdot \vec{b}+3\vec{b}\cdot \vec{c}+3\vec{c}\cdot \vec{a}$, which can be rewritten as:

$\vec{a}\cdot \vec{b}+3(\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a})$

Substituting the obtained values:

$\frac{11}{2}+3\left(-\frac{21}{2}\right)=\frac{11-63}{2}=-\frac{52}{2}=-26$

Answer: $-26$