Question:
Let the vectors $\vec{a}, \vec{b}, \vec{c}$ be such that $|\vec{a}|=2,|\vec{b}|=4$ and $|\vec{c}|=4$. If the projection of $\vec{b}$ on $\vec{a}$ is equal to the projection of $\vec{c}$ on $\vec{a}$ and $\vec{b}$ is perpendicular to $\vec{c}$, then the value of $|\vec{a}+\vec{b}-\vec{c}|$ is_________.
Solution:
$\because$ Projection of $\vec{b}$ on $\vec{a}=$ Projection of $\vec{c}$ on $\vec{a}$
$\therefore \vec{a} \cdot \vec{b}=\vec{a} \cdot \vec{c}$
Given, $\vec{b} \cdot \vec{c}=0$
$\because|\vec{a}+\vec{b}-\vec{c}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2 \vec{a} \cdot \vec{b}-2 \vec{b} \cdot \vec{c}-2 \vec{a} \cdot \vec{c}$
$=4+16+16=36$
$\Rightarrow|\vec{a}+\vec{b}-\vec{c}|^{2}=6$