Prove the following

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, $\vec{b} \cdot \vec{c}=0$

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

$=4+16+16=36$

$\Rightarrow|\vec{a}+\vec{b}-\vec{c}|^{2}=6$

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