# Let the vectors vector a, vector b, vector c be such that |vector a|=2,|vector b|=4 and |vector c|=4.

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:

Projection of $\overrightarrow{\mathrm{b}}$ on $\overrightarrow{\mathrm{a}}=$ projection of $\overrightarrow{\mathrm{c}}$ on $\overrightarrow{\mathrm{a}}$

$\Rightarrow \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|}=\frac{\vec{c} \cdot \vec{a}}{|\vec{a}|} \Rightarrow \vec{b} \cdot \vec{a}=\vec{c} \cdot \vec{a}$

$\because \overrightarrow{\mathrm{b}}$ is perpendicular to $\overrightarrow{\mathrm{c}} \Rightarrow \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=0$

Let $|\vec{a}+\vec{b}-\vec{c}|=k$

Square both sides

$\mathrm{k}^{2}=\overrightarrow{\mathrm{a}}^{2}+\overrightarrow{\mathrm{b}}^{2}+\overrightarrow{\mathrm{c}}^{2}+2 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}-2 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}-2 \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}$

$\mathrm{k}^{2}=\overrightarrow{\mathrm{a}}^{2}+\overrightarrow{\mathrm{b}}^{2}+\overrightarrow{\mathrm{c}}^{2}=36$

$\mathrm{k}=6=|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}}|$