Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Given $|\vec{a}|=2$, $|\vec{b}|=4$, $|\vec{c}|=4$.

The projection of $\vec{b}$ on $\vec{a}$ is equal to the projection of $\vec{c}$ on $\vec{a}$:

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

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

$\Rightarrow \vec{a}\cdot \vec{b}-\vec{a}\cdot \vec{c}=0$

Since $\vec{b}$ is perpendicular to $\vec{c}$, we have $\vec{b}\cdot \vec{c}=0$.

Now, expanding the square of the required magnitude:

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

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

Substituting the known values:

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

$|\vec{a}+\vec{b}-\vec{c}{|}^2=4+16+16=36$

Taking the square root:

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

Answer: 6