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...

We have three vectors $$\vec{a},\;\vec{b},\;\vec{c}$$ whose magnitudes are given as $$|\vec{a}| = 2,\;|\vec{b}| = 4,\;|\vec{c}| = 4.$$

The problem states that the projection of $$\vec{b}$$ on $$\vec{a}$$ is equal to the projection of $$\vec{c}$$ on $$\vec{a}$$. The scalar projection (sometimes called the component) of a vector $$\vec{u}$$ on another vector $$\vec{v}$$ is defined by the formula

$$\text{proj}_{\vec{v}}(\vec{u}) \;=\; \frac{\vec{v}\!\cdot\!\vec{u}}{|\vec{v}|}.$$

Applying this to the two given projections, we get

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

Because $$|\vec{a}| \neq 0$$, we can multiply both sides by $$|\vec{a}|$$ and conclude

$$\vec{a}\!\cdot\!\vec{b} \;=\; \vec{a}\!\cdot\!\vec{c}.$$

Let us denote this common dot product by $$k$$, so that

$$\vec{a}\!\cdot\!\vec{b} = k \quad\text{and}\quad \vec{a}\!\cdot\!\vec{c} = k.$$

Next, we are told that $$\vec{b}$$ is perpendicular to $$\vec{c}$$. Two vectors are perpendicular precisely when their dot product is zero, so

$$\vec{b}\!\cdot\!\vec{c} = 0.$$

Our goal is to find the magnitude of the vector $$\vec{a} + \vec{b} - \vec{c}.$$ A convenient way is to compute its squared magnitude first. For any vector $$\vec{u}$$, we have $$|\vec{u}|^2 = \vec{u}\!\cdot\!\vec{u}.$$ Hence,

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

We now expand this dot product term by term, remembering the distributive and commutative properties of the dot product:

$$$\begin{aligned} (\vec{a} + \vec{b} - \vec{c}) \!\cdot\! (\vec{a} + \vec{b} - \vec{c}) &= \vec{a}\!\cdot\!\vec{a} \;+\; \vec{b}\!\cdot\!\vec{b} \;+\; (-\vec{c})\!\cdot\!(-\vec{c}) \\ &\quad+ 2\bigl(\vec{a}\!\cdot\!\vec{b}\bigr) \;-\; 2\bigl(\vec{a}\!\cdot\!\vec{c}\bigr) \;-\; 2\bigl(\vec{b}\!\cdot\!\vec{c}\bigr). \end{aligned}$$$

Let us evaluate each of these terms one by one.

First, the three pure magnitude terms are simply the squares of the magnitudes already known:

$$\vec{a}\!\cdot\!\vec{a} = |\vec{a}|^2 = 2^2 = 4,$$ $$\vec{b}\!\cdot\!\vec{b} = |\vec{b}|^2 = 4^2 = 16,$$ $$(-\vec{c})\!\cdot\!(-\vec{c}) = |\vec{c}|^2 = 4^2 = 16.$$

Next, for the mixed terms we use the relations established earlier:

$$2\bigl(\vec{a}\!\cdot\!\vec{b}\bigr) = 2k,$$ $$-\,2\bigl(\vec{a}\!\cdot\!\vec{c}\bigr) = -\,2k,$$ $$-\,2\bigl(\vec{b}\!\cdot\!\vec{c}\bigr) = -\,2(0) = 0.$$

Notice that $$2k - 2k = 0,$$ so the mixed terms cancel out completely.

Putting everything together, we obtain

$$$\begin{aligned} |\vec{a} + \vec{b} - \vec{c}|^2 &= 4 + 16 + 16 + 0 \\ &= 36. \end{aligned}$$$

Finally, taking the positive square root (because magnitude is always non-negative), we get

$$|\vec{a} + \vec{b} - \vec{c}| = \sqrt{36} = 6.$$

So, the answer is $$6$$.