Magnitude in complex sum of vectors If $\vec{a}=\hat{i}+2\hat{j}+3\hat{k}$ and $\vec{b}=2\hat{i}+3\hat{j}+2\hat{k}$ and $\vec{c}$ be a vector non collinear with $\vec{b}.$ Find magnitude of
$$
\frac{\vec{a} \cdot (\vec{b}\times \vec{c})}{|\vec{b}\times \vec{c}|^2}(\vec{b}\times \vec{c})+\frac{(\vec{a}\cdot \vec{b})\vec{b}}{|\vec{b}|^2}+\frac{(\vec{a}\cdot \vec{c})\vec{c}}{|\vec{c}|^2}
$$

What I tried:
Vector component of $\vec{a}$ along $\displaystyle \vec{c}=(|\vec{a}|\cos \alpha)\hat{c}=\frac{(\vec{a}\cdot \vec{c})}{|\vec{c}|^2}\vec{c}$
Vector component of $\vec{a}$ along $\displaystyle \vec{b}=(|\vec{a}|\cos \beta)\hat{b}=\frac{(\vec{a}\cdot \vec{b})}{|\vec{b}|^2}\vec{b}$
Vector component of $\vec{a}$ along $$\displaystyle \vec{b }\times\vec{c}=(|\vec{a}|\cos \gamma)(\hat{b}\times \hat{c})=\frac{(\vec{a}\cdot (\vec{b}\times\vec{c}))}{|\vec{b}\times\vec{c}|^2}{(\vec{b}\times \vec{c})}$$
How do I solve it?
Help me please.
As @MMM says, the result is not pretty in general. To illustrate this, I wrote a Mathematica script which assumes $\vec{c}=x\hat{i}+y\hat{j}+z\hat{k}$ where $x,y,z$ are real numbers.

a = {1, 2, 3}; b = {2, 3, 2}; c = {x, y, z};
d = a.Cross[b,c]/Norm[Cross[b,c]]^2 Cross[b,c] + a.b/Norm[b]^2 b + a.c/Norm[c]^2 c;


And then

Assuming[x\[Element]Reals &amp;&amp; y\[Element]Reals &amp;&amp; z\[Element]Reals, Simplify[d]]


which yields
$$\vec{d}=\left(\frac{x (x+2 y+3 z)}{x^2+y^2+z^2}+\frac{(2 y-3 z) (5 x-4 y+z)}{13 x^2-4 x (3 y+2 z)+8 y^2-12 y z+13 z^2}+\frac{28}{17}\right)\hat{i}\\+\left(\frac{y (x+2 y+3 z)}{x^2+y^2+z^2}-\frac{2 (x-z) (5 x-4 y+z)}{(3 x-2 y)^2+4 (x-z)^2+(2 y-3 z)^2}+\frac{42}{17}\right)\hat{j}\\+\left(\frac{z (x+2 y+3 z)}{x^2+y^2+z^2}+\frac{(3 x-2 y) (5 x-4 y+z)}{(3 x-2 y)^2+4 (x-z)^2+(2 y-3 z)^2}+\frac{28}{17}\right)\hat{k}$$
It is evident that the magnitude of such vector is far from pretty. But just to be sure,

Assuming[x\[Element]Reals &amp;&amp; y\[Element]Reals &amp;&amp; z\[Element]Reals, Simplify[Norm[d]]]


results in
$$|\vec{d}|=\sqrt{\frac{P(x,y,z)}{Q(x,y,z)}}$$
where
$$P=3922 x^4+3728 y^4+6738 z^4+x^3 (2256 z-476 y)+x^2 \left(5297 y^2-2004 y z+8052 z^2\right)-2 x \left(1876 y^3+5331 y^2 z+5954 y z^2-136 z^3\right)-776 y^3 z+3225 y^2 z^2+1044 y z^3,\text{ and}\\
Q=17 \left(x^2+y^2+z^2\right) \left(13 x^2-4 x (3 y+2 z)+8 y^2-12 y z+13 z^2\right)$$
HINT

Let $\vec{z} = \vec{b} \times \vec{c}$. Note that $\vec{b}, \vec{c}$ are not collinear, so $\vec{z} \perp \vec{b}$ and $\vec{z} \perp \vec{c}$ and so $\mathcal{B} = \{\vec{z}, \vec{b}, \vec{c}\}$ is a basis for $\mathbb{R}^3$.

Each component of your sum is basically the projection of $\vec{a}$ on each of the elements of that basis $\mathcal{B}$, so the end result should be very related to $a$. Can you finish the problem now?

UPDATE

If $a,b$ are some vectors, then the projection of $a$ onto $b$ is given by
$$
p_b(a) = \frac{a \cdot b}{|b|^2} b.
$$
Note that $p_b(a)$ is always in the direction of $b$.

Now if you have a basis for some vector space, you can write any element of that vector space as a linear combination of the elements of the basis. For example, consider the standard basis in $\mathbb{R}^2$ and note that I can write any vector $(u,v)^T$ as
$$
\begin{bmatrix} u \\ v \end{bmatrix} 
 = \begin{bmatrix} u \\ 0 \end{bmatrix} 
 + \begin{bmatrix} 0 \\ v \end{bmatrix} 
 = u \begin{bmatrix} 1 \\ 0 \end{bmatrix} 
 + v \begin{bmatrix} 0 \\ 1 \end{bmatrix} 
$$
so each element $(u,0)^T$ and $(0,v)^T$ is the projection of the original vector $(u,v)^T$ onto one of the elements of the standard basis.

Finally, in a feature unique to $\mathbb{R}^3$, if $x,y \in \mathbb{R}^3$ are non-collinear, then their cross product $x \times y$ is perpendicular to them both. Thus, the vectors $\{x,y, x \times y\}$ form a basis of $\mathbb{R}^3$.



Your problem sums three different terms, each of which is a projection of one vector onto the 3 elements of a basis of $\mathbb{R}^3$, constructed exactly through such a cross product.

When you sum the sizes of the projections, what should you get?