Magnitude in complex sum of vectors

Question

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.

Answer 1

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 $\mathbb{R}^3$.

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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?

Answer 2

Simple case (Good case): $\vec{c}$ is perpendicular to $\vec{b}$.

In that case, the answer is $\sqrt 14$. The complicated looking expression is another way of expressing the vector $\mathbb{a}$ in terms of the new basis of $\mathbb{R^3}$ that is {$\vec{b} \times \vec{c}, \vec{b}, \vec{c}$}. In this case, these three vectors are orthogonal to each other. So, we can form an orthonormal basis, and my solution will work.

General Case: $\vec{c}$ is not necesarly perpendicular to $\vec{b}$.

In this case, there should not be any good looking/simple answer. If someone does some ugly computation, s/he will get an ugly expression that depends on the choice of $\mathbb{c}$. I believe that should not be a good/interesting question.

Conclusion: There is no good looking/simple expression for the magnitude of this complicated vector which is independent of $c$.

Answer 3

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 && y\[Element]Reals && 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 && y\[Element]Reals && z\[Element]Reals, Simplify[Norm[d]]]

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

Answer 4

Here we do not take the given values of $\vec{a}$ and $\vec{b}$, but look instead at the structural representation in a real vector space which might help to better see what's going on. For convenience only we use angle brackets $\langle \ldots ,\ldots\rangle$ to denote the dot-product.

We calculate the square of the magnitude and obtain \begin{align*} &\color{blue}{\left|\left\langle\vec{a},\frac{\vec{b}\times \vec{c}}{\left|\vec{b}\times\vec{c}\right|}\right\rangle \frac{\vec{b}\times \vec{c}}{\left|\vec{b}\times\vec{c}\right|} +\left\langle\vec{a},\frac{\vec{b}}{\left|\vec{b}\right|}\right\rangle\frac{\vec{b}}{\left|\vec{b}\right|} +\left\langle\vec{a},\frac{\vec{c}}{\left|\vec{c}\right|}\right\rangle\frac{\vec{c}}{\left|\vec{c}\right|}\right|^{\,2}}\\ &=\left\langle \left\langle\vec{a},\frac{\vec{b}\times \vec{c}}{\left|\vec{b}\times\vec{c}\right|}\right\rangle \frac{\vec{b}\times \vec{c}}{\left|\vec{b}\times\vec{c}\right|} +\left\langle\vec{a},\frac{\vec{b}}{\left|\vec{b}\right|}\right\rangle\frac{\vec{b}}{\left|\vec{b}\right|} +\left\langle\vec{a},\frac{\vec{c}}{\left|\vec{c}\right|}\right\rangle\frac{\vec{c}}{\left|\vec{c}\right|},\right.\\ &\qquad\left.\left\langle\vec{a},\frac{\vec{b}\times \vec{c}}{\left|\vec{b}\times\vec{c}\right|}\right\rangle \frac{\vec{b}\times \vec{c}}{\left|\vec{b}\times\vec{c}\right|} +\left\langle\vec{a},\frac{\vec{b}}{\left|\vec{b}\right|}\right\rangle\frac{\vec{b}}{\left|\vec{b}\right|} +\left\langle\vec{a},\frac{\vec{c}}{\left|\vec{c}\right|}\right\rangle\frac{\vec{c}}{\left|\vec{c}\right|} \right\rangle\tag{1}\\ &=\left\langle\vec{a},\frac{\vec{b}\times \vec{c}}{\left|\vec{b}\times\vec{c}\right|}\right\rangle^2 \left\langle\frac{\vec{b}\times \vec{c}}{\left|\vec{b}\times\vec{c}\right|},\frac{\vec{b}\times \vec{c}}{\left|\vec{b}\times\vec{c}\right|}\right\rangle\\ &\qquad+\left\langle\vec{a},\frac{\vec{b}}{\left|\vec{b}\right|}\right\rangle^2 \left\langle\frac{\vec{b}}{\left|\vec{b}\right|},\frac{\vec{b}}{\left|\vec{b}\right|}\right\rangle +\left\langle\vec{a},\frac{\vec{b}}{\left|\vec{b}\right|}\right\rangle \left\langle\vec{a},\frac{\vec{c}}{\left|\vec{c}\right|}\right\rangle \left\langle\frac{\vec{b}}{\left|\vec{b}\right|},\frac{\vec{c}}{\left|\vec{c}\right|}\right\rangle\\ &\qquad+\left\langle\vec{a},\frac{\vec{c}}{\left|\vec{c}\right|}\right\rangle \left\langle\vec{a},\frac{\vec{b}}{\left|\vec{b}\right|}\right\rangle \left\langle\frac{\vec{c}}{\left|\vec{c}\right|},\frac{\vec{b}}{\left|\vec{b}\right|}\right\rangle +\left\langle\vec{a},\frac{\vec{c}}{\left|\vec{c}\right|}\right\rangle^2 \left\langle\frac{\vec{c}}{\left|\vec{c}\right|},\frac{\vec{c}}{\left|\vec{c}\right|}\right\rangle\tag{2}\\ &\,\,\color{blue}{=\left\langle\vec{a},\frac{\vec{b}\times \vec{c}}{\left|\vec{b}\times\vec{c}\right|}\right\rangle^2 +\left\langle\vec{a},\frac{\vec{b}}{\left|\vec{b}\right|}\right\rangle^2 +2\left\langle\vec{a},\frac{\vec{b}}{\left|\vec{b}\right|}\right\rangle \left\langle\vec{a},\frac{\vec{c}}{\left|\vec{c}\right|}\right\rangle \left\langle\frac{\vec{b}}{\left|\vec{b}\right|},\frac{\vec{c}}{\left|\vec{c}\right|}\right\rangle +\left\langle\vec{a},\frac{\vec{c}}{\left|\vec{c}\right|}\right\rangle^2}\tag{3} \end{align*}

Comment:

• In (1) we write the square of the magnitude as dot-product: $|\vec{z}|^2=\langle \vec{z},\vec{z}\rangle$.

• In (2) we multiply out according to \begin{align*} &\langle \vec{x}+\vec{y}+\vec{z},\vec{x}+\vec{y}+\vec{z} \rangle\\ &\quad=\langle \vec{x},\vec{x}\rangle +\langle \vec{x},\vec{y} \rangle +\langle \vec{x},\vec{z} \rangle\\ &\quad\quad+\langle \vec{y},\vec{x} \rangle +\langle \vec{y},\vec{y} \rangle +\langle \vec{y},\vec{z} \rangle\\ &\qquad\qquad+\langle \vec{z},\vec{x} \rangle +\langle \vec{z},\vec{y} \rangle +\langle \vec{z},\vec{z} \rangle \end{align*} and use the orthogonal property of the cross product \begin{align*} \langle \vec{b}\times \vec{c},\vec{b}\rangle=\langle \vec{b}\times \vec{c},\vec{c}\rangle =\langle\vec{b}, \vec{b}\times \vec{c}\rangle=\langle\vec{c}, \vec{b}\times \vec{c}\rangle=\vec{0}\\ \end{align*} which significantly reduces the number of terms.

• In (3) we use that the magnitude of unit vectors $\left|\frac{\vec{z}}{\left|\vec{z}\right|}\right|=1$ reducing the number of factors. We also use the symmetry of the dot-product $\langle \vec{x},\vec{y}\rangle=\langle\vec{y},\vec{x}\rangle$.