Confusion about vector triple product The question is simple. Say I have two vector $\mathbf{a}$ and $\mathbf{b}$, and I want to simplify the product $(\mathbf{a} \times \mathbf{b}) \times \mathbf{b}$. According to this, it can be rewritten as $-(\mathbf{b}\cdot\mathbf{b})\mathbf{a}+(\mathbf{b}\cdot\mathbf{a})\mathbf{b}$. This product is written in terms of both $\mathbf{a}$ and $\mathbf{b}$.
However, this cross product is absolutely perpendicular to $\mathbf{b}$ so it should not have any component on $\mathbf{b}$. I realized that $\mathbf{a}$ and $\mathbf{b}$ may not be orthogonal but I can't continue from there to justify the presence of $\mathbf{b}$.
Can anyone explain this in simple terms why $\mathbf{b}$ is there?
 A: If $a$ and $b$ are orthogonal to each other, then you are right that $(a\times b)\times b$ has to be a scalar of $a$. Otherwise, you can draw a picture and see that $(a\times b)\times b$ is a vector in the plane spanned by $a$ and $b$ and orthogonal to $b$. But since $a$ and $b$ are not orthogonal, so need both $a$ and $b$ to write $(a\times b)\times b$ as a linear combination of $a$ and $b$.
A: Let's write it in a more systematic manner. Given the equality
$$
(a \times b) \times b = -(b\cdot b)a + (b \cdot a) b,
$$
you correctly claim that the left hand side should be perpendicular to $b$. Now let's check if the right hand side is perpendicular to $b$:
$$
\text{RHS}\cdot b = -\left( (b\cdot b)a + (b \cdot a) b \right) \cdot b = - (b \cdot b)(a \cdot b) + (b \cdot a)(b \cdot b) = 0.
$$
As you can see the right hand side is also perpendicular to $b$, and there is no contradiction.
Your confusion roots from assuming that the existence of $b$ in the right hand side prevents it from being perpendicular to $b$. This is not always correct, unless $a$ and $b$ are perpendicular. But as you can see in this particular case, $a\cdot b$ would be zero, and $\text{RHS}=-(b \cdot b)a$. So, still no contradiction.
A: In fact there is no component along with $b$. $b$ can be decomposed to two perpendicular components: the image of $a$ on $b$ and the remaining. In other words$$\vec{b}=k\vec{a}+\vec{c}$$where$$\vec{b}\cdot\vec{c}=0$$here the image of $\vec{a}$ on $\vec{b}$ is $$\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\vec{b}$$
A: As long as $a$ and $b$ are linearly independent, they span a plane $P$,
and $a\times b$ is perpendicular to $P$. But $(a\times b)\times b$
is perpendicular to $a\times b$, so is in the plane $P$. That is,
$(a\times b)\times b=\mu a+\nu b$ for some $\mu$ and $\nu$.
That is how $b$ enters.
To confirm that $\mu=-b\cdot b$ and $\nu=a\cdot b$, all one has to
do is to prove that $c=-(b\cdot b)a+(a\cdot b)b$ has the same dot products
with $a$ and $b$ that $(a\times b)\times b$ has. But $$c\cdot b
=-(b\cdot b)(a\cdot b)+(a\cdot b)(b\cdot b)=0$$ which is good
as $(a\times b)\times b$ is orthogonal to $b$. On the other hand
$$c\cdot a=-(a\cdot a)(b\cdot b)+(a\cdot b)^2=|a|^2|b|^2(-1+\cos^2t)
=-|a|^2|b|^2\sin^2t$$
where $t$ is the angle between vectors $a$ and $b$. But
$$((a\times b)\times b)\cdot a =(a\times b)\cdot(b\times a)
=-|a\times b|^2=-|a|^2|b|^2\sin^2t.$$
Indeed then $c=(a\times b)\times b$.
A: $\frac{(a\cdot b)\,b}{b\cdot b}$ is the projection of $a$ onto the direction of $b$.
$\frac{a\times b}{|b|}$ is the projection of $a$ onto the plane perpendicular to $b$ then rotated $\frac\pi2$ about the direction of $b$. Since $\frac{a\times b}{|b|}$ is perpendicular to $b$, the cross product with $\frac{b}{|b|}$ simply rotates it another $\frac\pi2$ about the direction of $b$. That is, $-\frac{(a\times b)\times b}{b\cdot b}$ is the projection of $a$ onto the plane perpendicular to $b$.
Adding these two orthogonal projections yields
$$
\frac{(a\cdot b)\,b}{b\cdot b}-\frac{(a\times b)\times b}{b\cdot b}=a\tag1
$$
Solving for $(a\times b)\times b$ gives
$$
(a\times b)\times b=(a\cdot b)\,b-(b\cdot b)\,a\tag2
$$

Although it appears that there is a component of $b$ on the right-hand side of $(2)$, all that the $(a\cdot b)\,b$ serves to do is to cancel the part of $(b\cdot b)\,a$ that is in the direction of $b$. Note that
$$
((a\cdot b)\,b-(b\cdot b)\,a)\cdot b=0\tag3
$$
So there is indeed no component in the direction of $b$.
