Question

Prove that, if a, b, c are non-zero and

(a x b) x c = a x (b x c)

then either

(i) b is perpendicular to both a and c, or

(ii) a and c are parallel or anti-parallel

Note: "x" is cross product

Answer 1

$$\vec a \times (\vec b\times \vec c)=\vec b(\vec a\cdot \vec c)-\vec c(\vec a\cdot \vec b)$$

$$(\vec a\times \vec b)\times \vec c=-\vec c \times(\vec a\times \vec b) =-\vec a(\vec c\cdot \vec b)+\vec b(\vec c\cdot \vec a)$$

Given $(\vec a\times \vec b)\times \vec c=\vec a \times (\vec b\times \vec c).$

Then

$$-\vec a(\vec c\cdot \vec b)+\vec b(\vec c\cdot \vec a)=\vec b(\vec a\cdot \vec c)-\vec c(\vec a\cdot \vec b)$$

Use that $\vec b(\vec c\cdot \vec a)=\vec b(\vec a\cdot \vec c), \forall\vec a,\vec b,\vec c.$

Then

$$-\vec a(\vec c\cdot \vec b)=-\vec c(\vec a\cdot \vec b)$$

$$\vec a(\vec c\cdot \vec b)=\vec c(\vec a\cdot \vec b)$$

(i)

Suppose that the nonzero vectors $\vec a$ and $\vec c$ are not collinear.

Now suppose that $\vec b$ is not perpendicular to $\vec a$ . It means that $(\vec a\cdot \vec b)\not=0.$

Then $(\vec c\cdot \vec b)\not=0$ and

$$\vec a=\dfrac{(\vec a\cdot \vec b)}{(\vec c\cdot \vec b)}\vec c$$

If two nonzero vectors $\vec a$ and $\vec c$ are not collinear, then we cannot find the number $k\not=0$ such that the $\vec a=k\vec c.$ Hence we have a contradiction.

Therefore if vectors $\vec a$ and $\vec c$ are not collinear then

$$\vec c\cdot \vec b=0\ and \ \vec a\cdot \vec b=0$$

These mean that $\vec b$ is perpendicular to both $\vec a$ and $\vec c.$

(ii)

$$\vec c\cdot \vec b\not=0\ or \ \vec a\cdot \vec b\not=0$$

Zero vector $\vec 0=0\cdot\vec v$ is considered to be parallel to every other vector $\vec v.$

Then the vectors $\vec a$ and $\vec b$ are collinear. These mean that $\vec a$ and $\vec c$ are parallel or anti-parallel.