Question

Let $a,b,c$ be three vectors. Then show that
$(\overline{a}\times \overline{b})\times \overline{c}=(\overline{a}.\overline{c})\overline{b}-(\overline{b}.\overline{c})\overline{a}$

Answer 1

Let $a,b,c$ be three vectors then we have to show that

Proof

Let $\overrightarrow{r}=\left(\overrightarrow{a}\times \overrightarrow{b}\right)\times \overrightarrow{c}$

since, cross product of two vectors is a vector perpendicular to both the vectors therefore

$\overrightarrow{r}=\left(\overrightarrow{a}\times \overrightarrow{b}\right)\times \overrightarrow{c}$

$\Rightarrow \overrightarrow{r}\perp \left(\overrightarrow{a}\times \overrightarrow{b}\right)and\overrightarrow{r}\perp \overrightarrow{c}$

But $\overrightarrow{c}$ is a vector perpendicular to the plane of $\overrightarrow{a}$ and $\overrightarrow{b}$ therefore

$\overrightarrow{r}\perp \overrightarrow{c}$

$\Rightarrow \overrightarrow{r}$ lies in the plane $\overrightarrow{a}$ and $\overrightarrow{b}$

$\Rightarrow Thereexistscalerx,y$

such that

$\overrightarrow{r}=x\overrightarrow{a}+y\overrightarrow{b}-----\left(i\right)$

Now

$\Rightarrow \left(x\overrightarrow{a}+y\overrightarrow{b}\right).\overrightarrow{a}=0$

$\Rightarrow x\left(\overrightarrow{b}.\overrightarrow{a}\right)+y\left(\overrightarrow{c}.\overrightarrow{a}\right)=0$

$\Rightarrow x\left(\overrightarrow{a}.\overrightarrow{b}\right)+y\left(\overrightarrow{a}.\overrightarrow{c}\right)=0$

$\Rightarrow x\left(\overrightarrow{a}.\overrightarrow{b}\right)=-y\left(\overrightarrow{a}.\overrightarrow{c}\right)$

$\Rightarrow \frac{x}{\left(\overrightarrow{a}.\overrightarrow{c}\right)}=\frac{y}{-\left(\overrightarrow{a}.\overrightarrow{b}\right)}=\lambda \left(say\right)$

$\Rightarrow x=\lambda \left(\overrightarrow{a}.\overrightarrow{c}\right)$

$y=-\lambda \left(\overrightarrow{a}.\overrightarrow{b}\right)$

Substituting the vector of $x$ and $y$ in (i) we get

$\overrightarrow{r}=\lambda \left(\overrightarrow{a}.\overrightarrow{c}\right)\overrightarrow{b}-\lambda \left(\overrightarrow{a}.\overrightarrow{b}\right)\overrightarrow{c}$

$\Rightarrow \overrightarrow{r}=\lambda \left[\left(\overrightarrow{a}.\overrightarrow{c}\right)\overrightarrow{b}-\lambda \left(\overrightarrow{a}.\overrightarrow{b}\right)\overrightarrow{c}\right]$

$\Rightarrow \overrightarrow{a}\times \overrightarrow{b}\times \overrightarrow{c}=\lambda \left[\left(\overrightarrow{a}.\overrightarrow{c}\right)\overrightarrow{b}-\lambda \left(\overrightarrow{a}.\overrightarrow{b}\right)\overrightarrow{c}\right]$

This is vector identity and so it is true $\forall \overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$

$\therefore \left(\overrightarrow{a}\times \overrightarrow{b}\right)\times \overrightarrow{c}=\left(\overrightarrow{a}.\overrightarrow{c}\right)\overrightarrow{b}-\left(\overrightarrow{b}.\overrightarrow{c}\right)\overrightarrow{a}$

Hence, the given question is proved