If vec u and vec v be unit vectors. If vec w is a vector such that ve…

We have,

$\vec w+(\vec w×\vec u) =\vec v$  ...(i)

Taking scalar product with $\vec v$, we obtain

$\{\vec w+(\vec w×\vec u)\}.\vec v=\vec v.\vec v$

$⇒\vec w.\vec v+(\vec w×\vec u).\vec v=|\vec v|^2$

$⇒\vec w.\vec v+[\vec u\,\,\vec v\,\,\vec w]=1$  $[∵|\vec v|=1]$

$⇒[\vec u\,\,\vec v\,\,\vec w]=1-\vec v.\vec w$   ...(ii)

So, option (1) is correct.

Taking cross-product of (i) with $\vec u$, we obtain

$\vec u×\vec w+\vec u×(\vec w×\vec u) = \vec u×\vec v$

$⇒\vec u×\vec w+(\vec u.\vec u)\vec w-(\vec u.\vec w)\vec u= \vec u×\vec v$

$⇒\vec u×\vec w+\vec w-(\vec u.\vec w)\vec u= \vec u×\vec v$

Taking dot product with $\vec w$, we get

$\vec w. (\vec u×\vec w)+\vec w.\vec w-(\vec u.\vec w) (\vec w.\vec u) = \vec w. (\vec u×\vec v)$

$⇒|\vec w|^2-(\vec u.\vec w)^2=[\vec u\,\,\vec v\,\,\vec w]$  ...(iii)

So, option (c) is correct.

Taking scalar product of (i) with $\vec w$, we get

$\vec w.\vec w+\vec w.(\vec w×\vec u)=\vec w.\vec v$

$⇒|\vec w|^2+0=\vec v.\vec w$

$⇒\vec v.\vec w=|\vec w|^2$

Substituting the value of $\vec v.\vec w$ in (ii), we get

$[\vec u\,\,\vec v\,\,\vec w]=1-|\vec w|^2$

So, option (b) is correct.