Practicing Success
If $\vec u$ and $\vec v$ be unit vectors. If $\vec w$ is a vector such that $\vec w+(\vec w×\vec u) =\vec v$ then $\vec u.(\vec v×\vec w)$ will be equal to |
$1-\vec v.\vec w$ $1-|\vec w|$ $|\vec w|^2-(\vec v.\vec w)^2$ all of these |
all of these |
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. |