Practicing Success
If $\vec a,\vec b,\vec c$ are three non-coplanar vectors, then a vector $\vec r$ satisfying $\vec r.\vec a=\vec r.\vec b=\vec r.\vec c = 1$, is |
$\vec a × \vec b + \vec b × \vec c + \vec c × \vec a$ $\frac{1}{[\vec a\,\,\vec b\,\,\vec c]}\left\{\vec a × \vec b + \vec b × \vec c + \vec c × \vec a\right\}$ $[\vec a\,\,\vec b\,\,\vec c]\left\{\vec a × \vec b + \vec b × \vec c + \vec c × \vec a\right\}$ none of these |
$\frac{1}{[\vec a\,\,\vec b\,\,\vec c]}\left\{\vec a × \vec b + \vec b × \vec c + \vec c × \vec a\right\}$ |
Since $\vec a,\vec b,\vec c$ are three non-coplanar vectors. 3. Therefore, $\vec a×\vec b, \vec b×\vec c$ and $\vec c×\vec a$ are also non-coplanar vectors. Let $\vec r = x(\vec a×\vec b) + y (\vec b×\vec c) +z (\vec c×\vec a)$. Then, $\vec r.\vec a=1⇒1=y\left\{(\vec b×\vec c)×\vec a\right\}⇒y=\frac{1}{[\vec a\,\,\vec b\,\,\vec c]}$ Similarly, $\vec r. \vec b = 1$ and $\vec r . \vec c = 1$ will give $z=\frac{1}{[\vec a\,\,\vec b\,\,\vec c]}$ and $x=\frac{1}{[\vec a\,\,\vec b\,\,\vec c]}$ Hence, $\vec r=\frac{1}{[\vec a\,\,\vec b\,\,\vec c]}\left\{\vec a × \vec b + \vec b × \vec c + \vec c × \vec a\right\}$ is the required solution. |