Practicing Success
If the vectors $\vec a$ and $\vec b$ are perpendicular to each other, then a vector $\vec v$ in terms of $\vec a$ and $\vec b$ satisfying the equations $\vec v.\vec a=0, \vec v.\vec b=1$ and $[\vec v\,\, \vec a\,\, \vec b]=1$ is |
$\frac{\vec b}{|\vec b|^2}+\frac{\vec a×\vec b}{|\vec a×\vec b|^2}$ $\frac{\vec b}{|\vec b|}+\frac{\vec a×\vec b}{|\vec a×\vec b|^2}$ $\frac{\vec b}{|\vec b|^2}+\frac{\vec a×\vec b}{|\vec a×\vec b|}$ none of these |
$\frac{\vec b}{|\vec b|^2}+\frac{\vec a×\vec b}{|\vec a×\vec b|^2}$ |
If $\vec a,\vec b,\vec c$ are three non coplanar vectors, then any vector $\vec r$ can be expressed as $\vec r=\left\{\frac{\vec r.\vec a}{|\vec a|^2}\right\}\vec a+\left\{\frac{\vec r.\vec b}{|\vec b|^2}\right\}\vec b+\left\{\frac{\vec r.\vec c}{|\vec c|^2}\right\}\vec c$ Clearly, $\vec a,\vec b$ and $\vec a×\vec b$ are three non-coplanar vectors. $∴\vec v=\left\{\frac{\vec v.\vec a}{|\vec a|^2}\right\}\vec a+\left\{\frac{\vec v.\vec b}{|\vec b|^2}\right\}\vec b+\left\{\frac{\vec v.(\vec a×\vec b)}{|\vec a×\vec b|^2}\right\}(\vec a×\vec b)$ $⇒\vec v=0\vec a+\frac{1}{|\vec b|^2}.\vec b+\frac{1}{|\vec a×\vec b|^2}.(\vec a×\vec b)$ |