$\vec a$ and $\vec b$ are two given vectors. On these vectors as adjacent sides a parallelogram is constructed. The vector which is the altitude of the parallelogram and which is perpendicular to $\vec a$ is not equal to

$\frac{\vec a×(\vec b×\vec a)}{|\vec b|^2}$

Clearly,

AM = Projection of $\vec b$ on $\vec a =\frac{\vec a.\vec b}{|\vec a|}$

$∴\vec{AM}=\left\{\frac{\vec a.\vec b}{|\vec a|^2}\right\}\vec a$

In ΔADM, we have

$\vec{AD}=\vec{AM} + \vec{MD}$

$⇒ \vec{DM} = \vec{AM} - \vec{AD} ⇒\vec{DM}=\frac{(\vec a.\vec b)\vec a}{|\vec a|^2}-\vec b$

Also,

$\vec{DM}=\frac{1}{|\vec a|^2}\left\{(\vec a.\vec b)\vec a-|\vec a|^2\vec b\right\}=\frac{1}{|\vec a|^2}\{(\vec a.\vec b)\vec a-(\vec a.\vec a)\vec b\}$

$⇒ \vec{DM}=\frac{\vec a×(\vec b×\vec a)}{|\vec b|^2}$

Hence, vector in option (4) is equal to the required vector.