Maximized Parallelepiped

$1)$ $\vec{w}$ $+$ $(\vec{w}\ \times \hat{u})$ $=$ $\hat{v}$

Taking dot product on both sides by $\hat{v}$ ,

$2)$ $\hat{v} \cdot \vec{w}$ $+$ $\hat{v} \cdot (\vec{w} \times \hat{u})$ $= 1$

Taking dot product of equation $1)$ with $\hat{u}$ ,

$3)$ $\hat{u} \cdot \vec{w}$ $=$ $\hat{u} \cdot \hat{v}$

Taking cross product of equation $1)$ with $\hat{u}$ ,

$\hat{u} \times \vec{w}$ $+$ $\hat{u} \times (\vec{w} \times \hat{u})$ $=$ $\hat{u} \times \vec{w}$ $+$ $\vec{w}$ $-$ $(\hat{u} \cdot \vec{w})$ $\hat{u}$ $=$ $\hat{u} \times \hat{v}$ . The last step comes from opening the vector triple product $\hat{u} \times (\vec{w} \times \hat{u})$

Using equation $3)$ , we get

$\hat{u} \times \vec{w}$ $+$ $\vec{w}$ $-$ $(\hat{u} \cdot \hat{v})$ $\hat{u}$ = $\hat{u} \times \hat{v}$

Taking dot proguct with $\hat{v}$ ,

$\hat{v} \cdot (\hat{u} \times \vec{w})$ $+$ $\hat{v} \cdot \vec{w}$ $-$ $({\hat{v} \cdot \hat{u}})^2$ $= 0$

Using equation $2)$ ,

$\hat{v} \cdot (\hat{u} \times \vec{w})$ $+ 1 -$ $\hat{v} \cdot (\vec{w} \times \hat{u})$ $-$ $({\hat{u} \cdot \hat{v}})^2$ $=$ $0$

$\Rightarrow$ 2 $(\hat{u} \times \hat{v}) \cdot \vec{w}$ $=$ $1$ $-$ $({\hat{u} \cdot \hat{v}})^2$

Now it is easy to note that $|(\hat{u} \times \hat{v}) \cdot \vec{w}|$ is maximized when $\hat{u} \cdot \hat{v}$ is zero or $\theta = 90^\circ$ and its maximum value is $M = 0.5$ . Hence $\theta + M = \boxed{90.5}$ .

If you have any queries, please ask in the comment box.

$\left| (\hat{u} \times \hat{v} ) \cdot \vec{w} \right| = \left| uv \sin\theta w \right| = \left| w \sin \theta \right|$ .... where $w$ is magnitude of $\vec{w}$ .

This is maximum when $\sin \theta$ is maximum, and it is at $\theta = 90^\circ$ .

In the given first equation, think of the magnitudes on $LHS$ and $RHS$ .

$\left| (\vec{w} \times \hat{u}) \right| u = w \sin \theta_1$ , let direction vector be $\hat{x}$ , $\theta_1$ is angle between vectors w and u.

$| \hat{v} |= 1$

Thus we have in LHS, magnitude of vector in LHS is given by

$\sqrt {w^2 + w^2 \sin^2 \theta_1 + 2w^2 \sin\theta_1 \cos\theta_2}$

and see that it is maximum at $\sin \theta_1 = \cos \theta_2 =1$ , for which we get $| w| = \frac{1}{2}$

Thus in asked values, maximum $M$ is $0.5$ and angle in degrees is $90^\circ$ . Answer being $\boxed{90.5}$