Let two non-collinear unit vectors $\hat a$ and $\hat b$ form an acute angle. A point P moves so that at any time t the position vector $\vec{OP}$ (where O is the origin) is given by $\hat a\cos t +\hat b \sin t$. When P is farthest from the origin O, let M be the length of $\vec{OP}$ and $\vec u$ be the unit vector along $\vec{OP}$. Then,

$\hat u=\frac{\hat a+\hat b}{|\hat a+\hat b|}$ and $M=(1+\hat a.\hat b)^{1/2}$

We have,

$\vec{OP} = \hat a \cos t+\hat b \sin t$

$⇒|\vec{OP}|^2 = (\hat a \cos t+\hat b \sin t). (\hat a \cos t+\hat b \sin t)$

$⇒|\vec{OP}|^2 =\cos^2 t+(\hat a.\hat a)+\sin^2 t(\hat b.\hat b)+2\cos t\, \sin t(\hat a.\hat b)$

$⇒|\vec{OP}|^2 =\cos^2 t+\sin^2 t+\sin 2t(\hat a.\hat b)$

$⇒|\vec{OP}|=\sqrt{1+\sin 2t(\hat a.\hat b)}$

Clearly, $\hat a.\hat b$ is constant. So, $|\vec{OP}|$ is maximum when $\sin 2t$ is maximum i.e. at $t=\frac{π}{4}$

For $t=\frac{π}{4}$, we have

$|\vec{OP}|=[(1+\hat a.\hat b)^{1/2}M=(1+\hat a.\hat b)^{1/2}]$

Also,

$\hat u=\frac{\vec{OP}}{|\vec{OP}|}=\frac{\hat a\cos\frac{π}{4}+\hat b\sin\frac{π}{4}}{\left|\hat a\cos\frac{π}{4}+\hat b\sin\frac{π}{4}\right|}=\frac{\hat a+\hat b}{|\hat a+\hat b|}$