Let $\vec u,\vec v,\vec w$ be three vectors such that $|\vec u|=1, |\vec v|=2,|\vec w| = 3$. If the projection $\vec v$ of along $\vec u$ is equal to that of $\vec w$ along $\vec u$ and $\vec v,\vec w$ are perpendicular to each other, then $|\vec u-\vec v +\vec w|$ a equals
Answer & explanation
Correct answer: option 3
We have,
Projection of $\vec v$ along $\vec u$ = Projection of $\vec w$ along $\vec u$
$⇒\frac{\vec v.\vec u}{|\vec u|}=\frac{\vec w.\vec u}{|\vec u|}⇒\vec v.\vec u=\vec w.\vec u$ ...(i)
Also, $\vec v$ and $\vec w$ are perpendicular to each other.
$∴\vec v.\vec w=0$ ...(ii)
Now,
$|\vec u-\vec v +\vec w|^2=|\vec u|^2+|\vec v|^2+|\vec w|^2-2(\vec u.\vec v)-2(\vec v.\vec w)+2(\vec u.\vec w)$
$⇒|\vec u-\vec v +\vec w|^2=1+4+9$ [Using (i) and (ii)]
$⇒|\vec u-\vec v +\vec w|=\sqrt{14}$