$\left.\begin{array}{l}{[\text { Angles between lines and planes }]} \\ {[\text { Orthogonal projection (other) }]}\end{array}\right]$

A line $l$ forms an angle $\alpha$ with a plane $P$. Find the orthogonal projection on the plane $P$ of a segment of length $d$ located on the line $l$.
Let $AB = d$ be a segment located on a given line $l$, intersecting the plane $P$ at point $O$; $A_1$ and $B_1$ are the orthogonal projections of points $A$ and $B$ onto this plane. Then points $A_1$ and $B_1$ lie on the orthogonal projection $l_1$ of the line $l$ onto the plane $P$. Consider the plane passing through the parallel lines $AA_1$ and $BB_1$. Drop the perpendicular $AM$ from point $A$ to the line $BB_1$. From the right triangle $AMB$, we find that $AM = AB \cos \angle MAB = AB \cos \angle BOB_1 = d \cos \alpha$.

Therefore, $A_1B_1 = AM = d \cos \alpha$.

## Answer

$d \cdot \cos \alpha$.