Foot of Perpendicular, Image, and Projection

Write the line as $\vec{r} = \vec{a} + \lambda\vec{d}$ (or in symmetric form $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c} = \lambda$). Then:

• Parametric foot: the foot is $F = \vec{a} + \lambda\vec{d}$, a single point depending on $\lambda$.
• Perpendicularity condition: $\overrightarrow{PF}\cdot\vec{d} = 0$, where $\overrightarrow{PF} = F - P$. This is one linear equation in $\lambda$ — solve it.
• Substitute back the $\lambda$ you found into $F = \vec{a} + \lambda\vec{d}$ to get the coordinates.

For a plane $ax + by + cz + d = 0$ the normal is $\vec{n} = (a,b,c)$. From $P(x_1,y_1,z_1)$:

• Normal-parametric line: $(x,y,z) = (x_1 + at,\; y_1 + bt,\; z_1 + ct)$.
• Hit the plane: substitute these into $ax + by + cz + d = 0$ — one linear equation in $t$. Solve for $t$.
• Foot: put that $t$ back into the parametric point.

Compactly, $t = -\dfrac{ax_1 + by_1 + cz_1 + d}{a^2 + b^2 + c^2}$, and $F = P + t\,\vec{n}$.

Reflecting a point in a plane is foot-on-a-plane plus one reflection step:

• Step 1 — find the foot $F = P + t\,\vec{n}$ using $t = -\dfrac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}$ (Concept 2).
• Step 2 — reflect: $F$ is the midpoint of $P$ and $P'$, so $P' = 2F - P$.

A useful shortcut: since $P' = 2F - P = P + 2t\,\vec{n}$, the image is reached by going twice as far along the normal as the foot.

To reflect $P$ in a line $\vec{r} = \vec{a} + \lambda\vec{d}$:

• Step 1 — foot: $F = \vec{a} + \lambda\vec{d}$ with $\overrightarrow{PF}\cdot\vec{d} = 0$ (Concept 1).
• Step 2 — reflect: $P' = 2F - P$.

Backward ("find $a,b$") variant: you are given $P$ and its image $P'$ in terms of unknowns. Two facts pin the unknowns down:

• the midpoint $\tfrac{P + P'}{2}$ lies on the line (its coordinates satisfy the symmetric equation), and
• the segment $\overrightarrow{PP'}$ is perpendicular to $\vec{d}$ (i.e. $\overrightarrow{PP'}\cdot\vec{d} = 0$).

Solve the resulting equations for the unknowns.

To reflect a line $\frac{x-x_0}{p} = \frac{y-y_0}{q} = \frac{z-z_0}{r}$ in a plane with normal $\vec{n}$:

• Check direction $\perp$ normal: if $\vec{d}\cdot\vec{n} = 0$ the line is parallel to the plane and its direction is unchanged by the reflection.
• Reflect one point: take any point $(x_0,y_0,z_0)$ on the line and find its image $P'$ using Concept 3 ($P' = 2F - P$).
• Reassemble: the image line is $\dfrac{x - x_0'}{p} = \dfrac{y - y_0'}{q} = \dfrac{z - z_0'}{r}$ — image point, same direction ratios $(p,q,r)$.

A close cousin asks for a plane through an image point containing a given line: reflect the point (Concept 3), then build the plane through that image point and the line.

The projection (length of the shadow) of the segment $\overrightarrow{AB}$ onto a line with direction $\vec{d}$ is

• Compute $\overrightarrow{AB} = B - A$.
• Dot it with the line's direction ratios $\vec{d}$.
• Divide by $|\vec{d}| = \sqrt{a^2+b^2+c^2}$; take the absolute value (a length is non-negative).

The projection of $\overrightarrow{AB}$ onto a plane with unit normal $\hat{n}$ is the part of $\overrightarrow{AB}$ lying in the plane. Its length is

• $|\overrightarrow{AB}|^2$ is the full squared length.
• $\overrightarrow{AB}\cdot\hat{n} = \dfrac{\overrightarrow{AB}\cdot\vec{n}}{|\vec{n}|}$ is the component ALONG the normal (the bit that gets flattened away).
• Subtract its square and take the root — the in-plane shadow.