Intersection, Coplanarity, and Skew Lines

A line through $A(x_1,y_1,z_1)$ with direction ratios $(a,b,c)$ is written symmetrically as $\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}=t$. Setting that common ratio to a parameter $t$ gives the general point:

• $x = x_1 + at,\quad y = y_1 + bt,\quad z = z_1 + ct$.

In vector form, $\vec{r} = \vec{a} + t\,\vec{d}$: a fixed position vector $\vec{a}$ plus a scalar $t$ times the direction $\vec{d}$. Every later technique starts here — substitute this moving point into whatever condition the question gives.

To find where the line $\vec{r} = \vec{a} + t\,\vec{d}$ meets the plane:

• Substitute the general point $(x_1+at,\,y_1+bt,\,z_1+ct)$ into the plane equation $Ax + By + Cz = D$.
• Solve the resulting linear equation for $t$.
• Back-substitute $t$ into the general point to get the coordinates.

For a coordinate plane, set the relevant coordinate to 0: XZ-plane $\Rightarrow y = 0$, XY-plane $\Rightarrow z = 0$, YZ-plane $\Rightarrow x = 0$. Solve for $t$ from that single equation. A variable plane in intercept form $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ meets the axes at $(a,0,0),(0,b,0),(0,0,c)$.

To find the intersection of $\vec{r}_1 = \vec{a}_1 + t\,\vec{d}_1$ and $\vec{r}_2 = \vec{a}_2 + s\,\vec{d}_2$:

• Write both general points and equate coordinate-by-coordinate $\Rightarrow$ three equations in $t, s$.
• Solve two of them for $t$ and $s$.
• Check the third equation is satisfied (consistency = they really intersect).
• Substitute $t$ (or $s$) into its line to get the point.

Two lines with points $A_1, A_2$ and directions $\vec{d}_1=(a_1,b_1,c_1)$, $\vec{d}_2=(a_2,b_2,c_2)$ are coplanar (they intersect or are parallel) iff:

• For NON-parallel lines, coplanar $\Leftrightarrow$ they intersect. So 'find $k$ for which the lines intersect' uses the SAME determinant $=0$.
• If the unknown appears in both direction rows, expanding gives a quadratic $\Rightarrow$ TWO answers.
• A non-zero value $\Rightarrow$ the lines are skew (do not intersect).

Points $A, B, C, D$ are coplanar iff the three vectors from $A$ have zero scalar triple product:

For lines $\vec{r}_1=\vec{a_1}+\lambda\vec{d_1}$ and $\vec{r}_2=\vec{a_2}+\mu\vec{d_2}$, the shortest distance is

• The numerator is $\big|[\,\vec{a_2}-\vec{a_1},\,\vec{d_1},\,\vec{d_2}\,]\big|$ — the coplanarity determinant in absolute value.
• $d = 0 \Leftrightarrow$ coplanar (intersecting or parallel).
• 'SD given, find a parameter' reverses it: set $d$ equal to the given value and solve — often a quadratic, giving two values whose SUM the question may ask for.

Planes $\vec{r}\cdot\vec{n_1}=d_1$ and $\vec{r}\cdot\vec{n_2}=d_2$ intersect in a line whose direction is

To find where a line of given direction $(l,m,n)$ meets two lines at $A$ and $B$:

• Write $A$ as a general point of line 1 (parameter $\lambda$) and $B$ of line 2 (parameter $\mu$).
• Force $\vec{AB}\parallel (l,m,n)$: the components of $\vec{AB}$ are proportional to $(l,m,n)$, giving $\dfrac{(\vec{AB})_x}{l}=\dfrac{(\vec{AB})_y}{m}=\dfrac{(\vec{AB})_z}{n}$.
• Solve the two independent proportion equations for $\lambda, \mu$; substitute back for $A$ and $B$.

The line through $P_0$ with direction $\vec{d}$ lies in the plane $Ax+By+Cz+D=0$ (normal $\vec{n}=(A,B,C)$) iff BOTH:

• Direction perpendicular to normal: $\vec{d}\cdot\vec{n}=0$ (line is parallel to the plane), AND
• Point on plane: $P_0$ satisfies the plane equation.

Either condition alone is not enough — $\vec{d}\cdot\vec{n}=0$ without the point gives a line parallel to but OUTSIDE the plane.