Advanced Topics in
3D Geometry
Coplanarity of lines, angle between two planes, distance of a point from a plane, and angle between a line and a plane — all in one place.
Coplanarity of Two Lines
Two lines are coplanar if they lie in the same plane. Lines can be coplanar by being:
- Parallel — lie in infinitely many common planes (all parallel to each other)
- Intersecting — lie in exactly one common plane
If lines are skew, they are NOT coplanar.
For lines $\vec{r}=\vec{a_1}+\lambda\vec{b_1}$ and $\vec{r}=\vec{a_2}+\mu\vec{b_2}$:
For lines $\dfrac{x-x_1}{a_1}=\dfrac{y-y_1}{b_1}=\dfrac{z-z_1}{c_1}$ and $\dfrac{x-x_2}{a_2}=\dfrac{y-y_2}{b_2}=\dfrac{z-z_2}{c_2}$:
If the lines are coplanar, the plane containing them has normal $\vec{n}=\vec{b_1}\times\vec{b_2}$ and passes through $\vec{a_1}$:
$$(\vec{r}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2})=0$$Angle Between Two Planes
The angle between two planes equals the angle between their normal vectors.
For planes $\vec{r}\cdot\vec{n_1}=d_1$ and $\vec{r}\cdot\vec{n_2}=d_2$:
Modulus gives the acute angle between planes
For planes $a_1x+b_1y+c_1z+d_1=0$ and $a_2x+b_2y+c_2z+d_2=0$:
| Condition | Vector | Cartesian |
|---|---|---|
| Planes Perpendicular ($\theta=90°$) | $\vec{n_1}\cdot\vec{n_2}=0$ | $a_1a_2+b_1b_2+c_1c_2=0$ |
| Planes Parallel ($\theta=0°$) | $\vec{n_1}\times\vec{n_2}=\vec{0}$ | $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$ |
Distance of a Point from a Plane
The perpendicular distance from point $P(x_0, y_0, z_0)$ to the plane $ax+by+cz+d=0$ is:
Distance from point with position vector $\vec{p}$ to plane $\vec{r}\cdot\hat{n}=d$ (normal form):
$$\text{Distance} = |\vec{p}\cdot\hat{n} - d|$$For general vector plane $\vec{r}\cdot\vec{n}=d'$ (where $\vec{n}$ is not necessarily a unit vector):
$$\text{Distance} = \frac{|\vec{p}\cdot\vec{n}-d'|}{|\vec{n}|}$$Setting $(x_0,y_0,z_0)=(0,0,0)$:
$$\text{Distance from origin to plane }ax+by+cz+d=0 \;=\; \frac{|d|}{\sqrt{a^2+b^2+c^2}}$$Angle Between a Line and a Plane
The angle $\phi$ between a line and a plane is the complement of the angle between the line's direction and the plane's normal.
If $\theta$ is the angle between direction vector $\vec{b}$ and normal $\vec{n}$, then $\phi = 90°-\theta$. Since $\sin\phi=\cos\theta$:
Note: We use $\sin\phi$, not $\cos\phi$!
For line with DRs $(a,b,c)$ and plane $px+qy+rz+s=0$ (normal $(p,q,r)$):
| Relation | Condition | Angle $\phi$ |
|---|---|---|
| Line lies in the plane | $\vec{b}\cdot\vec{n}=0$ AND $\vec{a}\cdot\vec{n}=d$ | $\phi=0°$ |
| Line parallel to plane | $\vec{b}\cdot\vec{n}=0$ | $\phi=0°$ |
| Line perpendicular to plane | $\vec{b}=k\vec{n}$, i.e., $\dfrac{a}{p}=\dfrac{b}{q}=\dfrac{c}{r}$ | $\phi=90°$ |
Foot of Perpendicular & Image in a Plane
To find the foot of perpendicular from $P(x_0,y_0,z_0)$ to plane $ax+by+cz+d=0$:
- Write the line through $P$ along the normal: $\dfrac{x-x_0}{a}=\dfrac{y-y_0}{b}=\dfrac{z-z_0}{c}=t$
- General point on this line: $(x_0+at,\; y_0+bt,\; z_0+ct)$
- Substitute into the plane equation and solve for $t$
- Put $t$ back to get the foot coordinates
If $F$ is the foot of perpendicular, the image $Q$ of $P$ satisfies: $F$ = midpoint of $P$ and $Q$.
$$Q = 2F - P$$i.e., $Q_x=2F_x-P_x$, $Q_y=2F_y-P_y$, $Q_z=2F_z-P_z$
For point $P(x_0,y_0,z_0)$ and plane $ax+by+cz+d=0$, the image $Q(x',y',z')$ is given by:
Master Summary Table — All of Chapter 11
| Topic | Vector Form | Cartesian Form |
|---|---|---|
| DC relation | $l^2+m^2+n^2=1$ | |
| DR → DC | $l=\dfrac{a}{\sqrt{a^2+b^2+c^2}}$ | |
| Line (pt+dir) | $\vec{r}=\vec{a}+\lambda\vec{b}$ | $\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}$ |
| Angle between lines | $\cos\theta=\left|\dfrac{\vec{b_1}\cdot\vec{b_2}}{|\vec{b_1}||\vec{b_2}|}\right|$ | $\cos\theta=\left|\dfrac{\sum a_1a_2}{\sqrt{\sum a_1^2}\sqrt{\sum a_2^2}}\right|$ |
| Shortest distance | $\dfrac{|(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2})|}{|\vec{b_1}\times\vec{b_2}|}$ | $3\times3$ determinant formula |
| Coplanarity | $(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2})=0$ | $3\times3$ determinant $=0$ |
| Plane equation | $\vec{r}\cdot\vec{n}=d$ | $ax+by+cz+d=0$ |
| Angle between planes | $\cos\theta=\left|\dfrac{\vec{n_1}\cdot\vec{n_2}}{|\vec{n_1}||\vec{n_2}|}\right|$ | $\cos\theta=\left|\dfrac{\sum a_1a_2}{\sqrt{\sum a_1^2}\sqrt{\sum a_2^2}}\right|$ |
| Distance: pt from plane | $\dfrac{|\vec{p}\cdot\vec{n}-d'|}{|\vec{n}|}$ | $\dfrac{|ax_0+by_0+cz_0+d|}{\sqrt{a^2+b^2+c^2}}$ |
| Angle: line & plane | $\sin\phi=\left|\dfrac{\vec{b}\cdot\vec{n}}{|\vec{b}||\vec{n}|}\right|$ | $\sin\phi=\left|\dfrac{\sum ap}{\sqrt{\sum a^2}\sqrt{\sum p^2}}\right|$ |
Solved Examples
Find the angle between the planes $2x-y+z=7$ and $x+y+2z=9$.
Normals: $\vec{n_1}=(2,-1,1)$, $\vec{n_2}=(1,1,2)$
$$\cos\theta = \left|\frac{(2)(1)+(-1)(1)+(1)(2)}{\sqrt{4+1+1}\;\sqrt{1+1+4}}\right| = \left|\frac{2-1+2}{\sqrt{6}\cdot\sqrt{6}}\right| = \frac{3}{6} = \frac{1}{2}$$ $$\theta = \cos^{-1}\!\left(\frac{1}{2}\right) = \boxed{60°}$$Find the distance from the point $(2, 3, -5)$ to the plane $x+2y-2z=9$.
Plane: $x+2y-2z-9=0$. Point: $(2,3,-5)$. Normal magnitude: $\sqrt{1+4+4}=3$.
$$d = \frac{|(1)(2)+(2)(3)+(-2)(-5)-9|}{3} = \frac{|2+6+10-9|}{3} = \frac{|9|}{3} = \boxed{3}$$Find the angle between the line $\dfrac{x-1}{2}=\dfrac{y+3}{-1}=\dfrac{z-4}{2}$ and the plane $2x+y-2z=5$.
Direction of line: $(2,-1,2)$. Normal of plane: $(2,1,-2)$.
$$\sin\phi = \left|\frac{(2)(2)+(-1)(1)+(2)(-2)}{\sqrt{4+1+4}\;\sqrt{4+1+4}}\right| = \left|\frac{4-1-4}{3\cdot 3}\right| = \frac{|-1|}{9} = \frac{1}{9}$$ $$\phi = \sin^{-1}\!\left(\frac{1}{9}\right)$$Find the foot of perpendicular from $P(1, 3, 4)$ to the plane $2x-y+z+3=0$.
$a=2,b=-1,c=1,d=3$; Point $P=(1,3,4)$
$$t = -\frac{(2)(1)+(-1)(3)+(1)(4)+3}{4+1+1} = -\frac{2-3+4+3}{6} = -\frac{6}{6} = -1$$ $$\text{Foot} = (1+2(-1),\; 3+(-1)(-1),\; 4+(1)(-1)) = (-1,\; 4,\; 3)$$Check if the lines $\dfrac{x-1}{-3}=\dfrac{y-2}{2}=\dfrac{z-4}{1}$ and $\dfrac{x-3}{2}=\dfrac{y-4}{-1}=\dfrac{z-5}{2}$ are coplanar. If yes, find the equation of the plane.
$(x_1,y_1,z_1)=(1,2,4)$, $(a_1,b_1,c_1)=(-3,2,1)$; $(x_2,y_2,z_2)=(3,4,5)$, $(a_2,b_2,c_2)=(2,-1,2)$
$$\Delta = \begin{vmatrix}3-1 & 4-2 & 5-4\\-3&2&1\\2&-1&2\end{vmatrix} = \begin{vmatrix}2&2&1\\-3&2&1\\2&-1&2\end{vmatrix}$$ $$=2[(2)(2)-(-1)(1)]-2[(-3)(2)-(2)(1)]+1[(-3)(-1)-(2)(2)]$$ $$=2[4+1]-2[-6-2]+1[3-4]=10+16-1=25\neq 0$$Since $\Delta \neq 0$, the lines are skew (not coplanar).
Find the image of the point $(1, 2, 3)$ in the plane $x+2y+4z=38$.
Plane: $x+2y+4z-38=0$. Point $P=(1,2,3)$. $(a,b,c)=(1,2,4)$, $a^2+b^2+c^2=21$.
$$t = -\frac{(1)(1)+(2)(2)+(4)(3)-38}{21} = -\frac{1+4+12-38}{21} = -\frac{-21}{21} = 1$$ $$\text{Foot }F = (1+1,\; 2+2,\; 3+4) = (2,4,7)$$ $$\text{Image }Q = 2F-P = (4-1,\; 8-2,\; 14-3) = \boxed{(3,6,11)}$$PYQ Patterns & Tips
| Question Type | Marks | Key Formula |
|---|---|---|
| Find distance from point to plane | 2–3 | $\dfrac{|ax_0+by_0+cz_0+d|}{\sqrt{a^2+b^2+c^2}}$ |
| Find angle between two planes | 3–4 | $\cos\theta = |\dfrac{\sum a_1a_2}{\ldots}|$ |
| Angle between line and plane | 3–4 | $\sin\phi = |\dfrac{\vec{b}\cdot\vec{n}}{|\vec{b}||\vec{n}|}|$ |
| Foot of perpendicular to plane | 5 | Parametrize normal line, substitute |
| Coplanarity + equation of plane | 5 | $3\times3$ det $=0$, then find plane |
| S.D. between skew lines | 5 | Cross-product formula |
- Distance formula: plug in, absolute value, divide — always show working clearly
- For angle between planes: the formula is identical to angle between lines (both use normals/directions)
- Angle between line and plane uses $\sin\phi$, NOT $\cos\phi$ — most common error in exams
- Foot of perpendicular: write parametric line along normal, substitute into plane, solve $t$
- Perpendicular planes condition is the same as perpendicular lines: dot product of normals $= 0$
- For MCQs on distance, compute $ax_0+by_0+cz_0+d$ carefully — sign errors are common
- Image in a plane: compute $t$ using the direct formula, don't redo from scratch each time
- JEE Advanced may ask: "Plane containing line $L$ and parallel to line $M$" — use family of planes with constraint
- If a line is perpendicular to a plane, its DRs are proportional to the plane's normal
- If a line is parallel to a plane, its direction is perpendicular to the plane's normal: $\vec{b}\cdot\vec{n}=0$
- Combine: foot of perp + distance + image is one complete JEE question
- Angle line-plane: Using $\cos\phi$ instead of $\sin\phi$ — the angle is with the plane, not the normal
- Distance formula: Forgetting absolute value signs in the numerator
- Normal form: Using DRs instead of DCs (divide by $\sqrt{a^2+b^2+c^2}$ first!)
- Coplanarity: Skipping the determinant check and assuming lines are coplanar
- S.D.: Mixing up the skew line formula with the parallel line formula