JEE Corner — 3D Geometry for JEE Main and Advanced

The Plane — Dropped by NCERT, Kept by JEE

The rationalized NCERT chapter stops at lines, but the JEE syllabus retains the plane in full, and every JEE Main session uses it. Here is the complete formula sheet.

Equations of a plane

1. Normal form (vector): $\vec{r}\cdot\hat{n} = d$, where $\hat{n}$ is the unit normal and $d \geq 0$ is the distance from the origin.
2. Cartesian general form: $ax + by + cz + d = 0$ — the coefficients $(a, b, c)$ are direction ratios of the normal. Reading normals off plane equations is the single most used skill.
3. Point-normal form: plane through $(x_1, y_1, z_1)$ with normal $(a, b, c)$: $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$.
4. Three-point form: determinant of $\left(\vec{r} - \vec{a}\right), \left(\vec{b} - \vec{a}\right), \left(\vec{c} - \vec{a}\right)$ set to zero (a scalar triple product).
5. Intercept form: $\dfrac{x}{p} + \dfrac{y}{q} + \dfrac{z}{r} = 1$ with intercepts $p, q, r$ on the axes.

The distance and angle kit

1. Point to plane: distance from $(x_1, y_1, z_1)$ to $ax + by + cz + d = 0$ is

$$D = \frac{\left|ax_1 + by_1 + cz_1 + d\right|}{\sqrt{a^2 + b^2 + c^2}}$$

2. Parallel planes $ax + by + cz = d_1$ and $ax + by + cz = d_2$: distance $\dfrac{\left|d_1 - d_2\right|}{\sqrt{a^2 + b^2 + c^2}}$ — but first scale the equations so the normals match exactly.
3. Angle between planes = angle between normals: $\cos\theta = \dfrac{\left|\vec{n}_1\cdot\vec{n}_2\right|}{\left|\vec{n}_1\right|\left|\vec{n}_2\right|}$.
4. Angle between a line and a plane (the one that flips): the line makes angle $\theta$ with the plane, which is complementary to its angle with the normal:

$$\sin\theta = \frac{\left|\vec{b}\cdot\vec{n}\right|}{\left|\vec{b}\right|\left|\vec{n}\right|}$$

Key Point: line-plane angle uses sine; plane-plane and line-line use cosine. Mixing these up is the most common JEE 3D error in existence. A line is parallel to a plane exactly when $\vec{b}\cdot\vec{n} = 0$, and lies in the plane if additionally its anchor point satisfies the plane's equation.

The Speed Toolkit

1. Line-plane intersection: substitute the line's parametric point $(x_1 + \lambda a, \ y_1 + \lambda b, \ z_1 + \lambda c)$ into the plane equation — one linear equation in $\lambda$, one substitution back.

2. Foot of perpendicular from $P$ to a plane: walk along the normal: $Q = P + t\,\vec{n}$; substitute into the plane to find $t$. Image: $P' = P + 2t\,\vec{n}$ with the same $t$ — no second computation.

3. Foot on coordinate planes for free: the image of $(x_1, y_1, z_1)$ in the $xy$-plane is $(x_1, y_1, -z_1)$; in the plane $x = 0$ it is $(-x_1, y_1, z_1)$ — sign-flip the one coordinate.

4. Family of planes: every plane through the intersection line of $P_1 = 0$ and $P_2 = 0$ is $P_1 + \lambda P_2 = 0$. One extra condition (a point, a parallelism, a distance) fixes $\lambda$. This replaces three-variable solving with one-variable solving.

5. Origin-distance readout: for $ax + by + cz + d = 0$, the origin's distance is $\dfrac{|d|}{\sqrt{a^2 + b^2 + c^2}}$ — no substitution needed.

6. Coplanarity of two lines (from Section 4) plus line-in-plane conditions cover most "which plane contains…" stems: direction $\perp$ normal, anchor on plane.

7. Intercept speed-reads: the plane $ax + by + cz = k$ cuts the axes at $\dfrac{k}{a}, \dfrac{k}{b}, \dfrac{k}{c}$ — divide, mind the signs.

8. Projection of a segment on a plane: if a segment of length $L$ makes angle $\theta$ with a plane, its shadow has length $L\cos\theta$ — pair with the $\sin\theta$ line-plane formula.

JEE-Pattern Worked Examples

Example 1: Point-normal to Cartesian

Find the equation of the plane through $(1, 2, 3)$ with normal along $\hat{i} + 2\hat{j} - 5\hat{k}$.

Solution:

1. Point-normal form: $1(x - 1) + 2(y - 2) - 5(z - 3) = 0$.
2. Simplify: $x + 2y - 5z + 10 = 0$.

Answer: $x + 2y - 5z + 10 = 0$ — the normal's ratios become the coefficients, the point fixes the constant.

Example 2: Distance from a point to a plane

Find the distance of $(2, 5, -3)$ from the plane $6x - 3y + 2z - 4 = 0$.

Solution:

1. Formula: $D = \dfrac{\left|6(2) - 3(5) + 2(-3) - 4\right|}{\sqrt{36 + 9 + 4}} = \dfrac{\left|12 - 15 - 6 - 4\right|}{7}$.

Answer: $D = \dfrac{13}{7}$.

Example 3: Line-plane angle

Find the angle between the line $\dfrac{x + 1}{2} = \dfrac{y}{3} = \dfrac{z - 3}{6}$ and the plane $10x + 2y - 11z = 3$.

Solution:

1. Sine formula: $\vec{b} = (2, 3, 6)$, $\vec{n} = (10, 2, -11)$.
2. Compute: $\left|\vec{b}\cdot\vec{n}\right| = |20 + 6 - 66| = 40$; $|\vec{b}| = 7$, $|\vec{n}| = 15$.
3. Angle: $\sin\theta = \dfrac{40}{105} = \dfrac{8}{21}$.

Answer: $\theta = \sin^{-1}\dfrac{8}{21}$ — sine, because one object is a line and the other a plane.

Example 4: Distance to an intersection point

Find the distance of the point $(-1, -5, -10)$ from the point where the line $\dfrac{x - 2}{3} = \dfrac{y + 1}{4} = \dfrac{z - 2}{12}$ meets the plane $x - y + z = 5$.

Solution:

1. Parametrise and substitute: $(2 + 3\lambda) - (-1 + 4\lambda) + (2 + 12\lambda) = 5 \Rightarrow 5 + 11\lambda = 5 \Rightarrow \lambda = 0$.
2. Intersection point: $(2, -1, 2)$.
3. Distance: $\sqrt{(2+1)^2 + (-1+5)^2 + (2+10)^2} = \sqrt{9 + 16 + 144} = 13$.

Answer: $13$ — a two-step chain: intersect, then measure.

Example 5: Foot of the perpendicular from the origin

Find the foot of the perpendicular from the origin to the plane $2x + 3y + 6z = 49$.

Solution:

1. Walk along the normal: $Q = t(2, 3, 6)$.
2. Substitute: $t(4 + 9 + 36) = 49$, so $t = 1$.

Answer: foot $(2, 3, 6)$, at distance $\sqrt{49} = 7$ from the origin — the numbers were designed so $|n|^2 = 49$; recognising that saves the algebra.

Example 6: Image of a point in a plane

Find the image of $(1, 2, 3)$ in the plane $x + 2y + 4z = 38$.

Solution:

1. Walk along the normal: $Q = (1 + t, \ 2 + 2t, \ 3 + 4t)$; substituting, $17 + 21t = 38$, so $t = 1$ and the foot is $(2, 4, 7)$.
2. Reflect: $P' = 2Q - P = (3, 6, 11)$ — equivalently $P + 2t\vec{n}$.

Answer: $(3, 6, 11)$.

Example 7: Plane through three points

Find the plane through $(1, 0, 0)$, $(0, 1, 0)$ and $(0, 0, 1)$.

Solution:

1. Spot the intercepts: the three points are the axis-intercepts at $1$ each.
2. Intercept form: $\dfrac{x}{1} + \dfrac{y}{1} + \dfrac{z}{1} = 1$.

Answer: $x + y + z = 1$ — three-point problems whose points sit on the axes are intercept-form one-liners; the general case uses the triple-product determinant.

Example 8: Distance between parallel planes

Find the distance between the planes $2x + y + 2z = 8$ and $4x + 2y + 4z + 5 = 0$.

Solution:

1. Match the normals: divide the second by $2$: $2x + y + 2z + \dfrac{5}{2} = 0$, i.e. $2x + y + 2z = -\dfrac{5}{2}$.
2. Constant gap over normal length: $D = \dfrac{\left|8 - \left(-\dfrac{5}{2}\right)\right|}{\sqrt{4 + 1 + 4}} = \dfrac{21/2}{3}$.

Answer: $D = \dfrac{7}{2}$. Skipping the normal-matching step (using $8$ and $-5$ directly) is the designed trap.