5. Three-Dimensional Geometry

Vector equation of a plane

The vector equation of a plane mathematically models a flat two-dimensional surface extending infinitely in three-dimensional space using position vectors and normal vectors.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Vector Equation of a Plane (Normal Form)

A plane passing through a point with position vector $\mathbf{a}$ and perpendicular to a given non-zero normal vector $\mathbf{n}$ is represented by the vector equation $$(\mathbf{r} - \mathbf{a}) \cdot \mathbf{n} = 0 \implies \mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}$$ where $\mathbf{r}$ is the position vector of an arbitrary point on the plane.

📐 Geometric Interpretation

The vector $\mathbf{r} - \mathbf{a}$ lies entirely within the plane, and since $\mathbf{n}$ is perpendicular to the plane, it must be orthogonal to every vector lying in the plane. Thus, their dot product vanishes.

⚙️ Algebraic & Structural Meaning

Translates geometric orthogonality into a linear algebraic equation in three variables, corresponding directly to the Cartesian linear equation $Ax + By + Cz + D = 0$.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Non-zero Normal Vector	`\mathbf{n} \neq \mathbf{0}`	Strictly Mandatory / Necessary	If $\mathbf{n} = \mathbf{0}$, the direction of orientation is undefined, collapsing the equation into a trivial statement.
Linear Independence of Direction Vectors (for parametric form)	`\mathbf{b} \times \mathbf{c} \neq \mathbf{0}`	Strictly Mandatory	If direction vectors are collinear, they span only a line instead of a plane.
Non-collinearity of Three Given Points	`Points \mathbf{a}, \mathbf{b}, \mathbf{c} not collinear`	Essential for unique plane determination	Collinear points define an infinite family of planes rather than a unique plane.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Conversion from Cartesian to Vector Normal Form

$$Ax + By + Cz + D = 0 \implies \mathbf{r} \cdot (A\hat{i} + B\hat{j} + C\hat{k}) = -D$$

Rule: Extract coefficients of $x, y, z$ directly as components of normal vector $\mathbf{n}$, and constant term as right-hand side scalar.

Application: Directly tested in UP PGT for quick identification of normal vectors and distance calculations.

Equation of Plane Passing Through Three Points

$$[\mathbf{r} - \mathbf{a} \quad \mathbf{b} - \mathbf{a} \quad \mathbf{c} - \mathbf{a}] = 0$$

Rule: Use scalar triple product of vectors formed by taking one point as reference and others as terminal points.

Application: Frequently asked in KVS PGT and DSSSB PGT to bypass lengthy cross-product calculations.

Perpendicular Distance from a Point to a Plane

$$d = \frac{|\mathbf{a} \cdot \mathbf{n} - d_0|}{|\mathbf{n}|}$$

Rule: Substitute position vector $\mathbf{a}$ into vector plane equation and divide by magnitude of normal vector.

Angle Between Two Planes

$$\cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1| |\mathbf{n}_2|}$$

Rule: Angle between planes is equal to the angle between their respective normal vectors.

Equation of Plane Passing Through Line of Intersection

$$(\mathbf{r} \cdot \mathbf{n}_1 - d_1) + \lambda (\mathbf{r} \cdot \mathbf{n}_2 - d_2) = 0$$

Rule: Use the family of planes formula with parameter $\lambda$ determined by an extra condition.

Coplanarity Condition for Four Points

$$[\mathbf{b}-\mathbf{a}, \mathbf{c}-\mathbf{a}, \mathbf{d}-\mathbf{a}] = 0$$

Rule: Four points are coplanar if and only if their scalar triple product determinant equals zero.

Application: High frequency in UP TGT vector algebra sections.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Confusing Normal Vector with Direction Vector in Parametric Form

Common Pitfall: Using direction vectors parallel to the plane directly as normal vectors $\mathbf{n}$ without taking their cross product.

Counterexample: \text{Given plane } \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} + \mu \mathbf{c}

Why it fails: Direction vectors $\mathbf{b}$ and $\mathbf{c}$ lie in the plane, whereas the normal vector must be $\mathbf{b} \times \mathbf{c}$. Using $\mathbf{b}$ or $\mathbf{c}$ as normal yields incorrect orthogonality.

🛡️ Exam Defense: Always compute the cross product $\mathbf{n} = \mathbf{b} \times \mathbf{c}$ when converting parametric form to normal form.

⚠️ COMMON ERROR #02

Ignoring Magnitude Normalization for Distance Formula

Common Pitfall: Applying point-to-plane distance formula without dividing by $|\mathbf{n}|$.

Counterexample: \text{Distance } d = |\mathbf{a} \cdot \mathbf{n} - d_0| \text{ (missing denominator)}

Why it fails: Fails dimensional homogeneity and yields values scaled by the magnitude of the normal vector.

🛡️ Exam Defense: Always divide the evaluated dot product expression by the magnitude of the normal vector.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Normal Form	Point & Normal Vector	(\mathbf{r} - \mathbf{a}) \cdot \mathbf{n} = 0	Plane defined by point and perpendicular vector	Given point and perpendicular direction
Cartesian Form	Coefficients A, B, C	Ax + By + Cz + D = 0	Linear equation in 3 variables representing flat surface	Standard algebraic coordinate form
Parametric Form	Point & Two Direction Vectors	\mathbf{r} = \mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c}	Plane generated by spanning vectors from base point	Given point and two parallel vectors
Intercept Form	Non-zero intercepts a, b, c	\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1	Plane intersecting coordinate axes at distinct points	Given intercepts on axes

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the vector equation of the plane passing through the point with position vector $2\hat{i} - \hat{j} + \hat{k}$ and normal to the vector $3\hat{i} + 2\hat{j} - \hat{k}$.

(A) $\mathbf{r} \cdot (3\hat{i} + 2\hat{j} - \hat{k}) = 3$(B) $\mathbf{r} \cdot (3\hat{i} + 2\hat{j} - \hat{k}) = 5$(C) $\mathbf{r} \cdot (3\hat{i} + 2\hat{j} - \hat{k}) = 7$(D) $\mathbf{r} \cdot (3\hat{i} + 2\hat{j} - \hat{k}) = 1$

📚 Method A: Formal Textbook Derivation

Use equation ($\mathbf{r} - \mathbf{a}$) $\cdot \mathbf{n} = 0$. Here $\mathbf{a} = 2\hat{i} - \hat{j} + \hat{k}$ and $\mathbf{n} = 3\hat{i} + 2\hat{j} - \hat{k}$. Thus $\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n} = (2)(3) + (-1)(2) + (1)(-1) = 6 - 2 - 1 = 3$. Hence $\mathbf{r} \cdot (3\hat{i} + 2\hat{j} - \hat{k}) = 3$.

⚡ Method B: 25-Second Exam Shortcut

Compute dot product of given point vector and normal vector directly to get right-hand side constant.

💡 Tip: Check scalar products carefully to avoid sign calculation errors.

CANONICAL PROBLEM 02 Dual-Method Breakdown

The distance of the plane $\mathbf{r} \cdot (2\hat{i} - 3\hat{j} + 6\hat{k}) = 14$ from the origin is:

(A) $2$(B) $14$(C) $\frac{14}{7}$(D) $7$

📚 Method A: Formal Textbook Derivation

Standard distance formula from origin to plane $\mathbf{r} \cdot \mathbf{n} = d$ is $\frac{|d|}{|\mathbf{n}|}$. Here $d = 14$ and $|\mathbf{n}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$. Distance = $\frac{14}{7} = 2$.

⚡ Method B: 25-Second Exam Shortcut

Divide constant term by magnitude of normal vector: $14 / \sqrt{4+9+36} = 14/7 = 2$.

💡 Tip: Quickly compute vector magnitude using Pythagorean triples (e.g., 2, 3, 6 or 3, 4, 12).

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: The vector equation of a plane passing through the origin and parallel to vectors $\mathbf{b}$ and $\mathbf{c}$ is:

A $\mathbf{r} = \lambda\mathbf{b} + \mu\mathbf{c}$

B $\mathbf{r} \cdot (\mathbf{b} \times \mathbf{c}) = 1$

C $(\mathbf{r}-\mathbf{b}) \cdot \mathbf{c} = 0$

D $\mathbf{r} \cdot \mathbf{b} = \mathbf{r} \cdot \mathbf{c}$

Derivation & Analysis:

Since the plane passes through the origin, position vector of point $\mathbf{a} = \mathbf{0}$. The parametric form becomes $\mathbf{r} = \lambda\mathbf{b} + \mu\mathbf{c}$.

MCQ 2: If the vectors $2\hat{i} - \hat{j} + \hat{k}$, $\hat{i} + 2\hat{j} - 3\hat{k}$, and $3\hat{i} + a\hat{j} + 5\hat{k}$ are coplanar, then the value of $a$ is:

A $-2$

B $2$

C $-4$

D $4$

Derivation & Analysis:

For coplanarity, the scalar triple product determinant must be zero: $\begin{vmatrix} 2 & -1 & 1 \\ 1 & 2 & -3 \\ 3 & a & 5 \end{vmatrix} = 0$. Expanding gives $2(10 + 3a) - (-1)(5 + 9) + 1(a - 6) = 0 \implies 20 + 6a + 14 + a - 6 = 0 \implies 7a + 28 = 0 \implies a = -4$.

MCQ 3: The angle between the planes $\mathbf{r} \cdot (2\hat{i} - \hat{j} + 2\hat{k}) = 3$ and $\mathbf{r} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) = 4$ is:

A $\cos^{-1}\left(\frac{16}{21}\right)$

B $\cos^{-1}\left(\frac{5}{21}\right)$

C $\frac{\pi}{2}$

D $\frac{\pi}{3}$

Derivation & Analysis:

$\cos \theta = \frac{\mathbf{n}_1 \cdot \mathbf{n}_2}{|\mathbf{n}_1| |\mathbf{n}_2|} = \frac{(2)(3) + (-1)(-6) + (2)(2)}{\sqrt{4+1+4} \sqrt{9+36+4}} = \frac{6 + 6 + 4}{3 \times 7} = \frac{16}{21}$.

MCQ 4: The Cartesian equation corresponding to the vector equation $\mathbf{r} \cdot (\hat{i} + 2\hat{j} - 2\hat{k}) = 5$ is:

A $x + 2y - 2z = 5$

B $x - 2y + 2z = 5$

C $2x + y - 2z = 5$

D $x + 2y + 2z = 5$

Derivation & Analysis:

Substitute $\mathbf{r} = x\hat{i} + y\hat{j} + z\hat{k}$ into the vector equation to get $(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{i} + 2\hat{j} - 2\hat{k}) = 5 \implies x + 2y - 2z = 5$.

MCQ 5: The intercept form of the plane whose vector equation is $\mathbf{r} \cdot (2\hat{i} + 3\hat{j} + 6\hat{k}) = 12$ is:

A $\frac{x}{6} + \frac{y}{4} + \frac{z}{2} = 1$

B $2x + 3y + 6z = 12$

C $\frac{x}{2} + \frac{y}{3} + \frac{z}{6} = 1$

D $6x + 4y + 2z = 1$

Derivation & Analysis:

Cartesian equation is $2x + 3y + 6z = 12$. Dividing by 12 yields $\frac{2x}{12} + \frac{3y}{12} + \frac{6z}{12} = 1 \implies \frac{x}{6} + \frac{y}{4} + \frac{z}{2} = 1$.

MCQ 6: The equation of the plane passing through $(1, 2, 3)$ and parallel to the plane $\mathbf{r} \cdot (\hat{i} - 2\hat{j} + 3\hat{k}) = 4$ is:

A $\mathbf{r} \cdot (\hat{i} - 2\hat{j} + 3\hat{k}) = 6$

B $\mathbf{r} \cdot (\hat{i} - 2\hat{j} + 3\hat{k}) = 4$

C $\mathbf{r} \cdot (\hat{i} - 2\hat{j} + 3\hat{k}) = 1$

D $\mathbf{r} \cdot (\hat{i} - 2\hat{j} + 3\hat{k}) = -6$

Derivation & Analysis:

Parallel planes share the same normal vector. Thus equation is $\mathbf{r} \cdot (\hat{i} - 2\hat{j} + 3\hat{k}) = d$. Substituting point $(1, 2, 3)$, $1(1) - 2(2) + 3(3) = 1 - 4 + 9 = 6$. Hence $d = 6$.

MCQ 7: The vector equation of the plane through three points $\mathbf{a}, \mathbf{b}, \mathbf{c}$ can be expressed in non-parametric form as:

A $(\mathbf{r}-\mathbf{a}) \cdot [(\mathbf{b}-\mathbf{a}) \times (\mathbf{c}-\mathbf{a})] = 0$

B $\mathbf{r} = \mathbf{a} + \lambda(\mathbf{b}-\mathbf{a})$

C $\mathbf{r} \cdot (\mathbf{a} + \mathbf{b} + \mathbf{c}) = 0$

D $(\mathbf{r} \cdot \mathbf{a})(\mathbf{b} \cdot \mathbf{c}) = 0$

Derivation & Analysis:

The normal to the plane is given by $(\mathbf{b}-\mathbf{a}) \times (\mathbf{c}-\mathbf{a})$, and the plane passes through $\mathbf{a}$. Substituting into $(\mathbf{r}-\mathbf{a}) \cdot \mathbf{n} = 0$ gives the result.

MCQ 8: Assertion (A): Two planes are perpendicular if their normal vectors are orthogonal. Reason (R): The dot product of two non-zero orthogonal vectors is zero.

A Both (A) and (R) are true and (R) is the correct explanation of (A)

B Both (A) and (R) are true but (R) is not the correct explanation

C (A) is true but (R) is false

D (A) is false but (R) is true

Derivation & Analysis:

Angle between planes is determined by their normal vectors. Orthogonality of normals directly implies perpendicular planes via zero dot product.

MCQ 9: The projection of the vector $\hat{i} + \hat{j} + \hat{k}$ on the normal to the plane $\mathbf{r} \cdot (2\hat{i} - \hat{j} + 2\hat{k}) = 5$ is:

A $\frac{3}{3} = 1$

B $\frac{5}{3}$

C $3$

D $1$

Derivation & Analysis:

Projection of vector $\mathbf{v} = \hat{i} + \hat{j} + \hat{k}$ on normal $\mathbf{n} = 2\hat{i} - \hat{j} + 2\hat{k}$ is $\frac{\mathbf{v} \cdot \mathbf{n}}{|\mathbf{n}|} = \frac{2 - 1 + 2}{\sqrt{4+1+4}} = \frac{3}{3} = 1$.

Q10

MCQ 10: The distance between the parallel planes $\mathbf{r} \cdot (2\hat{i} - 2\hat{j} + \hat{k}) = 5$ and $\mathbf{r} \cdot (4\hat{i} - 4\hat{j} + 2\hat{k}) = 15$ is:

A $\frac{5}{6}$

B $\frac{5}{2}$

C $5$

D $\frac{10}{3}$

Derivation & Analysis:

Normalize the second equation by dividing by 2: $\mathbf{r} \cdot (2\hat{i} - 2\hat{j} + \hat{k}) = 7.5$. Distance between $\mathbf{r} \cdot \mathbf{n} = d_1$ and $\mathbf{r} \cdot \mathbf{n} = d_2$ is $\frac{|d_2 - d_1|}{|\mathbf{n}|} = \frac{7.5 - 5}{\sqrt{4+4+1}} = \frac{2.5}{3} = \frac{5}{6}$ ... Wait, magnitude is 3, $7.5 - 5 = 2.5 = 5/2$, so $5/(2 \times 3) = 5/6$. Let's re-verify: $d_1 = 5$, scaled $d_2 = 15/2 = 7.5$, difference is $2.5$. Magnitude of normal is 3. Distance = $2.5 / 3 = 5/6$. Option correction: Correct is $5/6$.

Common questions

Use the normal form when a point and a perpendicular direction (or Cartesian equation) are given. Use the parametric form when a point and two spanning direction vectors (or three points) are provided.

Sign errors during dot and cross product calculations, and forgetting to divide by the magnitude of the normal vector when computing distances.

Memorize standard conversion shortcuts between Cartesian and vector forms, and leverage scalar triple product properties for coplanarity and point-plane relations.

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