Is it correct to say: Reading the direction ratios of (x − 1)/2 = (3 − y)/1 = z/4 as 2, 1, 4?

Rewrite (3 − y)/1 as (y − 3)/(−1). The direction ratios are 2, −1, 4.

Is it correct to say: Using cos for the angle between a line and a plane?

Use sin θ = |b · n|/(|b||n|). cos is used for line–line and plane–plane angles.

Is it correct to say: Substituting into the point–plane distance formula with the plane written as ax + by + cz = d?

Bring it to ax + by + cz + d₀ = 0 first. For 2x − y + 2z = 5, d₀ = −5.

Is it correct to say: Applying the skew-line formula to parallel lines?

If b₁ × b₂ = 0 the formula divides by zero. Use d = |(a₂ − a₁) × b|/|b|.

Is it correct to say: Writing the plane through the intersection as r · (n₁ + λn₂) = d₁ + d₂?

λ multiplies the whole second equation, including its constant: d₁ + λd₂.

Is it correct to say: Dropping the modulus and reporting a negative distance or an obtuse angle?

Distances are non-negative and the angle between lines or planes is taken as acute — keep the absolute value.

📄 Formula sheet (PDF)

Line and Plane

Vector and Cartesian equations of a line, distance of a point from a line, skew and parallel lines, equations of a plane, angles between lines and planes, coplanarity and distance of a point from a plane — Maharashtra HSC Maths Part I Ch 6

📐 16 formulas✏️ 4 examples🎯 6 practice🏫 Maharashtra State Board📚 Class 12✓ 2025–26 syllabus

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Board Exam Tips

→Before using the skew-line formula, check that b₁ and b₂ are not parallel. For parallel lines use |(a₂ − a₁) × b|/|b| instead.
→Read direction ratios only after the coefficients of x, y and z are 1: (3 − y)/2 means the direction ratio for y is −2.
→For the angle between a line and a plane use sin, not cos — the normal makes the complementary angle with the line.
→For a plane through the intersection of two planes, write both as '… = 0', add λ times the second to the first, and use the extra condition to find λ.
→Write scalar triple products as determinants — it is the quickest route for coplanarity, shortest distance and the plane through three points.
→Substitute any point you find (foot of a perpendicular, point of intersection) back into the equations as a check.

📐 Formulas(16)

Line through a Point, Parallel to a Vector

r = a + λ b

Symbol	Meaning
$a$	Position vector of a known point A on the line
$b$	Direction vector of the line
$λ$	Scalar parameter (λ ∈ R)

Cartesian Form of a Line

\frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = \frac{z - z _{1}}{c}

Symbol	Meaning
$(x_{1}, y_{1}, z_{1})$	A point on the line
$a, b, c$	Direction ratios of the line

Line through Two Points

r = a + λ (b - a), \frac{x - x _{1}}{x _{2} - x _{1}} = \frac{y - y _{1}}{y _{2} - y _{1}} = \frac{z - z _{1}}{z _{2} - z _{1}}

Angle between Two Lines

cos θ = \frac{∣ b _{1} \cdot b _{2} ∣}{∣ b _{1} ∣ ∣ b _{2} ∣} = \frac{∣ a _{1} a _{2} + b _{1} b _{2} + c _{1} c _{2} ∣}{a _{1}^{2} + b _{1}^{2} + c _{1}^{2} a _{2}^{2} + b _{2}^{2} + c _{2}^{2}}

Distance of a Point from a Line★ Board fav

d = \frac{∣ ( α - a ) \times b ∣}{∣ b ∣} = ∣ α - a ∣^{2} - [(α - a) \cdot b^]^{2}

Symbol	Meaning
$α$	Position vector of the given point P
$b^$	Unit vector along the line

Shortest Distance between Skew Lines★ Board fav

d = \frac{∣ ( a _{2} - a _{1} ) \cdot ( b _{1} \times b _{2} ) ∣}{∣ b _{1} \times b _{2} ∣}

Distance between Parallel Lines

d = \frac{∣ ( a _{2} - a _{1} ) \times b ∣}{∣ b ∣}

Condition for Two Lines to Intersect (Coplanarity)★ Board fav

(a_{2} - a_{1}) \cdot (b_{1} \times b_{2}) = 0 ⟺ x_{2} - x_{1} a_{1} a_{2} y_{2} - y_{1} b_{1} b_{2} z_{2} - z_{1} c_{1} c_{2} = 0

Plane in Normal Form

r \cdot n^= p ⟺ l x + m y + n z = p

Symbol	Meaning
$n^$	Unit vector normal to the plane
$p$	Distance of the plane from the origin

Plane through a Point, Perpendicular to a Vector

(r - a) \cdot n = 0 ⟺ a (x - x_{1}) + b (y - y_{1}) + c (z - z_{1}) = 0

Plane through Three Non-collinear Points★ Board fav

x - x_{1} x_{2} - x_{1} x_{3} - x_{1} y - y_{1} y_{2} - y_{1} y_{3} - y_{1} z - z_{1} z_{2} - z_{1} z_{3} - z_{1} = 0

Plane through the Intersection of Two Planes

r \cdot (n_{1} + λ n_{2}) = d_{1} + λ d_{2}

Angle between Two Planes

cos θ = \frac{∣ n _{1} \cdot n _{2} ∣}{∣ n _{1} ∣ ∣ n _{2} ∣}

Angle between a Line and a Plane

sin θ = \frac{∣ b \cdot n ∣}{∣ b ∣ ∣ n ∣}

Symbol	Meaning
$b$	Direction vector of the line
$n$	Normal vector of the plane

Line Parallel or Perpendicular to a Plane

b \cdot n = 0 (line parallel to plane), b = k n (line perpendicular to plane)

Distance of a Point from a Plane★ Board fav

d = \frac{∣ α \cdot n - p ∣}{∣ n ∣} = \frac{∣ a x _{1} + b y _{1} + c z _{1} + d _{0} ∣}{a ^{2} + b ^{2} + c ^{2}}

Symbol	Meaning
$(x_{1}, y_{1}, z_{1})$	The given point, with position vector α
$d_{0}$	Constant term of the plane ax + by + cz + d₀ = 0

✏️ Solved Examples

1Solved Exampleeasy3 steps

Find the vector and Cartesian equations of the line through A(1, −2, 3) and B(3, 1, 2).

Direction vector = AB = b − a.

b - a = (3 - 1) i^+ (1 + 2) j^+ (2 - 3) k^= 2 i^+ 3 j^- k^

2Solved Exampleboard5 steps

Find the equation of the plane through A(1, 1, 0), B(1, 2, 1) and C(−2, 2, −1).

Two vectors in the plane.

A B = j^+ k^, A C = - 3 i^+ j^- k^

3Solved Exampleboard5 steps

Find the foot of the perpendicular and the perpendicular distance from P(3, −1, 4) to the line r = (î + k̂) + λ(î + 2ĵ + 2k̂).

A general point of the line is M(1 + λ, 2λ, 1 + 2λ). Then PM is:

P M = (λ - 2) i^+ (2 λ + 1) j^+ (2 λ - 3) k^

4Solved ExampleHOTS4 steps

Show that the lines (x − 3)/2 = (y − 3)/1 = (z − 1)/2 and (x + 1)/2 = (y − 5)/(−3) = (z + 3)/2 intersect, and find their point of intersection.

Here A₁(3, 3, 1), b₁ = (2, 1, 2), A₂(−1, 5, −3), b₂ = (2, −3, 2). Use the coplanarity determinant.

- 4 22 21 - 3 - 4 22 = - 4 (2 + 6) - 2 (4 - 4) - 4 (- 6 - 2) = - 32 - 0 + 32 = 0

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Reading the direction ratios of (x − 1)/2 = (3 − y)/1 = z/4 as 2, 1, 4
✓Rewrite (3 − y)/1 as (y − 3)/(−1). The direction ratios are 2, −1, 4.
2
Using cos for the angle between a line and a plane
✓Use sin θ = |b · n|/(|b||n|). cos is used for line–line and plane–plane angles.
3
Substituting into the point–plane distance formula with the plane written as ax + by + cz = d
✓Bring it to ax + by + cz + d₀ = 0 first. For 2x − y + 2z = 5, d₀ = −5.
4
Applying the skew-line formula to parallel lines
✓If b₁ × b₂ = 0 the formula divides by zero. Use d = |(a₂ − a₁) × b|/|b|.
5
Writing the plane through the intersection as r · (n₁ + λn₂) = d₁ + d₂
✓λ multiplies the whole second equation, including its constant: d₁ + λd₂.
6
Dropping the modulus and reporting a negative distance or an obtuse angle
✓Distances are non-negative and the angle between lines or planes is taken as acute — keep the absolute value.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
Find the distance of the point (1, −2, 3) from the plane 2x − y + 2z + 4 = 0.
Q2
Find the equation of the plane through the intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 that passes through (1, 1, 1).
Q3
Find the angle between the line (x − 1)/1 = (y + 2)/2 = (z − 3)/1 and the plane 2x + y − z = 5.
Q4
Find the shortest distance between the lines r = (î − ĵ) + λ(2î + ĵ + k̂) and r = (2ĵ + k̂) + μ(î − ĵ + 2k̂).
Q5
Find the distance between the parallel lines r = (î + 2ĵ − k̂) + λ(î − 2ĵ + 2k̂) and r = (3î + 2k̂) + μ(î − 2ĵ + 2k̂).
Q6
Reduce 2x − y + 2z = 6 to normal form and find the distance of the plane from the origin.

📝 Notes

Line and Plane

This chapter applies the vector tools of Chapter 5 — dot product, cross product and scalar triple product — to lines and planes in space. Almost every formula is one of these products divided by a magnitude, so learn what each product measures rather than memorising the formulas separately.

Lines

A line needs a point and a direction: $\vec{r} = \vec{a} + \lambda\vec{b}$ , or in Cartesian form with direction ratios in the denominators. To find a foot of perpendicular or a point of intersection, write the general point of the line in terms of λ and impose the condition. Between two lines:

Angle — dot product of the directions.
Skew lines — shortest distance $\frac{|(\vec{a}_2 - \vec{a}_1)\cdot(\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$ .
Parallel lines — use the cross product with the common direction instead.
Intersecting lines — the scalar triple product is zero (the lines are coplanar).

Planes

A plane needs a point and a normal. The normal comes from the given data: directly, from the cross product of two vectors in the plane (three points), or from $\vec{n}_1 + \lambda\vec{n}_2$ for a plane through the intersection of two planes. The normal form $\vec{r}\cdot\hat{n} = p$ shows the distance from the origin directly.

Angles and distances with planes

The angle between two planes is the angle between their normals (cos). The angle between a line and a plane uses sin, because the line makes the complementary angle with the normal. The distance of $(x_1, y_1, z_1)$ from $ax + by + cz + d_0 = 0$ is $\frac{|ax_1 + by_1 + cz_1 + d_0|}{\sqrt{a^2 + b^2 + c^2}}$ — always rewrite the plane with every term on one side first.