Real-World Hook: Landing an Aircraft
When a pilot approaches a runway for landing, they need to calculate the angle between the flight path (line) and the runway (plane). The ideal approach angle is typically 3° - too steep and the landing is hard, too shallow and you might overshoot! Air traffic controllers use these line-plane calculations constantly.
Angle Between a Line and a Plane
Formula
If line has direction ratios (l, m, n) and plane has equation ax + by + cz + d = 0 (normal (a, b, c)), then the angle θ between them satisfies:
$$\boxed{\sin\theta = \frac{|al + bm + cn|}{\sqrt{a^2+b^2+c^2} \cdot \sqrt{l^2+m^2+n^2}}}$$Interactive Demo: Visualize Line-Plane Relations
Explore how lines interact with planes in 3D space. Adjust the plane coefficients and line parameters to see:
- Intersection point calculation
- Angle between line and plane
- Whether the line is parallel, perpendicular, or intersecting the plane
Why Sine (Not Cosine)?
- Angle between line and plane = Angle between line and normal to plane
- If φ = angle between line and normal, then θ = 90° - φ
- Since sin θ = cos(90° - θ), we get sin θ = cos φ
- And cos φ is the dot product formula!
Memory Trick 🧠
“Line-Plane uses SINE, Line-Line uses COSINE”
- Between two lines: cos θ formula
- Between line and plane: sin θ formula
- Both have same numerator (dot product), different trig function!
Common Mistake ⚠️
Using cosine instead of sine:
Line: (x-1)/2 = (y-2)/3 = (z-3)/4, DR: (2,3,4)
Plane: 2x + 3y + 4z = 5, Normal: (2,3,4)
❌ cos θ = |2(2)+3(3)+4(4)|/(√29·√29) = 29/29 = 1 → θ = 0°
[This is angle between line and NORMAL]
✓ sin θ = |2(2)+3(3)+4(4)|/(√29·√29) = 1 → θ = 90°
[Correct angle between line and PLANE]
Special Cases: Parallel and Perpendicular
Line Parallel to Plane (θ = 0°)
Line is parallel to plane when sin θ = 0, which means:
$$\boxed{al + bm + cn = 0}$$Interpretation: Direction vector of line is perpendicular to normal of plane.
Line Perpendicular to Plane (θ = 90°)
Line is perpendicular to plane when sin θ = 1, which means:
$$\boxed{\frac{l}{a} = \frac{m}{b} = \frac{n}{c}}$$Interpretation: Direction vector of line is parallel to normal of plane.
Distance from Point to Plane
Formula
Distance of point P(x₁, y₁, z₁) from plane ax + by + cz + d = 0:
$$\boxed{D = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}}$$Memory Trick 🧠
“Substitute and Normalize”
- Substitute point coordinates in LHS of plane equation
- Take absolute value
- Divide by magnitude of normal vector √(a² + b² + c²)
Derivation (Optional)
The perpendicular distance from P to plane equals the projection of AP on the normal direction, where A is any point on the plane.
If A is chosen such that plane becomes ax + by + cz + d = 0, the formula follows from vector projection.
Distance Between Parallel Planes
Formula
Distance between parallel planes ax + by + cz + d₁ = 0 and ax + by + cz + d₂ = 0:
$$\boxed{D = \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}}$$Important: Planes must have same normal (coefficients of x, y, z proportional).
Memory Trick 🧠
“Difference of Constants Over Normal Magnitude”
- Numerator: |d₁ - d₂| (difference of constant terms)
- Denominator: √(a² + b² + c²) (magnitude of normal)
Common Mistake ⚠️
Forgetting to check if planes are parallel:
Planes: 2x + 3y + 4z = 5 and 4x + 6y + 8z = 7
Check parallel: (2,3,4) and (4,6,8) → 2/4 = 3/6 = 4/8 = 1/2 ✓ PARALLEL
Rewrite second as: 2x + 3y + 4z = 7/2
Distance = |5 - 7/2|/√(4+9+16) = |3/2|/√29 = 3/(2√29) ✓
If you forgot to check and they weren't parallel, this formula gives nonsense!
Intersection of Line and Plane
Method
Step 1: Write line in parametric form: x = x₁ + lλ, y = y₁ + mλ, z = z₁ + nλ
Step 2: Substitute in plane equation ax + by + cz + d = 0
Step 3: Solve for λ
Step 4: Substitute λ back to get intersection point (x, y, z)
Special Cases
Case 1: If equation gives unique λ → Line intersects plane at one point
Case 2: If equation gives 0 = 0 (identity) → Line lies in the plane
Case 3: If equation gives 0 = k (k ≠ 0, contradiction) → Line is parallel to plane (no intersection)
Foot of Perpendicular from Point to Plane
Method
Step 1: Find normal to plane (a, b, c) from equation ax + by + cz + d = 0
Step 2: Write line through given point perpendicular to plane:
$$\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = \lambda$$Step 3: Parametric form: x = x₁ + aλ, y = y₁ + bλ, z = z₁ + cλ
Step 4: Substitute in plane equation to find λ
Step 5: Substitute λ back to get foot of perpendicular
Image of Point in a Plane
Method
Let P(x₁, y₁, z₁) be the point and Q be its image in plane ax + by + cz + d = 0.
Step 1: Find foot of perpendicular F using above method
Step 2: F is the midpoint of P and Q
Step 3: Use midpoint formula:
- x₂ = 2x_F - x₁
- y₂ = 2y_F - y₁
- z₂ = 2z_F - z₁
where (x₂, y₂, z₂) is the image Q.
Direct Formula (Advanced)
Image of point (x₁, y₁, z₁) in plane ax + by + cz + d = 0 is:
$$\boxed{\left(x_1 - \frac{2a(ax_1+by_1+cz_1+d)}{a^2+b^2+c^2}, \, y_1 - \frac{2b(ax_1+by_1+cz_1+d)}{a^2+b^2+c^2}, \, z_1 - \frac{2c(ax_1+by_1+cz_1+d)}{a^2+b^2+c^2}\right)}$$Worked Examples
Example 1: Angle Between Line and Plane (JEE Main Level)
Problem: Find angle between line (x-1)/1 = (y-2)/2 = (z-3)/3 and plane 2x + 3y + 4z = 5.
Solution: Line direction ratios: (1, 2, 3) Plane normal: (2, 3, 4)
$$\sin\theta = \frac{|1(2) + 2(3) + 3(4)|}{\sqrt{1^2+2^2+3^2} \cdot \sqrt{2^2+3^2+4^2}}$$ $$= \frac{|2 + 6 + 12|}{\sqrt{14} \cdot \sqrt{29}} = \frac{20}{\sqrt{406}}$$Answer: θ = sin⁻¹(20/√406) ≈ 82.4°
Example 2: Distance from Point to Plane (JEE Main Level)
Problem: Find distance of point (1, 2, 3) from plane 2x + 3y + 6z = 14.
Solution: Using formula:
$$D = \frac{|2(1) + 3(2) + 6(3) - 14|}{\sqrt{2^2 + 3^2 + 6^2}}$$ $$= \frac{|2 + 6 + 18 - 14|}{\sqrt{4 + 9 + 36}} = \frac{12}{\sqrt{49}} = \frac{12}{7}$$Answer: 12/7 units
Example 3: Intersection of Line and Plane (JEE Main Level)
Problem: Find point of intersection of line (x-1)/2 = (y-2)/3 = (z-3)/4 with plane 2x + y + z = 10.
Solution: Parametric form: x = 1 + 2λ, y = 2 + 3λ, z = 3 + 4λ
Substitute in plane equation:
$$2(1 + 2\lambda) + (2 + 3\lambda) + (3 + 4\lambda) = 10$$ $$2 + 4\lambda + 2 + 3\lambda + 3 + 4\lambda = 10$$ $$7 + 11\lambda = 10$$ $$\lambda = \frac{3}{11}$$Point of intersection:
- x = 1 + 2(3/11) = 1 + 6/11 = 17/11
- y = 2 + 3(3/11) = 2 + 9/11 = 31/11
- z = 3 + 4(3/11) = 3 + 12/11 = 45/11
Answer: (17/11, 31/11, 45/11)
Example 4: Foot of Perpendicular (JEE Advanced Level)
Problem: Find foot of perpendicular from point (1, 2, 3) to plane 2x + 3y + 4z = 20.
Solution: Normal to plane: (2, 3, 4)
Line through (1, 2, 3) perpendicular to plane:
$$\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} = \lambda$$Parametric: x = 1 + 2λ, y = 2 + 3λ, z = 3 + 4λ
Substitute in plane:
$$2(1+2\lambda) + 3(2+3\lambda) + 4(3+4\lambda) = 20$$ $$2 + 4\lambda + 6 + 9\lambda + 12 + 16\lambda = 20$$ $$20 + 29\lambda = 20$$ $$\lambda = 0$$Foot of perpendicular: x = 1 + 2(0) = 1 y = 2 + 3(0) = 2 z = 3 + 4(0) = 3
Answer: (1, 2, 3)
Interpretation: Point already lies on the plane! (Verify: 2(1) + 3(2) + 4(3) = 2 + 6 + 12 = 20 ✓)
Example 5: Image of Point (JEE Advanced Level)
Problem: Find image of point (1, 3, 4) in plane 2x - y + z + 3 = 0.
Solution: Step 1: Find foot of perpendicular
Line through (1, 3, 4) perpendicular to plane:
$$\frac{x-1}{2} = \frac{y-3}{-1} = \frac{z-4}{1} = \lambda$$Parametric: x = 1 + 2λ, y = 3 - λ, z = 4 + λ
Substitute in plane:
$$2(1+2\lambda) - (3-\lambda) + (4+\lambda) + 3 = 0$$ $$2 + 4\lambda - 3 + \lambda + 4 + \lambda + 3 = 0$$ $$6 + 6\lambda = 0$$ $$\lambda = -1$$Foot F:
- x_F = 1 + 2(-1) = -1
- y_F = 3 - (-1) = 4
- z_F = 4 + (-1) = 3
Step 2: F is midpoint of P(1, 3, 4) and image Q(x, y, z)
$$(-1, 4, 3) = \left(\frac{1+x}{2}, \frac{3+y}{2}, \frac{4+z}{2}\right)$$Solving:
- 1 + x = -2 → x = -3
- 3 + y = 8 → y = 5
- 4 + z = 6 → z = 2
Answer: Image is (-3, 5, 2)
Verification:
- Distance from P to plane = Distance from Q to plane ✓
- F is midpoint of P and Q ✓
Practice Problems
Level 1: JEE Main Basics
Find angle between line (x-1)/1 = (y-2)/1 = (z-3)/1 and plane x + y + z = 6.
Find distance of origin from plane 3x + 4y + 12z = 52.
Check if line (x-1)/2 = (y-3)/3 = (z-5)/4 is parallel to plane 2x + 3y + 4z = 7.
Find distance between planes 2x + 3y + 6z = 7 and 2x + 3y + 6z = 14.
Level 2: JEE Main Standard
Find point where line x/1 = y/2 = z/3 meets plane x + y + z = 6.
Find foot of perpendicular from (2, 3, 4) to plane 3x + 4y + 5z = 50.
Find the value of k if line (x-2)/3 = (y-3)/k = (z-4)/2 is parallel to plane 2x + 3y + 4z = 5.
Find distance of point (1, -2, 3) from plane 2x - y + 2z + 3 = 0.
Level 3: JEE Advanced
Find image of point (1, 6, 3) in the plane x + y + z = 5.
Find equation of plane through line (x-1)/1 = (y-2)/2 = (z-3)/3 and perpendicular to plane 2x + 3y + 4z = 5.
A variable plane is at a constant distance p from origin and meets the axes at A, B, C. Show that the locus of centroid of triangle ABC is x⁻² + y⁻² + z⁻² = p⁻².
Find shortest distance between line (x-1)/2 = (y-2)/3 = (z-3)/4 and plane 2x + 3y + 4z = 5.
Quick Reference Formulas
| Concept | Formula |
|---|---|
| Angle (line-plane) | sin θ = |al+bm+cn|/(√(a²+b²+c²)·√(l²+m²+n²)) |
| Line ⊥ plane | l/a = m/b = n/c |
| Line ∥ plane | al + bm + cn = 0 |
| Distance (point-plane) | D = |ax₁+by₁+cz₁+d|/√(a²+b²+c²) |
| Distance (parallel planes) | D = |d₁-d₂|/√(a²+b²+c²) |
Algorithm: Line-Plane Intersection
Step 1: Write line in parametric form
- x = x₁ + lλ, y = y₁ + mλ, z = z₁ + nλ
Step 2: Substitute in plane equation ax + by + cz + d = 0
Step 3: Solve for λ
- Unique λ → Single intersection point
- 0 = 0 → Line lies in plane
- 0 = k (k ≠ 0) → Line parallel to plane, no intersection
Step 4: Find point coordinates using λ
Cross-Links
- Previous Topic: Plane Equations - Different forms of plane
- Related: Line in Space - Line equations
- Related: Angle Between Lines - Similar angle calculations
- Next Topic: Skew Lines - Non-intersecting, non-parallel lines
- Foundation: Distance Formula - 3D distance concepts
Common Exam Patterns
JEE Main typically asks:
- Angle between line and plane (3 marks)
- Distance from point to plane (2-3 marks)
- Intersection point (3-4 marks)
- Checking parallel/perpendicular (2 marks)
JEE Advanced patterns:
- Foot of perpendicular and image (4-6 marks)
- Locus problems involving planes
- Combined with spheres and cones
- Optimization (shortest distance)
Pro Tip:
- For line-plane angle, use SINE (most common mistake is using cosine!)
- Always verify intersection by substituting back
- Image of point: Foot is always the midpoint
Last updated: November 2025