Three Dimensional Geometry — Class 12 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Mathematics chapter "Three Dimensional Geometry" — 8 important questions with detailed answers for CBSE board exam preparation.
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TL;DR: Free step-by-step NCERT solutions for Class 12 Mathematics chapter "Three Dimensional Geometry" — 8 important questions with detailed answers for CBSE…
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Key Questions Covered:
- Find the distance of the point (1, 2, 3) from the origin.
- Find the equation of the plane passing through points A(1, 2, 3), B(2, 3, 4),…
- Find the foot of perpendicular from point P(1, 3, 2) to the plane x + 2y - z …
- Find the distance between the parallel planes x + 2y - 2z = 3 and x + 2y - 2z…
- Find the equation of the line passing through A(1, 2, 3) and B(4, 5, 6).
- Find the angle between the planes x + 2y - z = 7 and 3x - 2y - z = 11.
- + 2 more questions in the full chapter
Solutions Summary:
| Question | Status |
|---|---|
| Find the distance of the point (1, 2, 3) from the origin. | ✓ Solved |
| Find the equation of the plane passing through points A(1… | ✓ Solved |
| Find the foot of perpendicular from point P(1, 3, 2) to t… | ✓ Solved |
| Find the distance between the parallel planes x + 2y - 2z… | ✓ Solved |
| Find the equation of the line passing through A(1, 2, 3) … | ✓ Solved |
| Find the angle between the planes x + 2y - z = 7 and 3x -… | ✓ Solved |
Showing 6 of 8 questions
Q1: Find the distance of the point (1, 2, 3) from the origin.
To find the distance from P(1, 2, 3) to origin O(0, 0, 0):
Step 1: Use the distance formula.
For points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂):
d = √[(x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²]
Step 2: Substitute values.
Here: P(1, 2, 3) and O(0, 0, 0)
d = √[(0-1)² + (0-2)² + (0-3)²]
= √[1 + 4 + 9]
= √14
Final Answer: Distance = √14 units
Q2: Find the equation of the plane passing through points A(1, 2, 3), B(2, 3, 4), and C(3, 1, 2).
To find the equation of plane through A, B, C:
Step 1: Find vectors AB and AC.
AB = (2-1, 3-2, 4-3) = (1, 1, 1)
AC = (3-1, 1-2, 2-3) = (2, -1, -1)
Step 2: Find the normal vector n = AB × AC.
n = |î ĵ k̂ |
|1 1 1 |
|2 -1 -1 |
= î[1×(-1) - 1×(-1)] - ĵ[1×(-1) - 1×2] + k̂[1×(-1) - 1×2]
= î(-1 + 1) - ĵ(-1 - 2) + k̂(-1 - 2)
= 0î + 3ĵ - 3k̂
= (0, 3, -3)
Simplify: n = (0, 1, -1)
Step 3: Use point-normal form.
Equation: 0(x - 1) + 1(y - 2) - 1(z - 3) = 0
0 + y - 2 - z + 3 ...
Q3: Find the foot of perpendicular from point P(1, 3, 2) to the plane x + 2y - z = 5.
To find the foot of perpendicular from P(1, 3, 2) to plane x + 2y - z = 5:
Step 1: Find the equation of the line perpendicular to the plane.
Normal to plane: n = (1, 2, -1)
Line through P parallel to n:
(x, y, z) = (1, 3, 2) + t(1, 2, -1)
x = 1 + t
y = 3 + 2t
z = 2 - t
Step 2: Find intersection with the plane.
Substitute the parametric equations into plane equation:
(1 + t) + 2(3 + 2t) - (2 - t) = 5
1 + t + 6 + 4t - 2 + t = 5
5 + 6t = 5
6t = 0
t = 0
Step 3: Find the foot.
x = 1 + 0 = 1
y = 3 ...
Q4: Find the distance between the parallel planes x + 2y - 2z = 3 and x + 2y - 2z = 9.
To find distance between parallel planes ax + by + cz = d₁ and ax + by + cz = d₂:
Step 1: Verify planes are parallel.
Both planes: x + 2y - 2z = d
Normals are identical: n = (1, 2, -2)
Planes are parallel ✓
Step 2: Use distance formula.
For parallel planes ax + by + cz = d₁ and ax + by + cz = d₂:
distance = |d₁ - d₂|/√(a² + b² + c²)
Step 3: Substitute values.
Here: d₁ = 3, d₂ = 9
a = 1, b = 2, c = -2
distance = |3 - 9|/√(1² + 2² + (-2)²)
= |-6|/√(1 + 4 + 4)
= 6/√9
= 6/3
= 2
Final Answer: Di...
Q5: Find the equation of the line passing through A(1, 2, 3) and B(4, 5, 6).
To find the equation of line through A(1, 2, 3) and B(4, 5, 6):
Step 1: Find direction vector.
d = B - A = (4-1, 5-2, 6-3) = (3, 3, 3)
Simplify: d = (1, 1, 1)
Step 2: Write the parametric form.
(x, y, z) = (1, 2, 3) + t(1, 1, 1)
x = 1 + t
y = 2 + t
z = 3 + t
Step 3: Write the symmetric form.
(x - 1)/1 = (y - 2)/1 = (z - 3)/1
Simplified: x - 1 = y - 2 = z - 3
Step 4: Verify with points.
At t = 0: (1, 2, 3) = A ✓
At t = 3: (1+3, 2+3, 3+3) = (4, 5, 6) = B ✓
Final Answer: x - 1 = y - 2 = z - 3 ...
Q6: Find the angle between the planes x + 2y - z = 7 and 3x - 2y - z = 11.
To find the angle between two planes:
Step 1: Extract normal vectors.
Plane 1: x + 2y - z = 7, normal n₁ = (1, 2, -1)
Plane 2: 3x - 2y - z = 11, normal n₂ = (3, -2, -1)
Step 2: Use the formula.
cos θ = |n₁ · n₂|/(|n₁| × |n₂|)
Step 3: Calculate dot product.
n₁ · n₂ = (1)(3) + (2)(-2) + (-1)(-1)
= 3 - 4 + 1
= 0
Step 4: Find magnitudes.
|n₁| = √(1² + 2² + (-1)²) = √6
|n₂| = √(3² + (-2)² + (-1)²) = √14
Step 5: Calculate angle.
cos θ = |0|/(√6 × √14) = 0
θ = cos⁻¹(0) = π/2 = 90°
Final Answer: T...
Showing 6 of 8 questions. Visit the full page for complete solutions.
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