Introduction to Three Dimensional Geometry — Class 11 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 11 Mathematics chapter "Introduction to Three Dimensional Geometry" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 11 Mathematics chapter "Introduction to Three Dimensional Geometry" — 8 important questions with detailed…
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NCERT Solutions for Class 11 Mathematics — Introduction to Three Dimensional Geometry. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Introduction to Three Dimensional Geometry — All 8 Questions Solved
Q1. Find the distance between the points A(1, 2, 3) and B(4, 5, 6) in 3D space.
Given points: A(1, 2, 3) and B(4, 5, 6)
Step 1: Use the distance formula in 3D:
d = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²]
Step 2: Substitute the coordinates:
d = √[(4 - 1)² + (5 - 2)² + (6 - 3)²]
d = √[3² + 3² + 3²]
d = √[9 + 9 + 9]
d = √27
d = 3√3
Step 3: Simplify:
d = 3√3 ≈ 5.196 units
Final Answer: Distance AB = 3√3 units
Q2. Find the coordinates of the point that divides the line segment joining A(2, 3, 4) and B(8, 9, 10) in the ratio 1:2 internally.
Given: A(2, 3, 4), B(8, 9, 10), and ratio m:n = 1:2
Step 1: Use the section formula for internal division:
P = [(m × x₂ + n × x₁)/(m + n), (m × y₂ + n × y₁)/(m + n), (m × z₂ + n × z₁)/(m + n)]
Step 2: Substitute m = 1, n = 2:
P = [(1 × 8 + 2 × 2)/(1 + 2), (1 × 9 + 2 × 3)/(1 + 2), (1 × 10 + 2 × 4)/(1 + 2)]
Step 3: Calculate each coordinate:
x-coordinate: (8 + 4)/3 = 12/3 = 4
y-coordinate: (9 + 6)/3 = 15/3 = 5
z-coordinate: (10 + 8)/3 = 18/3 = 6
Step 4: Verify:
Distance AP = √[(4-2)² + (5-3)² + (6-4)²] = √[4 + 4 + 4] = 2√3
Distance PB = √[(8-4)² + (9-5)² + (10-6)²] = √[16 + 16 + 16] = 4√3
Ratio AP:PB = 2√3:4√3 = 1:2 ✓
Final Answer: The dividing point is P(4, 5, 6)
Q3. Find the direction cosines of the line joining the points A(2, 3, 1) and B(6, 9, 7).
Given points: A(2, 3, 1) and B(6, 9, 7)
Step 1: Find the direction ratios:
Direction ratios = (6 - 2, 9 - 3, 7 - 1) = (4, 6, 6)
Step 2: Find the magnitude of direction ratios:
|d| = √(4² + 6² + 6²)
|d| = √(16 + 36 + 36)
|d| = √88
|d| = 2√22
Step 3: Find direction cosines by dividing each ratio by the magnitude:
Direction cosines = (4/(2√22), 6/(2√22), 6/(2√22))
= (2/√22, 3/√22, 3/√22)
Step 4: Rationalize:
= (2√22/22, 3√22/22, 3√22/22)
= (√22/11, 3√22/22, 3√22/22)
Step 5: Verify that l² + m² + n² = 1:
(4/(2√22))² + (6/(2√22))² + (6/(2√22))² = 16/88 + 36/88 + 36/88 = 88/88 = 1 ✓
Final Answer: Direction cosines are (2/√22, 3/√22, 3/√22) or (√22/11, 3√22/22, 3√22/22)
Q4. Find the angle between the lines whose direction cosines are (1/√3, 1/√3, 1/√3) and (1/√2, 1/√2, 0).
Given direction cosines:
Line 1: l₁ = 1/√3, m₁ = 1/√3, n₁ = 1/√3
Line 2: l₂ = 1/√2, m₂ = 1/√2, n₂ = 0
Step 1: Use the formula for angle between lines:
cos θ = |l₁l₂ + m₁m₂ + n₁n₂|
Step 2: Calculate the dot product:
l₁l₂ = (1/√3)(1/√2) = 1/√6
m₁m₂ = (1/√3)(1/√2) = 1/√6
n₁n₂ = (1/√3)(0) = 0
Step 3: Sum the products:
l₁l₂ + m₁m₂ + n₁n₂ = 1/√6 + 1/√6 + 0 = 2/√6 = 2√6/6 = √6/3
Step 4: Find the angle:
cos θ = |√6/3| = √6/3
θ = cos⁻¹(√6/3)
Step 5: Approximate:
√6 ≈ 2.449
√6/3 ≈ 0.816
θ ≈ cos⁻¹(0.816) ≈ 35.26°
Final Answer: cos θ = √6/3, or θ ≈ 35.26°
Q5. Find the centroid of the tetrahedron with vertices at A(1, 2, 3), B(4, 5, 6), C(7, 8, 9), and D(2, 3, 4).
Given vertices: A(1, 2, 3), B(4, 5, 6), C(7, 8, 9), D(2, 3, 4)
Step 1: Use the formula for centroid of a tetrahedron:
Centroid G = [(x₁ + x₂ + x₃ + x₄)/4, (y₁ + y₂ + y₃ + y₄)/4, (z₁ + z₂ + z₃ + z₄)/4]
Step 2: Calculate x-coordinate:
x = (1 + 4 + 7 + 2)/4 = 14/4 = 7/2 = 3.5
Step 3: Calculate y-coordinate:
y = (2 + 5 + 8 + 3)/4 = 18/4 = 9/2 = 4.5
Step 4: Calculate z-coordinate:
z = (3 + 6 + 9 + 4)/4 = 22/4 = 11/2 = 5.5
Step 5: Write the centroid:
G = (7/2, 9/2, 11/2) or (3.5, 4.5, 5.5)
Final Answer: Centroid G = (7/2, 9/2, 11/2) or (3.5, 4.5, 5.5)
Q6. Find the equation of the line passing through the point (1, 2, 3) and having direction ratios 2:3:4.
Given: Point P₀(1, 2, 3) and direction ratios a:b:c = 2:3:4
Step 1: Use the parametric form of a line:
x = x₀ + at
y = y₀ + bt
z = z₀ + ct
Where t is a parameter.
Step 2: Substitute the values:
x = 1 + 2t
y = 2 + 3t
z = 3 + 4t
Step 3: Alternatively, use the symmetric form:
(x - x₀)/a = (y - y₀)/b = (z - z₀)/c
(x - 1)/2 = (y - 2)/3 = (z - 3)/4
Step 4: Verify by choosing a value of parameter:
At t = 0: Point is (1, 2, 3) ✓
At t = 1: Point is (3, 5, 7)
Check: (3 - 1)/2 = 1, (5 - 2)/3 = 1, (7 - 3)/4 = 1 ✓
Final Answer: Parametric form: x = 1 + 2t, y = 2 + 3t, z = 3 + 4t
or Symmetric form: (x - 1)/2 = (y - 2)/3 = (z - 3)/4
Q7. The coordinates of three vertices of a triangle are A(1, 2, 3), B(4, 5, 6), and C(7, 2, 1). Find the area of the triangle ABC.
Given vertices: A(1, 2, 3), B(4, 5, 6), C(7, 2, 1)
Step 1: Find vectors AB and AC:
AB = B - A = (4 - 1, 5 - 2, 6 - 3) = (3, 3, 3)
AC = C - A = (7 - 1, 2 - 2, 1 - 3) = (6, 0, -2)
Step 2: Find the cross product AB × AC:
AB × AC = |i j k|
|3 3 3|
|6 0 -2|
Step 3: Calculate the cross product:
i component: 3(-2) - 3(0) = -6
j component: -(3(-2) - 3(6)) = -(-6 - 18) = 24
k component: 3(0) - 3(6) = -18
AB × AC = (-6, 24, -18)
Step 4: Find the magnitude of the cross product:
|AB × AC| = √[(-6)² + 24² + (-18)²]
|AB × AC| = √[36 + 576 + 324]
|AB × AC| = √936
|AB × AC| = √(36 × 26)
|AB × AC| = 6√26
Step 5: Find the area:
Area = (1/2)|AB × AC| = (1/2) × 6√26 = 3√26
Final Answer: Area of triangle ABC = 3√26 square units
Q8. The distance between two points in 3D space is 10 units. One point is at origin (0, 0, 0). If the other point is (a, b, 4) where a = b, find the values of a and b.
Given: First point P₁(0, 0, 0), second point P₂(a, b, 4)
Distance = 10 units, and a = b
Step 1: Use the distance formula:
d = √[(a - 0)² + (b - 0)² + (4 - 0)²] = 10
Step 2: Substitute a = b:
√[a² + a² + 16] = 10
√[2a² + 16] = 10
Step 3: Square both sides:
2a² + 16 = 100
2a² = 84
a² = 42
a = ±√42
Step 4: Since a = b:
b = ±√42
Step 5: Verify:
For a = b = √42:
d = √[42 + 42 + 16] = √100 = 10 ✓
For a = b = -√42:
d = √[42 + 42 + 16] = √100 = 10 ✓
Final Answer: a = √42 and b = √42, or a = -√42 and b = -√42
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.