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Class 12 Science (PCM)

Mathematics Part 2

Three Dimensional Geometry

Three Dimensional Geometry

Action PlanNotesQuestionsMock-Test
Study Action Plan (Sorted by Difficulty)
Line Perpendicular to Two Given Lines
Accuracy: 0%
Total Attempted1
Total Correct0
Total Incorrect1
Difficulty Level:
High
Shortest Distance Between Two Lines
Accuracy: 33%
Total Attempted6
Total Correct2
Total Incorrect4
Difficulty Level:
High
Collinearity of Points in 3D
Accuracy: 50%
Total Attempted6
Total Correct3
Total Incorrect3
Difficulty Level:
Medium
Converting Cartesian Line Equation to Vector Form
Accuracy: 50%
Total Attempted2
Total Correct1
Total Incorrect1
Difficulty Level:
Medium
Direction Angles and Directed Lines
Accuracy: 50%
Total Attempted4
Total Correct2
Total Incorrect2
Difficulty Level:
Medium
Direction Cosines of Coordinate Axes
Accuracy: 50%
Total Attempted4
Total Correct2
Total Incorrect2
Difficulty Level:
Medium
Distance Between Parallel Lines
Accuracy: 50%
Total Attempted4
Total Correct2
Total Incorrect2
Difficulty Level:
Medium
Equation of Line Through Two Points
Accuracy: 50%
Total Attempted2
Total Correct1
Total Incorrect1
Difficulty Level:
Medium
Skew Lines
Accuracy: 50%
Total Attempted2
Total Correct1
Total Incorrect1
Difficulty Level:
Medium
Direction Ratios and Direction Cosines from Two Points
Accuracy: 60%
Total Attempted5
Total Correct3
Total Incorrect2
Difficulty Level:
Medium
Direction Cosines and Ratios
Accuracy: 75%
Total Attempted8
Total Correct6
Total Incorrect2
Difficulty Level:
Low
Cartesian Equation of Line Through Point and Direction Ratios
Accuracy: 80%
Total Attempted5
Total Correct4
Total Incorrect1
Difficulty Level:
Low
Parametric Equation of a Line in 3D
Accuracy: 80%
Total Attempted5
Total Correct4
Total Incorrect1
Difficulty Level:
Low
Angle Between Two Lines
Accuracy: 83%
Total Attempted6
Total Correct5
Total Incorrect1
Difficulty Level:
Low
Equation of a Line in Space
Accuracy: 89%
Total Attempted9
Total Correct8
Total Incorrect1
Difficulty Level:
Low
Conversion from Direction Ratios to Direction Cosines
Accuracy: 100%
Total Attempted5
Total Correct5
Total Incorrect0
Difficulty Level:
Low
Parallel and Perpendicular Lines in 3D
Accuracy: 100%
Total Attempted3
Total Correct3
Total Incorrect0
Difficulty Level:
Low
Vector Equation of Line Through Point and Parallel Vector
Accuracy: 100%
Total Attempted3
Total Correct3
Total Incorrect0
Difficulty Level:
Low

Learning Objectives

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Learning Objectives

  • Understand the concept of direction cosines and direction ratios of a line in three-dimensional geometry.
  • Derive the equations of lines and planes in space under various conditions.
  • Calculate the angle between two lines, two planes, and a line and a plane.
  • Determine the shortest distance between two skew lines.
  • Find the distance of a point from a plane.
  • Translate vector results into Cartesian form for clearer geometric and analytic understanding.

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Revision Notes & Summary

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Chapter 11: Three Dimensional Geometry

11.1 Introduction

  • Focus on using vector algebra for three-dimensional geometry.
  • Topics covered include:
    • Direction cosines and direction ratios of a line joining two points.
    • Equations of lines and planes in space.
    • Angles between lines and planes.
    • Shortest distance between skew lines.
    • Distance of a point from a plane.

11.2 Direction Cosines and Direction Ratios of a Line

  • Definition: Direction cosines are the cosines of the angles made by a directed line with the coordinate axes.
  • If a line makes angles α, ß, and γ with the x, y, and z-axes respectively, then:
    • Direction cosines are given by:
      • cos α, cos ß, cos γ
  • Example: For the x-axis, direction cosines are (1, 0, 0).

11.3 Equation of a Line in Space

  • A line can be defined if:
    1. It passes through a given point and has a given direction.
    2. It passes through two given points.
  • Equation: For a line through point A with position vector a and parallel to vector b:
    • Vector equation: r = a + λb

11.5 Shortest Distance between Two Lines

  • If two lines intersect, the shortest distance is zero.
  • For parallel lines, the shortest distance is the perpendicular distance between them.
  • Skew Lines: Lines that are neither parallel nor intersecting.
  • Formula for Shortest Distance: For skew lines, the distance is given by:
    • d = |b₁ × b₂|

Important Notes

  • Direction ratios are proportional to direction cosines.
  • The angle between two lines can be calculated using:
    • cos Θ = (a₁a₂ + b₁b₂ + c₁c₂) / (√(a₁² + b₁² + c₁²) * √(a₂² + b₂² + c₂²))
  • The shortest distance between two skew lines is the length of the segment perpendicular to both lines.
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Practice Test – MCQs, True/False

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Multiple Choice Questions

A.

Yes, they are collinear.

B.

No, they are not collinear.

C.

They form a right triangle.

D.

They form an equilateral triangle.
Correct Answer: A

Solution:

The direction ratios of line ABAB are (−1,5,7)(-1, 5, 7) and for line BCBC are (2,10,−14)(2, 10, -14). These are proportional, hence AA, BB, and CC are collinear.

Chapter Concept:

Collinearity of Points in 3D

A.

l=350,m=450,n=550l = \frac{3}{\sqrt{50}}, m = \frac{4}{\sqrt{50}}, n = \frac{5}{\sqrt{50}}

B.

l=325,m=425,n=525l = \frac{3}{\sqrt{25}}, m = \frac{4}{\sqrt{25}}, n = \frac{5}{\sqrt{25}}

C.

l=335,m=435,n=535l = \frac{3}{\sqrt{35}}, m = \frac{4}{\sqrt{35}}, n = \frac{5}{\sqrt{35}}

D.

l=375,m=475,n=575l = \frac{3}{\sqrt{75}}, m = \frac{4}{\sqrt{75}}, n = \frac{5}{\sqrt{75}}
Correct Answer: A

Solution:

The direction cosines l,m,nl, m, n are proportional to the direction ratios 3,4,53, 4, 5. Therefore, l=332+42+52,m=432+42+52,n=532+42+52l = \frac{3}{\sqrt{3^2 + 4^2 + 5^2}}, m = \frac{4}{\sqrt{3^2 + 4^2 + 5^2}}, n = \frac{5}{\sqrt{3^2 + 4^2 + 5^2}}. Calculating, we get l=350,m=450,n=550l = \frac{3}{\sqrt{50}}, m = \frac{4}{\sqrt{50}}, n = \frac{5}{\sqrt{50}}.

Chapter Concept:

Cartesian Equation of Line Through Point and Direction Ratios

A.

The distance along the common perpendicular to both lines.

B.

The distance along one of the lines.

C.

The distance along the line joining their midpoints.

D.

The distance along the line parallel to the z-axis.
Correct Answer: A

Solution:

The shortest distance between two skew lines is along the common perpendicular to both lines.

Chapter Concept:

Shortest Distance Between Two Lines

A.

r=a+λ(b−a)\mathbf{r} = \mathbf{a} + \lambda(\mathbf{b} - \mathbf{a})

B.

r=b+λ(a−b)\mathbf{r} = \mathbf{b} + \lambda(\mathbf{a} - \mathbf{b})

C.

r=a+λb\mathbf{r} = \mathbf{a} + \lambda \mathbf{b}

D.

r=b+λa\mathbf{r} = \mathbf{b} + \lambda \mathbf{a}
Correct Answer: A

Solution:

The vector equation of a line passing through two points with position vectors a\mathbf{a} and b\mathbf{b} is given by r=a+λ(b−a)\mathbf{r} = \mathbf{a} + \lambda(\mathbf{b} - \mathbf{a}).

Chapter Concept:

Equation of Line Through Two Points

A.

∣(a2−a1)⋅(b1×b2)∣∣b1×b2∣\frac{|(\mathbf{a}_2 - \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2)|}{|\mathbf{b}_1 \times \mathbf{b}_2|}

B.

∣a2−a1∣|\mathbf{a}_2 - \mathbf{a}_1|

C.

∣b1×b2∣∣a2−a1∣\frac{|\mathbf{b}_1 \times \mathbf{b}_2|}{|\mathbf{a}_2 - \mathbf{a}_1|}

D.

∣b1×b2∣|\mathbf{b}_1 \times \mathbf{b}_2|
Correct Answer: A

Solution:

For skew lines, the shortest distance is given by ∣(a2−a1)⋅(b1×b2)∣∣b1×b2∣\frac{|(\mathbf{a}_2 - \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2)|}{|\mathbf{b}_1 \times \mathbf{b}_2|}, where b1\mathbf{b}_1 and b2\mathbf{b}_2 are direction vectors of the lines.

Chapter Concept:

Distance Between Parallel Lines

A.

r=a+λb\mathbf{r} = \mathbf{a} + \lambda \mathbf{b}

B.

r=b+λa\mathbf{r} = \mathbf{b} + \lambda \mathbf{a}

C.

r=a−λb\mathbf{r} = \mathbf{a} - \lambda \mathbf{b}

D.

r=λa+b\mathbf{r} = \lambda \mathbf{a} + \mathbf{b}
Correct Answer: A

Solution:

The vector equation of a line through a point with position vector a\mathbf{a} and parallel to vector b\mathbf{b} is given by r=a+λb\mathbf{r} = \mathbf{a} + \lambda \mathbf{b}.

Chapter Concept:

Equation of a Line in Space

A.

a1a2+b1b2+c1c2=0a_1a_2 + b_1b_2 + c_1c_2 = 0

B.

a1/a2=b1/b2=c1/c2a_1/a_2 = b_1/b_2 = c_1/c_2

C.

a1b2−a2b1=0a_1b_2 - a_2b_1 = 0

D.

a1c2+a2c1=0a_1c_2 + a_2c_1 = 0
Correct Answer: A

Solution:

Two lines are perpendicular if the dot product of their direction ratios is zero, i.e., a1a2+b1b2+c1c2=0a_1a_2 + b_1b_2 + c_1c_2 = 0.

Chapter Concept:

Angle Between Two Lines

A.

(14,−6,11)(14, -6, 11)

B.

(14,−7,11)(14, -7, 11)

C.

(10,−6,11)(10, -6, 11)

D.

(14,−6,10)(14, -6, 10)
Correct Answer: A

Solution:

Substituting λ=3\lambda = 3 into the vector equation r=(2,−3,5)+3(4i−j+2k)\mathbf{r} = (2, -3, 5) + 3(4i - j + 2k) gives the point (2+12,−3−3,5+6)=(14,−6,11)(2 + 12, -3 - 3, 5 + 6) = (14, -6, 11).

Chapter Concept:

Converting Cartesian Line Equation to Vector Form

A.

r=a0+λb\mathbf{r} = \mathbf{a}_0 + \lambda \mathbf{b}

B.

r=a1+λb\mathbf{r} = \mathbf{a}_1 + \lambda \mathbf{b}

C.

r=a1+λc\mathbf{r} = \mathbf{a}_1 + \lambda \mathbf{c}

D.

r=a0+λc\mathbf{r} = \mathbf{a}_0 + \lambda \mathbf{c}
Correct Answer: A

Solution:

The vector form of the line is r=a0+λb\mathbf{r} = \mathbf{a}_0 + \lambda \mathbf{b} where a0\mathbf{a}_0 is the position vector of the point (x1,y1,z1)(x_1, y_1, z_1) and b\mathbf{b} is the direction vector ai+bj+ckai + bj + ck.

Chapter Concept:

Converting Cartesian Line Equation to Vector Form

A.

335,435,535\frac{3}{\sqrt{35}}, \frac{4}{\sqrt{35}}, \frac{5}{\sqrt{35}}

B.

135,235,335\frac{1}{\sqrt{35}}, \frac{2}{\sqrt{35}}, \frac{3}{\sqrt{35}}

C.

350,450,550\frac{3}{\sqrt{50}}, \frac{4}{\sqrt{50}}, \frac{5}{\sqrt{50}}

D.

150,250,350\frac{1}{\sqrt{50}}, \frac{2}{\sqrt{50}}, \frac{3}{\sqrt{50}}
Correct Answer: A

Solution:

The direction ratios of the line are (4−1,6−2,8−3)=(3,4,5)(4-1, 6-2, 8-3) = (3, 4, 5). The magnitude is 32+42+52=50\sqrt{3^2 + 4^2 + 5^2} = \sqrt{50}. Thus, the direction cosines are 350,450,550\frac{3}{\sqrt{50}}, \frac{4}{\sqrt{50}}, \frac{5}{\sqrt{50}}.

Chapter Concept:

Parametric Equation of a Line in 3D

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True or False

Correct Answer: True

Solution:

Direction ratios of a line segment can be taken in either order as they represent proportional values indicating the direction of the line. Hence, both x2−x1,y2−y1,z2−z1x_2 - x_1, y_2 - y_1, z_2 - z_1 and x1−x2,y1−y2,z1−z2x_1 - x_2, y_1 - y_2, z_1 - z_2 are valid sets of direction ratios.

Chapter Concept :

Direction Ratios and Direction Cosines from Two Points

Correct Answer: True

Solution:

For parallel lines, the shortest distance is indeed the perpendicular distance between them.

Chapter Concept :

Distance Between Parallel Lines

Correct Answer: True

Solution:

Collinearity in 3D geometry is determined by the proportionality of direction ratios of the line segments formed by the points.

Chapter Concept :

Collinearity of Points in 3D

Correct Answer: True

Solution:

For parallel lines, the shortest distance is indeed the perpendicular distance, as it is the minimum distance between any two points on the lines.

Chapter Concept :

Distance Between Parallel Lines

Correct Answer: True

Solution:

Collinearity in 3D is determined by the proportionality of direction ratios of the line segments connecting the points.

Chapter Concept :

Collinearity of Points in 3D

Correct Answer: True

Solution:

Reversing the direction of a line changes the direction angles to their supplements, i.e., π−α\pi - \alpha, π−β\pi - \beta, and π−γ\pi - \gamma.

Chapter Concept :

Direction Angles and Directed Lines

Correct Answer: True

Solution:

The sum of the squares of the direction cosines of a line is always equal to 1, as they represent the cosines of the angles made by the line with the coordinate axes.

Chapter Concept :

Direction Cosines and Ratios

Correct Answer: True

Solution:

Direction ratios are the differences in the coordinates of the two points, which are x2−x1x_2 - x_1, y2−y1y_2 - y_1, and z2−z1z_2 - z_1.

Chapter Concept :

Direction Ratios and Direction Cosines from Two Points

Correct Answer: True

Solution:

The given equation correctly represents the Cartesian form of a line in three-dimensional space through a point (x1,y1,z1)(x_1, y_1, z_1) with direction ratios a,b,ca, b, c.

Chapter Concept :

Cartesian Equation of Line Through Point and Direction Ratios

Correct Answer: True

Solution:

The direction cosines are calculated by normalizing the direction ratios, which involves dividing each ratio by the magnitude of the vector formed by the direction ratios.

Chapter Concept :

Conversion from Direction Ratios to Direction Cosines

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