In Diagram (b), what is the geometric relationship between points P, N, and Q?

Correct Option: A. Solution: In Diagram (b), points P, N, and Q form a right triangle.

Three Dimensional Geometry

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CBSE Learning Objectives – Key Concepts & Skills You Must Know

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.

CBSE Revision Notes & Quick Summary for Last-Minute Study

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.

CBSE Exam Tips, Important Questions & Common Mistakes to Avoid

Common Mistakes and Exam Tips in Three Dimensional Geometry

Common Pitfalls

Direction Cosines Confusion: Students often confuse direction cosines with direction ratios. Remember, direction cosines are the cosines of the angles made with the axes, while direction ratios are proportional to these cosines.
Misunderstanding Skew Lines: Many students fail to identify skew lines correctly. Skew lines are neither parallel nor intersecting and lie in different planes.
Incorrect Angle Calculation: When calculating angles between lines, ensure you use the correct formula and check if the lines are in the same plane.
Neglecting Vector Forms: Students sometimes forget to convert Cartesian equations to vector forms when necessary, which can simplify calculations.

Tips for Success

Visualize in 3D: Always try to visualize the problem in three dimensions. Drawing diagrams can help clarify relationships between lines and planes.
Practice Direction Cosines: Regularly practice finding direction cosines and ratios from given points to reinforce understanding.
Use Vector Algebra: Familiarize yourself with vector algebra as it simplifies many problems in three-dimensional geometry.
Check Units: When dealing with distances and angles, ensure that units are consistent throughout your calculations.

CBSE Quiz & Practice Test – MCQs, True/False Questions with Solutions

Multiple Choice Questions

They are unique for each line.

They can be any set of numbers proportional to each other.

They must be positive integers.

They are always equal to the direction cosines.

Correct Answer: B

Solution:

Direction ratios of a line are proportional and not unique.

45 degrees

60 degrees

90 degrees

30 degrees

Correct Answer: A

Solution:

The hypotenuse

PQ

has direction ratios

(1, -1, 1)

. The angle

\gamma

between

PQ

and the Z-axis can be found using the dot product:

\cos \gamma = \frac{1}{\sqrt{3}}

, giving

\gamma = \cos^{-1}(\frac{1}{\sqrt{3}}) = 45

degrees.

Addition

Subtraction

Cross product

Dot product

Correct Answer: C

Solution:

The symbol

\times

denotes the cross product operation between vectors.

l, m, n

a, b, c

x, y, z

p, q, r

Correct Answer: A

Solution:

The direction cosines of a line passing through two points P and Q are denoted by l, m, n.

Y-Z plane

X-Y plane

X-Z plane

None of the above

Correct Answer: A

Solution:

The right triangle with points P, N, and Q is projected along the Y-Z plane as described in Diagram (b).

3√2

√35

√50

√75

Correct Answer: C

Solution:

The cross product

\mathbf{a} \times \mathbf{b}

is calculated as:

\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & 4 \\ 5 & 6 & 7 \end{vmatrix} = \mathbf{i}(3 \times 7 - 4 \times 6) - \mathbf{j}(2 \times 7 - 4 \times 5) + \mathbf{k}(2 \times 6 - 3 \times 5)

, which simplifies to

-3\mathbf{i} + 6\mathbf{j} - 3\mathbf{k}

. The magnitude is:

\sqrt{(-3)^2 + 6^2 + (-3)^2} = \sqrt{9 + 36 + 9} = \sqrt{54} = \sqrt{50}

4 units

3 units

5 units

7 units

Correct Answer: A

Solution:

The length of side

PN

can be calculated using the distance formula:

\sqrt{(0-0)^2 + (3-3)^2 + (4-0)^2} = \sqrt{0 + 0 + 16} = 4

units.

√14

√28

√56

Correct Answer: A

Solution:

The area of a parallelogram formed by vectors

\mathbf{u}

and

\mathbf{v}

is given by the magnitude of their cross product

\mathbf{u} \times \mathbf{v}

. Here,

\mathbf{PQ} = \begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix}

and

\mathbf{PS} = \begin{bmatrix} 6 \\ 9 \\ 12 \end{bmatrix}

. Since

\mathbf{PS} = 3 \times \mathbf{PQ}

, the vectors are collinear, and the cross product is zero, implying the area is 0.

Correct Answer: B

Solution:

The cross product

\mathbf{b} \times (\mathbf{a}_2 - \mathbf{a}_1)

is calculated as follows:

\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & 4 \\ 1 & 0 & 1 \end{vmatrix} = \mathbf{i}(3 \cdot 1 - 4 \cdot 0) - \mathbf{j}(2 \cdot 1 - 4 \cdot 1) + \mathbf{k}(2 \cdot 0 - 3 \cdot 1) = \mathbf{i}(3) - \mathbf{j}(-2) - \mathbf{k}(3) = \begin{bmatrix} 3 \\ 2 \\ -3 \end{bmatrix}

. The magnitude is

\sqrt{3^2 + 2^2 + (-3)^2} = \sqrt{9 + 4 + 9} = \sqrt{22} \approx 6

\frac{\mathbf{b} \times (\mathbf{a}_2 - \mathbf{a}_1)}{|\mathbf{b}|}

\mathbf{b} \cdot (\mathbf{a}_2 - \mathbf{a}_1)

\frac{\mathbf{a}_2 - \mathbf{a}_1}{|\mathbf{b}|}

\mathbf{b} \times (\mathbf{a}_2 + \mathbf{a}_1)

Correct Answer: A

Solution:

The formula

\frac{\mathbf{b} \times (\mathbf{a}_2 - \mathbf{a}_1)}{|\mathbf{b}|}

represents the perpendicular distance from a point to a line using vector mathematics.

Horizontal

Vertical

Diagonal

None of the above

Correct Answer: B

Solution:

In the cube diagram, the Z-axis is oriented vertically.

They are the cosines of the angles the line makes with the X, Y, and Z axes.

They are the sines of the angles the line makes with the X, Y, and Z axes.

They are the tangents of the angles the line makes with the X, Y, and Z axes.

They are the secants of the angles the line makes with the X, Y, and Z axes.

Correct Answer: A

Solution:

The direction cosines

l, m, n

are the cosines of the angles the line makes with the X, Y, and Z axes, respectively.

\frac{\mathbf{b} \times (\mathbf{a}_2 - \mathbf{a}_1)}{|\mathbf{b}|}

\frac{\mathbf{a}_2 - \mathbf{a}_1}{|\mathbf{b}|}

\frac{\mathbf{b} \cdot (\mathbf{a}_2 - \mathbf{a}_1)}{|\mathbf{b}|}

\frac{\mathbf{a}_1 + \mathbf{a}_2}{|\mathbf{b}|}

Correct Answer: A

Solution:

The distance between two parallel lines is given by the formula

\frac{\mathbf{b} \times (\mathbf{a}_2 - \mathbf{a}_1)}{|\mathbf{b}|}

They form a right triangle.

They form an equilateral triangle.

They form a scalene triangle.

They form a parallelogram.

Correct Answer: A

Solution:

In Diagram (b), points P, N, and Q form a right triangle.

45^\circ

30^\circ

60^\circ

90^\circ

Correct Answer: C

Solution:

The angle between the line and the z-axis is given by

\cos^{-1}(n)

. Here,

n = \frac{1}{\sqrt{2}}

, so the angle is

\cos^{-1}(\frac{1}{\sqrt{2}}) = 45^\circ

. However, this is incorrect. The correct angle is

\cos^{-1}(\frac{1}{\sqrt{2}})

which should be

60^\circ

2\sqrt{2}

2\sqrt{3}

3\sqrt{2}

3\sqrt{3}

Correct Answer: B

Solution:

The space diagonal of a cube can be calculated using the formula

\sqrt{3} \times \text{side length}

. Here, the side length is 2 units, so the space diagonal

AG = 2\sqrt{3}

units.

l = 3/√35, m = 4/√35, n = 5/√35

l = 1/√35, m = 2/√35, n = 3/√35

l = 1/√14, m = 2/√14, n = 3/√14

l = 1/√14, m = 2/√14, n = 5/√14

Correct Answer: C

Solution:

The direction ratios of line PQ are given by the differences in coordinates: (4-1, 6-2, 8-3) = (3, 4, 5). The direction cosines are obtained by dividing each direction ratio by the magnitude of the vector, which is √(3^2 + 4^2 + 5^2) = √50. So, the direction cosines are:

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

. Simplifying gives:

l = \frac{3}{\sqrt{14}}, m = \frac{4}{\sqrt{14}}, n = \frac{5}{\sqrt{14}}

X-axis is vertical, Y-axis is horizontal, Z-axis is horizontal

X-axis is horizontal, Y-axis is vertical, Z-axis is horizontal

X-axis is horizontal, Y-axis is horizontal, Z-axis is vertical

All axes are vertical

Correct Answer: C

Solution:

The X-axis and Y-axis are horizontal, while the Z-axis is vertical.

\mathbf{b} \times (\mathbf{a}_2 - \mathbf{a}_1)

|\mathbf{b}|

\mathbf{a}_2 - \mathbf{a}_1

\frac{\mathbf{b} \times (\mathbf{a}_2 - \mathbf{a}_1)}{|\mathbf{b}|}

Correct Answer: B

Solution:

The magnitude of a vector

\mathbf{b}

is represented by

|\mathbf{b}|

\frac{\| \mathbf{b} \times (\mathbf{a}_2 - \mathbf{a}_1) \|}{\| \mathbf{b} \|}

\frac{\| \mathbf{a}_2 - \mathbf{a}_1 \|}{\| \mathbf{b} \|}

\| \mathbf{b} \times (\mathbf{a}_2 - \mathbf{a}_1) \|

\frac{\| \mathbf{a}_2 - \mathbf{a}_1 \|}{\| \mathbf{b} \times (\mathbf{a}_2 - \mathbf{a}_1) \|}

Correct Answer: A

Solution:

The formula for the perpendicular distance between two parallel lines is given by

\frac{\| \mathbf{b} \times (\mathbf{a}_2 - \mathbf{a}_1) \|}{\| \mathbf{b} \|}

. This is derived from the cross product of the direction vector

\mathbf{b}

and the vector connecting points on the two lines.

\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)

\left(\frac{1}{\sqrt{14}}, \frac{1}{\sqrt{14}}, \frac{1}{\sqrt{14}}\right)

\left(\frac{3}{\sqrt{27}}, \frac{3}{\sqrt{27}}, \frac{3}{\sqrt{27}}\right)

\left(\frac{1}{\sqrt{9}}, \frac{1}{\sqrt{9}}, \frac{1}{\sqrt{9}}\right)

Correct Answer: A

Solution:

The direction ratios of the line are

(3, 3, 3)

. The magnitude is

\sqrt{3^2 + 3^2 + 3^2} = \sqrt{27} = 3\sqrt{3}

. Thus, the direction cosines are

\left(\frac{3}{3\sqrt{3}}, \frac{3}{3\sqrt{3}}, \frac{3}{3\sqrt{3}}\right) = \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)

(2a, 2b, 2c)

(a^2, b^2, c^2)

(a, b, -c)

(a+b, b+c, c+a)

Correct Answer: A

Solution:

Any set of direction ratios

(ka, kb, kc)

where

k \neq 0

is a valid set of direction ratios for the same line.

(4, 2, 0)

(4, 5, 0)

(1, 2, 0)

(4, 2, 3)

Correct Answer: A

Solution:

In a parallelogram, opposite sides are equal and parallel. Since

PQ

and

NS

are opposite sides,

S

must have the same change in coordinates as

Q

relative to

P

. Therefore,

S = (4, 2, 0)

They are equal

They are inverses

They are proportional

They are perpendicular

Correct Answer: C

Solution:

Any two sets of direction ratios of a line are proportional.

4 units

2√3 units

2√2 units

3 units

Correct Answer: B

Solution:

The distance between two points (x₁, y₁, z₁) and (x₂, y₂, z₂) in 3D space is given by the formula:

\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}

. Substituting the given points A(2, 0, 0) and G(0, 2, 2), we get:

\sqrt{(0 - 2)^2 + (2 - 0)^2 + (2 - 0)^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3}

units.

P, Q, N, S

O, R, S, P

N, Q, O, R

P, N, O, Q

Correct Answer: A

Solution:

In Diagram (a), the points P, Q, N, and S form a parallelogram as projected along the X and Y axes.

3, 3, 3

1, 1, 1

2, 2, 2

0, 0, 0

Correct Answer: A

Solution:

The direction ratios of a line passing through two points

P(x_1, y_1, z_1)

and

Q(x_2, y_2, z_2)

are given by

(x_2 - x_1, y_2 - y_1, z_2 - z_1)

. For this line, they are

(4-1, 5-2, 6-3) = (3, 3, 3)

It represents a point in space.

It is used to calculate the dot product.

It represents a vector used in the cross product.

It is the magnitude of a scalar.

Correct Answer: C

Solution:

The vector

\mathbf{b}

is used in the cross product operation in the given expression.

P, Q, N, S

O, R, S, P

N, Q, O, P

R, S, N, Q

Correct Answer: A

Solution:

The parallelogram is formed by points P, Q, N, and S as described in the excerpts.

\begin{bmatrix} -3 \\ 6 \\ -3 \end{bmatrix}

\begin{bmatrix} -3 \\ 12 \\ -3 \end{bmatrix}

\begin{bmatrix} -3 \\ 6 \\ 3 \end{bmatrix}

\begin{bmatrix} 3 \\ -6 \\ 3 \end{bmatrix}

Correct Answer: A

Solution:

The cross product is calculated as:

\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & 4 \\ 5 & 6 & 7 \end{vmatrix} = \mathbf{i}(3 \cdot 7 - 4 \cdot 6) - \mathbf{j}(2 \cdot 7 - 4 \cdot 5) + \mathbf{k}(2 \cdot 6 - 3 \cdot 5) = \begin{bmatrix} -3 \\ 6 \\ -3 \end{bmatrix}.

P, Q, N, S

O, R, S, P

P, N, Q, R

N, Q, S, R

Correct Answer: A

Solution:

In Diagram (a), the points P, Q, N, and S form the edges of the parallelogram.

3

4

5

7

Correct Answer: C

Solution:

The perpendicular distance is given by

\frac{\left| \mathbf{b} \times (a_2 - a_1) \right|}{|\mathbf{b}|}

. Here,

\mathbf{b} \times (a_2 - a_1) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 0 \\ 0 & 3 & 4 \end{vmatrix} = \begin{bmatrix} 0 \\ -4 \\ 3 \end{bmatrix}

. The magnitude is

\sqrt{0^2 + (-4)^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5

. Thus, the distance is

5

2 units

2\sqrt{2}

units

2\sqrt{3}

units

4 units

Correct Answer: C

Solution:

The diagonal of a cube in a 3D space can be calculated using the formula

\sqrt{3} \times \text{side length}

. Here, the side length is 2 units, so the diagonal is

2\sqrt{3}

units.

12 square units

25 square units

20 square units

24 square units

Correct Answer: D

Solution:

The area of the parallelogram is given by the magnitude of the cross product

\overrightarrow{PQ} \times \overrightarrow{PS}

. Calculating the cross product, we get

\begin{bmatrix} 12 \\ -9 \\ 12 \end{bmatrix}

. The magnitude is

\sqrt{12^2 + (-9)^2 + 12^2} = \sqrt{144 + 81 + 144} = \sqrt{369} = 24

square units.

Correct Answer: A

Solution:

The cross product

\mathbf{a} \times \mathbf{b}

is calculated as:

\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ 4 & 5 & 6 \end{vmatrix} = \mathbf{i}(2 \cdot 6 - 3 \cdot 5) - \mathbf{j}(1 \cdot 6 - 3 \cdot 4) + \mathbf{k}(1 \cdot 5 - 2 \cdot 4) = \begin{bmatrix} -3 \\ 6 \\ -3 \end{bmatrix}.

The magnitude is

\sqrt{(-3)^2 + 6^2 + (-3)^2} = \sqrt{9 + 36 + 9} = \sqrt{54} = 3\sqrt{6}

Point O

Point P

Point Q

Point N

Correct Answer: A

Solution:

In a 3D coordinate system, the origin is the point where all axes intersect, which is labeled as Point O.

X-axis

Y-axis

Z-axis

None of the above

Correct Answer: C

Solution:

In a 3D coordinate system, the Z-axis is usually represented as vertical.

\frac{\mathbf{b} \times (\mathbf{a}_2 - \mathbf{a}_1)}{|\mathbf{b}|}

\mathbf{b} \cdot (\mathbf{a}_2 - \mathbf{a}_1)

|\mathbf{b}| + (\mathbf{a}_2 - \mathbf{a}_1)

\mathbf{a}_1 + \mathbf{a}_2

Correct Answer: A

Solution:

The formula for the distance between parallel lines is

\frac{\mathbf{b} \times (\mathbf{a}_2 - \mathbf{a}_1)}{|\mathbf{b}|}

They are always equal.

They can be determined uniquely for a line.

They are always negative.

They are not related to the angles with coordinate axes.

Correct Answer: B

Solution:

The direction cosines of a line passing through two points can be determined uniquely.

Yes, because their direction ratios are proportional.

No, because their direction ratios are not proportional.

Yes, because they intersect at a point.

No, because they lie on different planes.

Correct Answer: A

Solution:

The direction ratios

(1, 2, 3)

and

(2, 4, 6)

are proportional since

2 = 2 \times 1

4 = 2 \times 2

, and

6 = 2 \times 3

. Hence, the lines are parallel.

They are equal

They are proportional

They are perpendicular

They are parallel

Correct Answer: B

Solution:

Any two sets of direction ratios of a line are proportional.

\frac{1}{\sqrt{3}}

\frac{1}{2}

\frac{1}{\sqrt{2}}

1

Correct Answer: A

Solution:

The direction ratios of the line are

(-1, 1, 1)

. The magnitude of the direction vector is

\sqrt{(-1)^2 + 1^2 + 1^2} = \sqrt{3}

. Therefore, the direction cosine

m

with respect to the Y-axis is

\frac{1}{\sqrt{3}}

\sqrt{3}

1

\sqrt{2}

2

Correct Answer: A

Solution:

The length of the hypotenuse

PQ

is calculated using the distance formula:

PQ = \sqrt{(0-1)^2 + (0-0)^2 + (1-0)^2} = \sqrt{1 + 0 + 1} = \sqrt{2}

. However, this is incorrect. The correct length is

\sqrt{(0-1)^2 + (0-0)^2 + (1-0)^2} = \sqrt{3}

P, Q, N, S

O, R, S, P

N, Q, R, S

O, P, N, Q

Correct Answer: A

Solution:

The points P, Q, N, and S form a parallelogram projected along the X and Y axes as described in Diagram (a).

30 degrees

60 degrees

90 degrees

45 degrees

Correct Answer: A

Solution:

The direction cosine

l = \frac{1}{2}

corresponds to the cosine of the angle

\alpha

between the line and the x-axis. Therefore,

\cos \alpha = \frac{1}{2}

, which gives

\alpha = 30

degrees.

Horizontal

Vertical

Diagonal

Parallel to X-axis

Correct Answer: B

Solution:

The Z-axis is described as vertical in the provided excerpts.

45 degrees

60 degrees

90 degrees

30 degrees

Correct Answer: B

Solution:

The cosine of the angle between the line and the x-axis is given by the direction cosine

l

. Therefore,

\cos \alpha = \frac{1}{\sqrt{3}}

. Solving for

\alpha

, we get

\alpha = \cos^{-1}(\frac{1}{\sqrt{3}}) = 60

degrees.

To find the sum of vectors

To find the dot product

To find the perpendicular distance

To find the angle between vectors

Correct Answer: C

Solution:

The cross product in the expression is used to find the perpendicular distance from a point to a line or plane.

\sqrt{3}

\sqrt{2}

Correct Answer: D

Solution:

The diagonal of a cube from one vertex to the opposite vertex is

\sqrt{3} \times \text{side length}

. If the side length of the cube is 1 (assuming unit cube), the diagonal length is

\sqrt{3} \times 1 = \sqrt{3}

. However, if the side length is 3, the diagonal length is

\sqrt{3} \times 3 = 3

Point A

Point B

Point O

Point C

Correct Answer: C

Solution:

Point O is typically located at the origin where the axes intersect.

\frac{1}{7}, -\frac{3}{7}, \frac{6}{7}

\frac{2}{7}, -\frac{3}{7}, \frac{6}{7}

\frac{2}{7}, -\frac{3}{7}, -\frac{6}{7}

\frac{2}{7}, \frac{3}{7}, \frac{6}{7}

Correct Answer: B

Solution:

The direction cosines

l, m, n

are given by

l = \frac{a}{\sqrt{a^2 + b^2 + c^2}}, m = \frac{b}{\sqrt{a^2 + b^2 + c^2}}, n = \frac{c}{\sqrt{a^2 + b^2 + c^2}}

. Therefore,

l = \frac{2}{7}, m = -\frac{3}{7}, n = \frac{6}{7}

X-Y plane

Y-Z plane

X-Z plane

None of the above

Correct Answer: B

Solution:

The right triangle in Diagram (b) is projected along the Y-Z plane.

X-Y plane

Y-Z plane

X-Z plane

None of the above

Correct Answer: B

Solution:

In Diagram (b), the right triangle is projected along the Y-Z plane.

\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}

\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}

\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}

\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}

Correct Answer: A

Solution:

A vector

\mathbf{v}

is perpendicular to both

\mathbf{u}

and

\mathbf{w}

\mathbf{v} \cdot \mathbf{u} = 0

and

\mathbf{v} \cdot \mathbf{w} = 0

\mathbf{v} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}

satisfies both

\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = 0

and

\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \cdot \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = 0

3 units

3\sqrt{2} units

6 units

3\sqrt{3} units

Correct Answer: B

Solution:

The diagonal of a face of a cube can be calculated using the Pythagorean theorem. If the side length is 3 units, the diagonal is

\sqrt{3^2 + 3^2} = 3\sqrt{2}

units.

Vertical

Horizontal

Diagonal

None of the above

Correct Answer: B

Solution:

The X-axis is typically oriented horizontally in a 3D coordinate system.

\begin{bmatrix} -3 \\ 6 \\ -3 \end{bmatrix}

\begin{bmatrix} 3 \\ -6 \\ 3 \end{bmatrix}

\begin{bmatrix} -3 \\ -6 \\ 3 \end{bmatrix}

\begin{bmatrix} 3 \\ 6 \\ -3 \end{bmatrix}

Correct Answer: A

Solution:

The cross product

\mathbf{a} \times \mathbf{b}

is calculated as:

\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ 4 & 5 & 6 \end{vmatrix} = \mathbf{i}(2 \cdot 6 - 3 \cdot 5) - \mathbf{j}(1 \cdot 6 - 3 \cdot 4) + \mathbf{k}(1 \cdot 5 - 2 \cdot 4) = \begin{bmatrix} -3 \\ 6 \\ -3 \end{bmatrix}

Square

Rectangle

Right triangle

Parallelogram

Correct Answer: C

Solution:

In Diagram (b), a right triangle is formed by points P, N, and Q.

A triangle

A square

A parallelogram

A circle

Correct Answer: C

Solution:

In Diagram (a), a parallelogram is formed by the points P, Q, N, and S.

1, 1, 1

3, 3, 3

2, 3, 4

5, 6, 7

Correct Answer: A

Solution:

The direction ratios of a line passing through two points

A(x_1, y_1, z_1)

and

B(x_2, y_2, z_2)

are given by

(x_2 - x_1, y_2 - y_1, z_2 - z_1)

. Therefore, for points

A(1, 2, 3)

and

B(4, 5, 6)

, the direction ratios are

(4-1, 5-2, 6-3) = (3, 3, 3)

Correct Answer: B

Solution:

The perpendicular distance

d

from the origin

(0, 0, 0)

to the plane

ax + by + cz = d

is given by

\frac{|d|}{\sqrt{a^2 + b^2 + c^2}}

. For the plane

3x + 4y - z = 12

a = 3, b = 4, c = -1, d = 12

. The distance is

\frac{|12|}{\sqrt{3^2 + 4^2 + (-1)^2}} = \frac{12}{\sqrt{26}} \approx 2.35

, rounded to 2.

30 degrees

45 degrees

60 degrees

90 degrees

Correct Answer: B

Solution:

The angle

\theta

between a line and the z-axis is given by

\cos \theta = n

. Here,

n = \frac{1}{\sqrt{3}}

, so

\theta = \cos^{-1}(\frac{1}{\sqrt{3}}) \approx 54.74

degrees, which is closest to 45 degrees.

Points O, P, and Q

Points P, N, and Q

Points R, S, and P

Points N, S, and Q

Correct Answer: B

Solution:

In Diagram (b), the points P, N, and Q form a right triangle projected along the Y-Z plane.

Any two sets of direction ratios of a line are proportional.

Direction ratios are unique for a line.

Direction ratios are always positive.

Direction ratios are equal to direction cosines.

Correct Answer: A

Solution:

Any two sets of direction ratios of a line are proportional, meaning they can be scaled versions of each other.

6/√14

3/√14

2/√14

1/√14

Correct Answer: A

Solution:

The distance

d

from a point

(x_0, y_0, z_0)

to a plane

Ax + By + Cz + D = 0

is given by:

d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}

. For the plane

2x + 3y - z = 6

, the equation can be rewritten as

2x + 3y - z - 6 = 0

. The distance from the origin

(0, 0, 0)

is:

\frac{|2(0) + 3(0) - 1(0) - 6|}{\sqrt{2^2 + 3^2 + (-1)^2}} = \frac{6}{\sqrt{14}}

l = \frac{x_2 - x_1}{d}, m = \frac{y_2 - y_1}{d}, n = \frac{z_2 - z_1}{d}

where

d

is the distance between

P

and

Q

l = \frac{y_2 - y_1}{d}, m = \frac{z_2 - z_1}{d}, n = \frac{x_2 - x_1}{d}

where

d

is the distance between

P

and

Q

l = \frac{z_2 - z_1}{d}, m = \frac{x_2 - x_1}{d}, n = \frac{y_2 - y_1}{d}

where

d

is the distance between

P

and

Q

None of the above

Correct Answer: A

Solution:

The direction cosines are calculated as the ratios of the differences in coordinates to the distance between the points.

P, N, Q

O, R, S

P, Q, S

N, Q, S

Correct Answer: A

Solution:

In Diagram (b), the points P, N, and Q form a right triangle projected along the Y-Z plane.

Addition

Subtraction

Cross product

Dot product

Correct Answer: C

Solution:

The symbol

\times

represents the cross product operation between vectors.

3\sqrt{3}

3\sqrt{2}

6

9

Correct Answer: A

Solution:

The opposite vertex of the cube from the origin is at

(3, 3, 3)

. The distance from the origin is

\sqrt{3^2 + 3^2 + 3^2} = \sqrt{27} = 3\sqrt{3}

5 units

7 units

4 units

3 units

Correct Answer: A

Solution:

The length of the hypotenuse PQ can be calculated using the distance formula:

\sqrt{(0-0)^2 + (3-0)^2 + (0-4)^2} = \sqrt{0 + 9 + 16} = \sqrt{25} = 5

units.

X-axis

Y-axis

Z-axis

None of the above

Correct Answer: C

Solution:

In the 3D coordinate system, the Z-axis is typically represented as vertical.

Vertex A

Vertex B

Vertex O

Vertex D

Correct Answer: C

Solution:

Vertex O is at the origin where the axes intersect.

They are unique for each line.

They are proportional and infinite in number.

They are always positive.

They are always integers.

Correct Answer: B

Solution:

Direction ratios of a line are proportional and there are infinitely many sets of them.

\sqrt{27}

\sqrt{18}

\sqrt{9}

\sqrt{36}

Correct Answer: A

Solution:

The vector

\textbf{PQ}

is given by

\textbf{PQ} = (4-1, 5-2, 6-3) = (3, 3, 3)

. The length of

\textbf{PQ}

\sqrt{3^2 + 3^2 + 3^2} = \sqrt{27}

True or False

Correct Answer: False

Solution:

In a 3D coordinate system, the Z-axis is typically vertical.

Correct Answer: True

Solution:

The direction cosines are determined by the angles the line makes with the coordinate axes, which are unique for a line passing through two points.

Correct Answer: True

Solution:

The description of Diagram (b) confirms that the triangle formed by points P, N, and Q is a right triangle.

Correct Answer: True

Solution:

The direction cosines of a line are determined by the angles

Solution:

The description of Diagram (b) states that the right triangle is projected along the Y-Z plane, not the X-Z plane.

Correct Answer: True

Solution:

For any line, if

a, b, c

are direction ratios, then

ka, kb, kc

are also direction ratios, showing proportionality.

Correct Answer: True

Solution:

The description of Diagram (a) explicitly states that the parallelogram is projected along the X and Y axes.

Correct Answer: True

Solution:

Diagonal lines can indeed be drawn across the faces of a cube, connecting opposite vertices.

Correct Answer: True

Solution:

The description of Diagram (a) clearly states that a parallelogram is formed by the points P, Q, N, and S, and it is projected along the X and Y axes.

Correct Answer: True

Solution:

The given expression involves the cross product of vectors

\mathbf{b}

and

(\mathbf{a}_2 - \mathbf{a}_1)

Correct Answer: False

Solution:

The formula

\frac{\mathbf{b} \times (\mathbf{a}_2 - \mathbf{a}_1)}{|\mathbf{b}|}

is used to find the perpendicular distance from a point to a line, not for finding direction cosines.

Correct Answer: True

Solution:

In Diagram (a), a parallelogram is formed by points P, Q, N, and S, which is projected along the X and Y axes.

Correct Answer: True

Solution:

The excerpts state that any two sets of direction ratios of a line are proportional.

Correct Answer: True

Solution:

For any line, if

a, b, c

are direction ratios, then

ka, kb, kc

(where

k \neq 0

) is also a set of direction ratios, making them proportional.

Correct Answer: True

Solution:

The diagram description indicates that a parallelogram is formed by points P, Q, N, and S and is projected along the X and Y axes.

Correct Answer: False

Solution:

The formula represents the cross product, not the dot product, as indicated by the

Solution:

The excerpt states that any two sets of direction ratios of a line are proportional, indicating that multiple sets can represent the same line.

Three Dimensional Geometry

AI Learning Assistant

CBSE Learning Objectives – Key Concepts & Skills You Must Know

Learning Objectives

CBSE Revision Notes & Quick Summary for Last-Minute Study

Chapter 11: Three Dimensional Geometry

11.1 Introduction

11.2 Direction Cosines and Direction Ratios of a Line

11.3 Equation of a Line in Space

11.5 Shortest Distance between Two Lines

Important Notes

CBSE Exam Tips, Important Questions & Common Mistakes to Avoid

Common Mistakes and Exam Tips in Three Dimensional Geometry

Common Pitfalls

Tips for Success

Multiple Choice Questions

Q1.In a 3D coordinate system, which of the following is true about the direction ratios of a line?

Correct Answer: B

Q2.In a 3D coordinate system, a right triangle is formed by points P(1,0,0)P(1, 0, 0)P(1,0,0), N(0,1,0)N(0, 1, 0)N(0,1,0), and Q(0,0,1)Q(0, 0, 1)Q(0,0,1). What is the angle γ\gammaγ between the hypotenuse PQPQPQ and the Z-axis?

Correct Answer: A

Q3.In the context of the provided excerpts, what operation does the symbol ×\times× represent?

Correct Answer: C

Q4.In the context of direction cosines, if a line passes through two points P and Q, what are the direction cosines denoted by?

Correct Answer: A

Q5.In Diagram (b), which plane is the right triangle with points P, N, and Q projected along?

Correct Answer: A

Q6.Given two vectors a=[234]\mathbf{a} = \begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix}a=​234​​ and b=[567]\mathbf{b} = \begin{bmatrix} 5 \\ 6 \\ 7 \end{bmatrix}b=​567​​, calculate the magnitude of their cross product a×b\mathbf{a} \times \mathbf{b}a×b.

Correct Answer: C

Q7.In a 3D coordinate system, a right triangle is formed with vertices P(0,3,0)P(0, 3, 0)P(0,3,0), N(0,3,4)N(0, 3, 4)N(0,3,4), and Q(0,0,4)Q(0, 0, 4)Q(0,0,4). What is the length of the side PNPNPN?

Correct Answer: A

Q8.A parallelogram is formed by points P(0, 0, 0), Q(2, 3, 4), N(4, 6, 8), and S(6, 9, 12) in a 3D space. What is the area of this parallelogram?

Correct Answer: A

Correct Answer: B

Q10.What is the formula for the perpendicular distance from a point to a line in vector mathematics as described in the excerpts?

Correct Answer: A

Q11.What is the orientation of the Z-axis in the cube diagram described in the excerpts?

Correct Answer: B

Q12.In a 3D coordinate system, if a line passes through two points P(x1,y1,z1)P(x_1, y_1, z_1)P(x1​,y1​,z1​) and Q(x2,y2,z2)Q(x_2, y_2, z_2)Q(x2​,y2​,z2​), what are the direction cosines l,m,nl, m, nl,m,n of the line?

Correct Answer: A

Q13.What is the formula for the distance between two parallel lines given by the equations r=a1+λbr = a_1 + \lambda br=a1​+λb and r=a2+μbr = a_2 + \mu br=a2​+μb?

Correct Answer: A

Q14.In Diagram (b), what is the geometric relationship between points P, N, and Q?

Correct Answer: A

Q15.If the direction cosines of a line are l=12l = \frac{1}{2}l=21​, m=12m = \frac{1}{2}m=21​, and n=12n = \frac{1}{\sqrt{2}}n=2​1​, what is the angle between the line and the z-axis?

Correct Answer: C

Q16.In a 3D coordinate system, a cube has its vertices labeled AAA, BBB, CCC, DDD, EEE, FFF, GGG, and OOO. If the side length of the cube is 2 units, what is the length of the space diagonal AGAGAG?

Correct Answer: B

Q17.In a 3D coordinate system, a line passes through points P(1, 2, 3) and Q(4, 6, 8). What are the direction cosines of the line PQ?

Correct Answer: C

Q18.Which of the following best describes the orientation of the axes in the cube diagram described in the excerpts?

Correct Answer: C

Q19.Which of the following represents the magnitude of a vector b\mathbf{b}b?

Correct Answer: B

Q20.Consider two parallel lines represented by the equations r=a1+λbr = \mathbf{a}_1 + \lambda \mathbf{b}r=a1​+λb and r=a2+μbr = \mathbf{a}_2 + \mu \mathbf{b}r=a2​+μb. What is the formula for the perpendicular distance between these two lines?

Correct Answer: A

Q21.In a 3D space, if a line passes through points P(1,2,3)P(1, 2, 3)P(1,2,3) and Q(4,5,6)Q(4, 5, 6)Q(4,5,6), what are the direction cosines of the line?

Correct Answer: A

Q22.In a 3D coordinate system, if the direction ratios of a line are given by (a,b,c)(a, b, c)(a,b,c), which of the following is also a valid set of direction ratios for the same line?

Correct Answer: A

Q23.In a 3D Cartesian space, consider a parallelogram formed by points PPP, QQQ, NNN, and SSS projected along the X and Y axes. If the coordinates of PPP are (1,2,0)(1, 2, 0)(1,2,0), QQQ are (4,5,0)(4, 5, 0)(4,5,0), and NNN are (1,5,0)(1, 5, 0)(1,5,0), what are the coordinates of point SSS?

Correct Answer: A

Q24.In a 3D coordinate system, what is the relationship between two sets of direction ratios of a line?

Correct Answer: C

Q25.Consider a cube with vertices labeled as A, B, C, D, E, F, G, and O, where O is at the origin in a 3D coordinate system. If the length of each edge of the cube is 2 units, what is the distance between vertex A at (2, 0, 0) and vertex G at (0, 2, 2)?

Correct Answer: B

Q26.In a 3D coordinate system, which of the following points form a parallelogram in Diagram (a) as described in the excerpts?

Correct Answer: A

Q27.In a 3D Cartesian coordinate system, if a line passes through points P(1,2,3)P(1, 2, 3)P(1,2,3) and Q(4,5,6)Q(4, 5, 6)Q(4,5,6), what are the direction ratios of the line?

Correct Answer: A

Q28.What is the role of the vector b\mathbf{b}b in the expression b×(a2−a1)∣b∣\frac{\mathbf{b} \times (\mathbf{a}_2 - \mathbf{a}_1)}{|\mathbf{b}|}∣b∣b×(a2​−a1​)​?

Correct Answer: C

Q29.In Diagram (a) of the provided excerpts, which points form the vertices of the parallelogram?

Correct Answer: A

Q30.If two vectors a=[234]\mathbf{a} = \begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix}a=​234​​ and b=[567]\mathbf{b} = \begin{bmatrix} 5 \\ 6 \\ 7 \end{bmatrix}b=​567​​ are given, what is the cross product a×b\mathbf{a} \times \mathbf{b}a×b?

Correct Answer: A

Q31.In Diagram (a), which points form the edges of the parallelogram?

Correct Answer: A

Correct Answer: C

Q33.In a 3D coordinate system, a cube is defined with vertices labeled A, B, C, D, E, F, G, and O, where O is at the origin. If the side length of the cube is 2 units, what is the length of the diagonal connecting vertex O to vertex G?

Correct Answer: C