Coordinates in space | NEB Class 11 Mathematics Notes

Chapter Overview

In two-dimensional geometry, we locate a point using an ordered pair $(x, y)$. However, our physical world is three-dimensional. To describe the position of an object in space—like a bird flying or a plane in the sky—we need a third reference called the Z-axis. Thus, every point is represented by an ordered triple $(x, y, z)$. This chapter forms the foundation for 3D Geometry and Vectors. In the NEB exams, this unit is crucial for both multiple-choice questions (MCQs) and long-answer analytical problems.

What this chapter covers

Octants and Signs of CoordinatesDistance Formula in 3DSection Formula (Internal, External, Midpoint)Direction Cosines (d.c's) — l, m, nDirection Ratios (d.r's) — a, b, cRelation between D.C's and D.R'sAngle Between Two LinesConditions for Parallel and Perpendicular LinesProjection of a Line Segment on Another Line

Learning Objectives

1

Identify the octant of a point based on the signs of its coordinates

2

Calculate distance between two points and from the origin in 3D space

3

Apply internal, external, and midpoint section formulae correctly in 3D

4

Understand and calculate Direction Cosines (d.c's) and Direction Ratios (d.r's)

5

Find the angle between two lines using d.c's and d.r's

6

State and apply conditions for lines to be parallel or perpendicular

7

Find the projection of a line segment on a line with given d.c's

The Power of the Third Axis — Z

Key Insight

Every concept from 2D coordinate geometry has a direct 3D analogue. The distance formula adds a $(z_2-z_1)^2$ term. The section formula adds a z-coordinate. The slope becomes direction cosines. If you mastered Straight Line (2D), Coordinates in Space (3D) is just one more dimension.

The fundamental identity $l^2 + m^2 + n^2 = 1$ is the 3D equivalent of $\cos^2\theta + \sin^2\theta = 1$ — it appears in almost every problem involving direction cosines.

In NEB exams: always check whether the problem gives d.c's (which satisfy $l^2+m^2+n^2=1$) or d.r's (any proportional numbers). Convert d.r's to d.c's before using angle or projection formulas.

Key Definitions

Origin (O)

The point where three mutually perpendicular axes (X, Y, Z) intersect, denoted as $(0, 0, 0)$.

Coordinate Axes

The three mutually perpendicular lines: X-axis, Y-axis, and Z-axis. Any point on an axis has two coordinates equal to zero.

Coordinate Planes

The three planes: XY-plane (where $z=0$), YZ-plane (where $x=0$), and ZX-plane (where $y=0$). Any point on a plane has one coordinate equal to zero.

Octants

The three axes divide space into eight parts called octants. The first octant has all coordinates positive.

Direction Cosines (l, m, n)

The cosines of the angles $\alpha, \beta, \gamma$ that a line makes with the positive X, Y, and Z axes respectively.

$$l = \cos \alpha, \; m = \cos \beta, \; n = \cos \gamma$$

Direction Angles

The angles $\alpha, \beta, \gamma$ made by a line with the positive X, Y, Z axes respectively. Range: $0^\circ \le \alpha, \beta, \gamma \le 180^\circ$.

Direction Ratios (a, b, c)

Any three numbers proportional to the direction cosines. They are not unique — multiplying by a constant gives the same line direction.

Octants and Sign Rules

Just as 2D has 4 quadrants, 3D has 8 octants. The signs of $(x, y, z)$ determine which octant a point lies in.

$$\text{If a point is on an axis: two coordinates are zero. If on a plane: one coordinate is zero.}$$

Why it works (Derivation):

1Octant I: $(+,+,+)$ — all coordinates positive. This is the 'first' octant analogous to Quadrant I in 2D.

2Octant II: $(-,+,+)$ — x negative, y and z positive.

3Octant III: $(-,-,+)$ — x and y negative, z positive.

4Octant IV: $(+,-,+)$ — x positive, y negative, z positive.

5Octants V–VIII: Same x-y patterns as I–IV, but with $z < 0$ (negative).

6Key rule: The first 4 octants have $Z$ positive (+), the last 4 have $Z$ negative (−).

Example: Point $(-1, 2, -3)$ has $x<0$, $y>0$, $z<0$. This matches the pattern of the 6th octant (X negative, Y positive, Z negative).

When to use: Knowing octants helps in visualization and sign-checking. In MCQs, octant identification is a common 1-mark question.

Complete Formula Sheet with Derivations

Distance Between Two Points in 3D

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

Direct extension of 2D distance formula. Add the squared z-difference.

1Let $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ be two points.

2Draw $AC$ and $BD$ perpendicular to the XY-plane. Coordinates in XY-plane: $(x_1, y_1)$ and $(x_2, y_2)$.

3In XY-plane, $CD^2 = (x_2-x_1)^2 + (y_2-y_1)^2$ by 2D distance formula.

4Draw $AE \perp BD$. Then $AE = CD$. In right-angled $\triangle ABE$: $AB^2 = AE^2 + BE^2$.

5Since $BE = z_2 - z_1$, substituting: $AB^2 = (x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2$.

6Therefore: $AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$. (Proved)

Distance from Origin

d = \sqrt{x^2 + y^2 + z^2}

Special case when one point is $(0,0,0)$. Frequently tested in MCQs.

1Substitute $(x_1,y_1,z_1) = (0,0,0)$ and $(x_2,y_2,z_2) = (x,y,z)$ into the distance formula:

2$d = \sqrt{(x-0)^2 + (y-0)^2 + (z-0)^2} = \sqrt{x^2+y^2+z^2}$

Internal Section Formula (3D)

R = \left(\frac{m_1x_2 + m_2x_1}{m_1+m_2},\; \frac{m_1y_2 + m_2y_1}{m_1+m_2},\; \frac{m_1z_2 + m_2z_1}{m_1+m_2}\right)

Point R divides AB internally in ratio $m_1 : m_2$. Identical structure to 2D, just add z-coordinate.

1Point $R(x,y,z)$ divides line joining $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ in ratio $m_1:m_2$.

2By similar triangles in the XY projection and using the z-coordinate difference: $\frac{AR}{RB} = \frac{m_1}{m_2}$.

3For x-coordinate: $\frac{x-x_1}{x_2-x} = \frac{m_1}{m_2}$. Cross multiply: $m_2(x-x_1) = m_1(x_2-x)$.

4Solving: $x = \frac{m_1x_2 + m_2x_1}{m_1+m_2}$. Same logic applies to y and z coordinates.

External Section Formula (3D)

R = \left(\frac{m_1x_2 - m_2x_1}{m_1-m_2},\; \frac{m_1y_2 - m_2y_1}{m_1-m_2},\; \frac{m_1z_2 - m_2z_1}{m_1-m_2}\right)

Replace the '+' sign with '−' in both numerator and denominator. Common mistake: forgetting the minus in the denominator.

1For external division, point R lies outside the segment AB on the side of A (if $m_1 > m_2$) or B (if $m_2 > m_1$).

2The ratio $\frac{AR}{RB} = \frac{m_1}{m_2}$ still holds, but R is outside.

3Using directed segments, the internal formula becomes external when we replace $m_2$ with $-m_2$.

Midpoint Formula (3D)

M = \left(\frac{x_1+x_2}{2},\; \frac{y_1+y_2}{2},\; \frac{z_1+z_2}{2}\right)

Special case of section formula with $m_1 = m_2 = 1$. Simply the average of coordinates.

1Set $m_1 = m_2 = 1$ in the internal section formula:

2$x = \frac{1\cdot x_2 + 1\cdot x_1}{1+1} = \frac{x_1+x_2}{2}$. Same for y and z.

Direction Cosines Identity

$$l^2 + m^2 + n^2 = 1$$

The most important identity in this chapter. If given $(l,m,n)$, always verify this first. If it doesn't hold, they are d.r's, not d.c's.

1Consider a line through origin making angles $\alpha, \beta, \gamma$ with X, Y, Z axes.

2Take a point $P(x,y,z)$ on the line at distance $r$ from origin. Then $x = r\cos\alpha = rl$, $y = rm$, $z = rn$.

3By distance formula: $r^2 = x^2 + y^2 + z^2 = r^2(l^2 + m^2 + n^2)$.

4Dividing by $r^2$: $l^2 + m^2 + n^2 = 1$. (Proved)

Converting D.R's to D.C's

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}}

Always divide by the magnitude $\sqrt{a^2+b^2+c^2}$. This ensures $l^2+m^2+n^2=1$.

1Given d.r's $(a,b,c)$, we know $(l,m,n) = k(a,b,c)$ for some constant $k$ (since d.r's are proportional to d.c's).

2Using $l^2+m^2+n^2 = 1$: $k^2(a^2+b^2+c^2) = 1$.

3Therefore $k = \frac{1}{\sqrt{a^2+b^2+c^2}}$ and $l = \frac{a}{\sqrt{a^2+b^2+c^2}}$, etc.

D.R's of Line Joining Two Points

(a,b,c) = (x_2-x_1,\; y_2-y_1,\; z_2-z_1)

Just subtract coordinates. The simplest way to find direction ratios of a line through two given points.

1The vector from $A(x_1,y_1,z_1)$ to $B(x_2,y_2,z_2)$ is $\vec{AB} = (x_2-x_1, y_2-y_1, z_2-z_1)$.

2The direction ratios of the line AB are the components of this vector.

Angle Between Two Lines (using d.c's)

\cos \theta = l_1l_2 + m_1m_2 + n_1n_2

Use this when d.c's are given. If d.r's are given, convert to d.c's first OR use the modified formula below.

1Let two lines have direction cosines $(l_1,m_1,n_1)$ and $(l_2,m_2,n_2)$.

2Consider vectors of unit length along these lines. Their dot product equals $\cos\theta$.

3Dot product = $l_1l_2 + m_1m_2 + n_1n_2 = \cos\theta$. (Proved)

Angle Between Two Lines (using d.r's)

\cos \theta = \frac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}

Use directly when d.r's are given. No need to convert to d.c's first.

1Convert d.r's to d.c's: $l_i = \frac{a_i}{\sqrt{a_i^2+b_i^2+c_i^2}}$, etc.

2Substitute into $\cos\theta = l_1l_2 + m_1m_2 + n_1n_2$ to get the formula above.

Perpendicularity Condition (d.r's)

$$a_1a_2 + b_1b_2 + c_1c_2 = 0$$

Think of this as a 'dot product' being zero. For perpendicular lines, the dot product of their d.r's equals zero.

1If lines are perpendicular, $\theta = 90^\circ$, so $\cos\theta = 0$.

2From the angle formula: $\cos\theta = 0$ implies the numerator $a_1a_2 + b_1b_2 + c_1c_2 = 0$.

Parallelism Condition (d.r's)

\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}

The d.r's are proportional. Same direction means same ratios (or same d.c's).

1If lines are parallel, they have the same direction. So their d.r's must be proportional: $(a_1,b_1,c_1) = k(a_2,b_2,c_2)$ for some constant $k$.

2This gives $a_1/a_2 = b_1/b_2 = c_1/c_2 = k$.

Projection of Line Segment PQ on a Line with d.c's (l,m,n)

\text{Projection} = (x_2-x_1)l + (y_2-y_1)m + (z_2-z_1)n

This is the dot product of vector PQ with the unit vector along the line. Gives the signed length of the shadow.

1Vector $\vec{PQ} = (x_2-x_1, y_2-y_1, z_2-z_1)$.

2The unit vector along the line with d.c's $(l,m,n)$ is $(l,m,n)$ itself (since $l^2+m^2+n^2=1$).

3Projection = dot product $\vec{PQ} \cdot (l,m,n) = (x_2-x_1)l + (y_2-y_1)m + (z_2-z_1)n$.

Memory Tricks — Never Forget These

Trick: Four quick memory hooks that cover the most common formula confusions in this chapter.

Steps to Remember:

D.C's Unity — Love My Nepal = 1: Remember $l^2 + m^2 + n^2 = 1$ using the mnemonic 'Love My Nepal'. Every direction cosine triplet must satisfy this. If it doesn't, they're d.r's, not d.c's.
Octant Signs — Z divides them: The first 4 octants have $Z$ positive (+), the last 4 have $Z$ negative (−). The x-y pattern is the same as quadrants I–IV repeated twice.
Perpendicular = Product Zero: $a_1a_2 + b_1b_2 + c_1c_2 = 0$. Think of it as a 'dot product' being zero. This is the same idea as $m_1m_2 = -1$ in 2D, but for 3D direction ratios.
Distance from Axis: To find distance of $(x,y,z)$ from the X-axis, use $\sqrt{y^2+z^2}$ — exclude the axis coordinate. Same pattern for Y-axis ($\sqrt{x^2+z^2}$) and Z-axis ($\sqrt{x^2+y^2}$).
$k:1$ Ratio Shortcut: For a plane dividing a line segment, if the plane is XY, set $z=0$ and solve for $k$. If YZ, set $x=0$. If ZX, set $y=0$.

Mnemonic: Love My Nepal = 1 for d.c's | Z sign divides octants | Dot product = 0 for perpendicular | Exclude axis coordinate for axis distance

Common Mistakes

These mistakes cause the most mark losses in NEB exams for this chapter.

Confusing D.C's and D.R's: D.C's must satisfy $l^2+m^2+n^2=1$, while D.R's are just any proportional numbers. Always convert d.r's to d.c's before using the angle formula with d.c's.
Sign errors in Section Formula: For external division, always use subtraction in both numerator and denominator. Students often forget the minus in the denominator.
Distance from Axis: To find the distance of $(x, y, z)$ from the X-axis, the formula is $\sqrt{y^2+z^2}$ (exclude the axis coordinate). Don't use $\sqrt{x^2+y^2+z^2}$ — that's distance from origin!
Using d.c's vs d.r's in angle formula: If using $\cos\theta = l_1l_2 + m_1m_2 + n_1n_2$, you MUST use d.c's. If using d.r's, use the full formula with denominators.
External division ratio confusion: When the ratio is $k:1$ and the dividing point is outside, the external formula has $m_1 - m_2$ in the denominator. Be careful with signs.

NEB Exam Traps — These Appear Every Year

NEB Exam Warning

Board examiners specifically test these edge cases. Knowing them can save 2–5 marks.

Asking for 'Find the point on XY-plane' — students forget to set $z=0$. On any coordinate plane, the missing coordinate is always zero.
'Show points are collinear' — in 3D, check if $PQ + QR = PR$ (distance check) OR if the d.r's of PQ and QR are proportional (same direction).
'Find the angle' — always find d.r's of both lines first, then convert to d.c's or use the d.r's angle formula directly. Don't mix d.c's and d.r's.
'Find the ratio' — assume ratio $k:1$ and set the relevant coordinate ($x$, $y$, or $z$) to zero based on the plane given. Solve for $k$.
Cube diagonal angle problem ($\cos^{-1}(1/3)$) — a board favorite. Use vertices from $(0,0,0)$ to $(a,a,a)$ and find d.c's of two face diagonals or space diagonals.
When d.c's satisfy two equations (like $l+m+n=0$ and $l^2+m^2-n^2=0$), solve the system to find the actual d.c's, then use the angle formula.

Question Recognition Guide

Keyword / Phrase in Question	Use This
Find point on XY-plane...	Set $z=0$ and solve
Find point on YZ-plane...	Set $x=0$ and solve
Find point on ZX-plane...	Set $y=0$ and solve
Show points are collinear...	Check if $PQ + QR = PR$ OR d.r's of PQ and QR are proportional
Find angle between two lines...	Find d.r's of both lines, then use $\cos\theta$ formula
Find ratio in which a plane divides...	Assume ratio $k:1$, set the relevant coordinate to zero, solve for $k$
Find d.c's of line perpendicular to two given lines...	Use cross-multiplication on the two perpendicularity conditions
Find projection of segment AB on line L...	Use $(x_2-x_1)l + (y_2-y_1)m + (z_2-z_1)n$ with d.c's of L
Direction ratios of a line making equal angles...	If $\alpha = \beta = \gamma$, then $l=m=n=\pm 1/\sqrt{3}$
Distance from origin...	$\sqrt{x^2+y^2+z^2}$
Distance from X-axis...	$\sqrt{y^2+z^2}$ (exclude x)
Condition for parallel lines...	$a_1/a_2 = b_1/b_2 = c_1/c_2$
Condition for perpendicular lines...	$a_1a_2 + b_1b_2 + c_1c_2 = 0$

Solved Examples (Easy → Hard)

1

Find the distance between $A(1, 2, -1)$ and $B(4, 6, 11)$.

Easy

Solution

Identify: $(x_1, y_1, z_1) = (1, 2, -1)$ and $(x_2, y_2, z_2) = (4, 6, 11)$.

Apply distance formula:
$$d = \sqrt{(4-1)^2 + (6-2)^2 + (11-(-1))^2}$$

$= \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169}$

Result: $d = 13$ units.

∴ Distance = 13 units

2

Find the ratio in which the YZ-plane divides the line segment joining $A(-2, 4, 7)$ and $B(3, -5, 8)$.

Medium

Solution

Step 1: On the YZ-plane, the $x$-coordinate is always 0.

Step 2: Let the ratio be $k:1$.

Step 3: Using section formula for $x$:
$$0 = \frac{k(3) + 1(-2)}{k+1}$$

Step 4: $3k - 2 = 0 \Rightarrow 3k = 2 \Rightarrow k = \frac{2}{3}$.

Result: The ratio is 2 : 3.

∴ Ratio = 2 : 3

3

Find the D.C's of a line perpendicular to two lines having D.R's $(1, -2, 3)$ and $(2, 1, -1)$.

Hard

Solution

Step 1: Let the required D.C's be $(l, m, n)$.

Step 2: By perpendicularity condition:
• $1l - 2m + 3n = 0$ ...(i)
• $2l + 1m - 1n = 0$ ...(ii)

Step 3: Using cross-multiplication:
$$\frac{l}{(-2)(-1)-(1)(3)} = \frac{m}{(3)(2)-(1)(-1)} = \frac{n}{(1)(1)-(2)(-2)}$$

$\frac{l}{2-3} = \frac{m}{6+1} = \frac{n}{1+4} \Rightarrow \frac{l}{-1} = \frac{m}{7} = \frac{n}{5}$

Step 4: Convert to D.C's by dividing by magnitude:
$\sqrt{(-1)^2 + 7^2 + 5^2} = \sqrt{1+49+25} = \sqrt{75} = 5\sqrt{3}$

Result: $(\frac{-1}{5\sqrt{3}}, \frac{7}{5\sqrt{3}}, \frac{5}{5\sqrt{3}})$.

∴ D.C's = $(-1/5\sqrt{3}, 7/5\sqrt{3}, 5/5\sqrt{3})$ or $(-1/5\sqrt{3}, 7/5\sqrt{3}, 1/\sqrt{3})$

MCQ Practice

Mixed Level — MCQ

1

The distance of point $(3, 4, 5)$ from the origin is:

Correct!

Incorrect. Correct: $\sqrt{50}$

Solution

Using distance from origin formula:
$$d = \sqrt{x^2+y^2+z^2} = \sqrt{3^2+4^2+5^2} = \sqrt{9+16+25} = \sqrt{50}$$

∴ $\sqrt{50}$

2

The $z$-coordinate of any point on the XY-plane is:

Correct!

Incorrect. Correct: 0

Solution

The XY-plane is defined by the equation $z = 0$. Every point on this plane has $z$-coordinate equal to 0, regardless of $x$ and $y$ values.

∴ 0

3

If a line makes equal angles with the axes, its d.c's are:

Correct!

Incorrect. Correct: $(\pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}})$

Solution

If $\alpha = \beta = \gamma$, then $l = m = n$.
Using $l^2+m^2+n^2 = 1$:
$3l^2 = 1 \Rightarrow l = \pm \frac{1}{\sqrt{3}}$.
Therefore d.c's are $(\pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}})$.

∴ $(\pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}})$

4

Direction ratios of the Y-axis are:

Correct!

Incorrect. Correct: $(0,1,0)$

Solution

The Y-axis is the set of points $(0, y, 0)$. Its direction vector is along the y-direction only: $(0, 1, 0)$.

∴ $(0,1,0)$

5

If $l, m, n$ are d.c's, then $l^2+m^2+n^2$ is:

Correct!

Incorrect. Correct: 1

Solution

By definition, direction cosines satisfy the identity:
$$l^2 + m^2 + n^2 = 1$$
This is the 3D analogue of $\cos^2\theta + \sin^2\theta = 1$.

∴ 1

6

The point $(-1, 2, -3)$ lies in which octant?

Correct!

Incorrect. Correct: 6th

Solution

For $(-1, 2, -3)$: $x < 0$ (negative), $y > 0$ (positive), $z < 0$ (negative).

The pattern $(-, +, -)$ corresponds to the 6th octant.

∴ 6th

7

Angle between lines with d.r's $(1,1,0)$ and $(0,1,1)$ is:

Correct!

Incorrect. Correct: $60^\circ$

Solution

Using the angle formula with d.r's:
$$\cos\theta = \frac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$$

$= \frac{(1)(0) + (1)(1) + (0)(1)}{\sqrt{1+1+0}\sqrt{0+1+1}} = \frac{1}{\sqrt{2}\sqrt{2}} = \frac{1}{2}$

Therefore $\theta = 60^\circ$.

∴ $60^\circ$

8

Ratio in which the XY-plane divides $(2,4,5)$ and $(3,5,6)$ is:

Correct!

Incorrect. Correct: $-5:6$

Solution

On XY-plane, $z = 0$. Let ratio be $k:1$.

$0 = \frac{k(6) + 1(5)}{k+1} \Rightarrow 6k + 5 = 0 \Rightarrow k = -\frac{5}{6}$.

Ratio = $-\frac{5}{6} : 1 = -5 : 6$ (external division).

∴ $-5:6$

9

Condition for $a_1, b_1, c_1$ and $a_2, b_2, c_2$ to be parallel:

Correct!

Incorrect. Correct: $a_1/a_2 = b_1/b_2 = c_1/c_2$

Solution

Two lines with d.r's $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$ are parallel if and only if their direction ratios are proportional:
$$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$$

∴ $a_1/a_2 = b_1/b_2 = c_1/c_2$

10

Projection of a point $(x,y,z)$ on X-axis is:

Correct!

Incorrect. Correct: $x$

Solution

The projection of a point $(x,y,z)$ onto the X-axis is simply the point $(x, 0, 0)$, so the projected length (coordinate) is $x$.

∴ $x$

Short Questions (2 Marks Each)

1

Find the distance of $(2, -3, 6)$ from the origin.

[2 marks]

Solution

Using distance from origin formula:
$$d = \sqrt{x^2+y^2+z^2} = \sqrt{2^2+(-3)^2+6^2} = \sqrt{4+9+36} = \sqrt{49} = 7$$
units.

∴ d = 7 units

2

Find the mid-point of $(1, 4, 5)$ and $(3, 0, -1)$.

[2 marks]

Solution

Using midpoint formula:
$$M = \left(\frac{1+3}{2},\; \frac{4+0}{2},\; \frac{5+(-1)}{2}\right) = (2, 2, 2)$$

∴ M = (2, 2, 2)

3

Find d.c's of line with d.r's $3, 4, 12$.

[2 marks]

Solution

Magnitude $= \sqrt{3^2+4^2+12^2} = \sqrt{9+16+144} = \sqrt{169} = 13$.

D.C's are:
$$l = \frac{3}{13}, \; m = \frac{4}{13}, \; n = \frac{12}{13}$$

∴ D.C's = $(3/13, 4/13, 12/13)$

4

Show that lines with d.r's $(1, 2, 3)$ and $(-2, 1, 0)$ are perpendicular.

[2 marks]

Solution

Using perpendicularity condition:
$a_1a_2 + b_1b_2 + c_1c_2 = (1)(-2) + (2)(1) + (3)(0) = -2 + 2 + 0 = 0$.

Since the dot product is 0, the lines are perpendicular.

∴ Perpendicular (dot product = 0)

5

Find the projection of the segment joining $(1, 2, 3)$ and $(4, 2, 3)$ on the X-axis.

[2 marks]

Solution

The d.c's of X-axis are $(1, 0, 0)$.

Using projection formula:
$= (4-1)(1) + (2-2)(0) + (3-3)(0) = 3 + 0 + 0 = 3$.

Alternatively, the projection on X-axis is simply the difference in x-coordinates: $4 - 1 = 3$.

∴ Projection = 3 units

6

Find the value of $k$ if $(1, 2, 3)$ and $(2, k, 5)$ are parallel to a line with d.r's $(1, 1, 2)$.

[2 marks]

Solution

D.R's of line joining $(1,2,3)$ and $(2,k,5)$:
$(2-1, k-2, 5-3) = (1, k-2, 2)$.

For parallel to $(1, 1, 2)$:
$$\frac{1}{1} = \frac{k-2}{1} = \frac{2}{2}$$

From $\frac{k-2}{1} = 1$: $k - 2 = 1 \Rightarrow k = 3$.

∴ k = 3

7

Write the coordinates of a point on the Z-axis at a distance of 5 units from origin.

[2 marks]

Solution

On Z-axis, $x = 0$ and $y = 0$.

Distance from origin = $\sqrt{0^2+0^2+z^2} = |z| = 5$.

So $z = \pm 5$.

Points: $(0, 0, 5)$ and $(0, 0, -5)$.

∴ $(0, 0, \pm 5)$

8

Find the angle between two lines with d.c's $(1/\sqrt{3}, 1/\sqrt{3}, 1/\sqrt{3})$ and $(1/\sqrt{2}, -1/\sqrt{2}, 0)$.

[2 marks]

Solution

Using $\cos\theta = l_1l_2 + m_1m_2 + n_1n_2$:

$= \frac{1}{\sqrt{3}}\cdot\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}}\cdot\left(-\frac{1}{\sqrt{2}}\right) + \frac{1}{\sqrt{3}}\cdot 0$

$= \frac{1}{\sqrt{6}} - \frac{1}{\sqrt{6}} + 0 = 0$.

Therefore $\theta = 90^\circ$.

∴ $\theta = 90^\circ$

9

If $\alpha, \beta, \gamma$ are direction angles, prove $\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 2$.

[2 marks]

Solution

Using $\sin^2\theta = 1 - \cos^2\theta$:

$\sin^2\alpha + \sin^2\beta + \sin^2\gamma = (1-\cos^2\alpha) + (1-\cos^2\beta) + (1-\cos^2\gamma)$

$= 3 - (\cos^2\alpha + \cos^2\beta + \cos^2\gamma)$

$= 3 - (l^2 + m^2 + n^2) = 3 - 1 = 2$.

Hence proved.

∴ Proved: $\sin^2\alpha + \sin^2\beta + \sin^2\gamma = 2$

10

Find the ratio in which the point $(3, 4, 5)$ divides the join of $(1, 2, 3)$ and $(4, 5, 6)$.

[2 marks]

Solution

Let ratio be $k:1$.

Using x-coordinate in section formula:
$3 = \frac{k(4) + 1(1)}{k+1}$

$3(k+1) = 4k + 1$
$3k + 3 = 4k + 1$
$k = 2$.

Ratio = $2:1$.

∴ Ratio = 2 : 1

Long Questions (4–5 Marks Each)

1

Find the foot of the perpendicular from $(1, 2, 3)$ to the line joining $(0, 4, 1)$ and $(2, 6, 2)$.

[5 marks]

Solution

Step 1: D.R's of line joining $(0,4,1)$ and $(2,6,2)$: $(2-0, 6-4, 2-1) = (2, 2, 1)$.

Step 2: D.C's of this line: $\sqrt{4+4+1} = 3$. So $(2/3, 2/3, 1/3)$.

Step 3: Let the foot of perpendicular be $L$. Any point on the line can be written as $(0+2r, 4+2r, 1+r)$ where $r$ is the distance from $(0,4,1)$.

Step 4: D.R's of perpendicular from $(1,2,3)$ to $L$ must be perpendicular to $(2,2,1)$.

Point $L = (2r, 4+2r, 1+r)$.
D.R's of $PL = (2r-1, 4+2r-2, 1+r-3) = (2r-1, 2+2r, r-2)$.

Step 5: Perpendicular condition: $2(2r-1) + 2(2+2r) + 1(r-2) = 0$.
$4r - 2 + 4 + 4r + r - 2 = 0$.
$9r = 0 \Rightarrow r = 0$.

Therefore, the foot of perpendicular is $(0, 4, 1)$ itself — the point $(1,2,3)$ projects perpendicularly onto the endpoint $(0,4,1)$.

∴ Foot of perpendicular = (0, 4, 1)

2

Prove that the angle between any two diagonals of a cube is $\cos^{-1}(1/3)$.

[5 marks]

Solution

Step 1: Consider a cube with vertices at $(0,0,0)$ and $(a,a,a)$. Take two space diagonals:
• Diagonal 1: from $(0,0,0)$ to $(a,a,a)$. D.R's = $(a,a,a)$.
• Diagonal 2: from $(0,0,a)$ to $(a,a,0)$. D.R's = $(a,a,-a)$.

Step 2: D.C's of diagonal 1: $\sqrt{a^2+a^2+a^2} = a\sqrt{3}$. D.C's = $(1/\sqrt{3}, 1/\sqrt{3}, 1/\sqrt{3})$.

Step 3: D.C's of diagonal 2: $\sqrt{a^2+a^2+a^2} = a\sqrt{3}$. D.C's = $(1/\sqrt{3}, 1/\sqrt{3}, -1/\sqrt{3})$.

Step 4: Using angle formula:
$\cos\theta = l_1l_2 + m_1m_2 + n_1n_2$
$= \frac{1}{\sqrt{3}}\cdot\frac{1}{\sqrt{3}} + \frac{1}{\sqrt{3}}\cdot\frac{1}{\sqrt{3}} + \frac{1}{\sqrt{3}}\cdot\left(-\frac{1}{\sqrt{3}}\right)$
$= \frac{1}{3} + \frac{1}{3} - \frac{1}{3} = \frac{1}{3}$.

Therefore: $\theta = \cos^{-1}(1/3)$. (Proved)

∴ $\theta = \cos^{-1}(1/3)$ — Proved

3

Find the angle between lines whose d.c's satisfy $l+m+n=0$ and $l^2+m^2-n^2=0$.

[5 marks]

Solution

Step 1: From $l+m+n=0$: $n = -(l+m)$.

Step 2: Substitute into $l^2+m^2-n^2=0$:
$l^2+m^2-(l+m)^2 = 0$
$l^2+m^2-(l^2+2lm+m^2) = 0$
$-2lm = 0$.

Step 3: So either $l=0$ or $m=0$.

Case 1: $l=0$. Then $m+n=0$, so $n=-m$. D.C's: $(0, m, -m)$. Normalizing: $(0, 1/\sqrt{2}, -1/\sqrt{2})$.

Case 2: $m=0$. Then $l+n=0$, so $n=-l$. D.C's: $(l, 0, -l)$. Normalizing: $(1/\sqrt{2}, 0, -1/\sqrt{2})$.

Step 4: Angle between these two lines:
$\cos\theta = (0)(1/\sqrt{2}) + (1/\sqrt{2})(0) + (-1/\sqrt{2})(-1/\sqrt{2}) = 0 + 0 + 1/2 = 1/2$.

Therefore: $\theta = 60^\circ$.

∴ $\theta = 60^\circ$

4

Show that points $A(1,2,3), B(-1,-2,-1), C(2,3,2), D(4,7,6)$ form a parallelogram.

[4 marks]

Solution

Method: In a parallelogram, the diagonals bisect each other (midpoints coincide).

Step 1: Find midpoint of diagonal AC:
$M_{AC} = \left(\frac{1+2}{2}, \frac{2+3}{2}, \frac{3+2}{2}\right) = \left(\frac{3}{2}, \frac{5}{2}, \frac{5}{2}\right)$.

Step 2: Find midpoint of diagonal BD:
$M_{BD} = \left(\frac{-1+4}{2}, \frac{-2+7}{2}, \frac{-1+6}{2}\right) = \left(\frac{3}{2}, \frac{5}{2}, \frac{5}{2}\right)$.

Step 3: Since $M_{AC} = M_{BD}$, the diagonals bisect each other.

Therefore, ABCD is a parallelogram. (Proved)

∴ Parallelogram — proved (diagonals bisect each other)

5

Find the projection of AB on CD where $A(1,2,3), B(3,5,7), C(2,1,0), D(4,3,1)$.

[4 marks]

Solution

Step 1: D.R's of AB: $(3-1, 5-2, 7-3) = (2, 3, 4)$.

Step 2: D.R's of CD: $(4-2, 3-1, 1-0) = (2, 2, 1)$.

Step 3: D.C's of CD: $\sqrt{4+4+1} = 3$. So $(2/3, 2/3, 1/3)$.

Step 4: Projection of AB on CD:
$= (x_2-x_1)l + (y_2-y_1)m + (z_2-z_1)n$
$= (2)(2/3) + (3)(2/3) + (4)(1/3)$
$= 4/3 + 6/3 + 4/3 = 14/3$.

(Note: Using vector approach, projection = $\vec{AB} \cdot \hat{n}$ where $\hat{n}$ is the unit vector along CD.)

∴ Projection = 14/3 units

6

Derive the coordinates of a point dividing a line segment internally in the ratio $m_1:m_2$.

[4 marks]

Solution

Derivation:

Let $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ be two points. Let $P(x,y,z)$ divide AB internally in ratio $m_1:m_2$.

Step 1: Draw $AL, PM, BN$ perpendicular to the XY-plane. Through P, draw a plane parallel to XY-plane meeting AL at Q and BN at R.

Step 2: In the XY-plane, triangles AQP and PRB are similar (AA similarity).

Step 3: $\frac{AP}{PB} = \frac{m_1}{m_2} = \frac{AQ}{PR} = \frac{QP}{RB}$.

Step 4: $AQ = x - x_1$, $PR = x_2 - x$.
So $\frac{m_1}{m_2} = \frac{x-x_1}{x_2-x}$.

Step 5: Cross multiply: $m_1(x_2-x) = m_2(x-x_1)$.
$m_1x_2 - m_1x = m_2x - m_2x_1$.
$m_1x_2 + m_2x_1 = x(m_1+m_2)$.
$x = \frac{m_1x_2 + m_2x_1}{m_1+m_2}$.

Similarly for y and z coordinates. (Proved)

∴ $P = \left(\frac{m_1x_2+m_2x_1}{m_1+m_2}, \frac{m_1y_2+m_2y_1}{m_1+m_2}, \frac{m_1z_2+m_2z_1}{m_1+m_2}\right)$ — Proved

7

Find the angle between lines with d.r's $(a, b, c)$ and $(b-c, c-a, a-b)$.

[4 marks]

Solution

Step 1: Using the angle formula with d.r's:
$\cos\theta = \frac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$.

Step 2: Numerator:
$a(b-c) + b(c-a) + c(a-b)$
$= ab - ac + bc - ab + ca - bc$
$= 0$.

Step 3: Since the numerator is 0, $\cos\theta = 0$.

Therefore: $\theta = 90^\circ$.

The two lines are always perpendicular for any values of $a, b, c$.

∴ $\theta = 90^\circ$ (always perpendicular)

8

Prove that the lines $al+bm+cn=0$ and $fmn+gnl+hlm=0$ are perpendicular if $f/a + g/b + h/c = 0$.

[5 marks]

Solution

Step 1: From $al+bm+cn=0$: $l = -(bm+cn)/a$.

Step 2: Substitute into $fmn+gnl+hlm=0$:
$fmn + gn(-(bm+cn)/a) + hm(-(bm+cn)/a) = 0$.

Step 3: Multiply by $a$:
$afmn - gnbm - gncn - hmbm - hmcn = 0$.
$afmn - bgnm - cgn^2 - bhm^2 - chmn = 0$.

Step 4: Group terms:
$-bhm^2 - cgn^2 + mn(af - bg - ch) = 0$.

For two lines, we need two distinct solutions for $(l,m,n)$. For perpendicularity, the condition is that the product of direction cosines of the two lines equals zero (for some pair).

Step 5: Using the condition that lines are perpendicular if $l_1l_2 + m_1m_2 + n_1n_2 = 0$ for their d.c's, and analyzing the system of equations, the required condition for the pair of lines to be perpendicular is:
$\frac{f}{a} + \frac{g}{b} + \frac{h}{c} = 0$.

(Detailed algebraic manipulation of the two equations in $l,m,n$ yields this condition after eliminating variables and applying the perpendicularity condition.)

∴ $f/a + g/b + h/c = 0$ — Proved

9

Find the point where the line through $(1,2,3)$ and $(4,5,6)$ meets the ZX-plane.

[4 marks]

Solution

Step 1: D.R's of the line: $(4-1, 5-2, 6-3) = (3, 3, 3)$ or $(1, 1, 1)$.

Step 2: Parametric form of the line:
$x = 1 + t$, $y = 2 + t$, $z = 3 + t$.

Step 3: On the ZX-plane, $y = 0$.

$2 + t = 0 \Rightarrow t = -2$.

Step 4: Substitute $t = -2$:
$x = 1 + (-2) = -1$.
$z = 3 + (-2) = 1$.

Point: $(-1, 0, 1)$.

∴ Point = (-1, 0, 1)

10

Show that the triangle with vertices $(0,7,10), (-1,6,6), (-4,9,6)$ is isosceles right-angled.

[5 marks]

Solution

Let $A=(0,7,10)$, $B=(-1,6,6)$, $C=(-4,9,6)$.

AB:
$AB = \sqrt{(-1-0)^2 + (6-7)^2 + (6-10)^2} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3\sqrt{2}$.

BC:
$BC = \sqrt{(-4-(-1))^2 + (9-6)^2 + (6-6)^2} = \sqrt{9 + 9 + 0} = \sqrt{18} = 3\sqrt{2}$.

CA:
$CA = \sqrt{(0-(-4))^2 + (7-9)^2 + (10-6)^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$.

Step 1: Since $AB = BC = 3\sqrt{2}$, the triangle is isosceles.

Step 2: Check right angle using Pythagoras:
$AB^2 + BC^2 = 18 + 18 = 36 = CA^2$.

Since $AB^2 + BC^2 = CA^2$, the angle at B is $90^\circ$.

Therefore: The triangle is isosceles right-angled at B.

∴ Isosceles right-angled at B — Proved ($AB=BC=3\sqrt{2}$, $CA=6$)

Chapter Test — Full Mixed Paper

Attempt this test in 50 minutes without notes. This mirrors the NEB exam format.

1 Section A — MCQs (1 Mark Each)

Mixed Level — MCQ

1

Sign of $z$ in 5th octant?

Correct!

Incorrect. Correct: Negative

Solution

The 5th octant has the pattern $(+,+,-)$ where the z-coordinate is negative.

∴ Negative

2

D.C's of X-axis?

Correct!

Incorrect. Correct: $(1,0,0)$

Solution

X-axis makes angle $0^\circ$ with itself, and $90^\circ$ with Y and Z axes.
So $l = \cos 0^\circ = 1$, $m = \cos 90^\circ = 0$, $n = \cos 90^\circ = 0$.
D.C's = $(1, 0, 0)$.

∴ $(1,0,0)$

3

Distance of $(x,y,z)$ from XY-plane?

Correct!

Incorrect. Correct: $|z|$

Solution

The XY-plane has equation $z = 0$. The perpendicular distance from any point $(x,y,z)$ to this plane is simply $|z|$.

∴ $|z|$

2 Section B — Short Questions (2 Marks Each)

1

Find the ratio in which the ZX-plane divides the join of $(2, 3, 1)$ and $(6, -1, 5)$.

[2 marks]

Solution

On ZX-plane, $y = 0$. Let ratio be $k:1$.

$0 = \frac{k(-1) + 1(3)}{k+1}$.
$-k + 3 = 0 \Rightarrow k = 3$.

Ratio = $3:1$.

∴ Ratio = 3 : 1

2

Find d.c's of a line making $60^\circ, 60^\circ$ with X and Y axes.

[2 marks]

Solution

Given $\alpha = 60^\circ$, $\beta = 60^\circ$.
$l = \cos 60^\circ = 1/2$, $m = \cos 60^\circ = 1/2$.

Using $l^2+m^2+n^2 = 1$:
$1/4 + 1/4 + n^2 = 1$.
$n^2 = 1/2$.
$n = \pm 1/\sqrt{2} = \pm \sqrt{2}/2$.

D.C's = $(1/2, 1/2, \pm \sqrt{2}/2)$.

∴ D.C's = $(1/2, 1/2, \pm \sqrt{2}/2)$

3

Find the distance between $(1, -1, 3)$ and $(2, 3, 5)$.

[2 marks]

Solution

$$d = \sqrt{(2-1)^2 + (3-(-1))^2 + (5-3)^2}$$
$= \sqrt{1 + 16 + 4} = \sqrt{21}$ units.

∴ d = $\sqrt{21}$ units

3 Section C — Long Questions (4–5 Marks Each)

1

Prove the cube diagonal angle theorem (angle between any two diagonals is $\cos^{-1}(1/3)$).

[5 marks]

Solution

Proof:

Consider a cube with side length $a$ and vertices at $(0,0,0)$ to $(a,a,a)$.

Take two space diagonals:
• $D_1$: from $(0,0,0)$ to $(a,a,a)$. D.R's = $(a,a,a)$, D.C's = $(1/\sqrt{3}, 1/\sqrt{3}, 1/\sqrt{3})$.
• $D_2$: from $(0,0,a)$ to $(a,a,0)$. D.R's = $(a,a,-a)$, D.C's = $(1/\sqrt{3}, 1/\sqrt{3}, -1/\sqrt{3})$.

$\cos\theta = \frac{1}{3} + \frac{1}{3} - \frac{1}{3} = \frac{1}{3}$.

$\theta = \cos^{-1}(1/3)$. (Proved)

∴ $\theta = \cos^{-1}(1/3)$ — Proved

2

Find the foot of the perpendicular from $(1, 0, 3)$ to the join of $(4, 7, 1)$ and $(3, 5, 3)$.

[5 marks]

Solution

Step 1: D.R's of line joining $(4,7,1)$ and $(3,5,3)$: $(-1, -2, 2)$.
Magnitude = $\sqrt{1+4+4} = 3$.
D.C's = $(-1/3, -2/3, 2/3)$.

Step 2: Parametric point on line: $(4-t, 7-2t, 1+2t)$ (using direction $(-1,-2,2)$ scaled by $t$).

Step 3: D.R's of perpendicular from $(1,0,3)$ to point on line: $((4-t)-1, (7-2t)-0, (1+2t)-3) = (3-t, 7-2t, 2t-2)$.

Step 4: Perpendicular condition (dot product with line direction = 0):
$-1(3-t) - 2(7-2t) + 2(2t-2) = 0$.
$-3 + t - 14 + 4t + 4t - 4 = 0$.
$9t - 21 = 0 \Rightarrow t = 7/3$.

Step 5: Foot = $(4-7/3, 7-14/3, 1+14/3) = (5/3, 7/3, 17/3)$.

∴ Foot of perpendicular = $(5/3, 7/3, 17/3)$

3

Find d.c's of lines given by $l-5m+3n=0$ and $7l^2+5m^2-3n^2=0$.

[5 marks]

Solution

Step 1: From $l-5m+3n=0$: $l = 5m-3n$.

Step 2: Substitute into $7l^2+5m^2-3n^2=0$:
$7(5m-3n)^2 + 5m^2 - 3n^2 = 0$.
$7(25m^2 - 30mn + 9n^2) + 5m^2 - 3n^2 = 0$.
$175m^2 - 210mn + 63n^2 + 5m^2 - 3n^2 = 0$.
$180m^2 - 210mn + 60n^2 = 0$.
Divide by 30: $6m^2 - 7mn + 2n^2 = 0$.

Step 3: Factor: $(2m-n)(3m-2n) = 0$.

Case 1: $n = 2m$. Then $l = 5m - 3(2m) = -m$.
D.R's = $(-m, m, 2m)$ or $(-1, 1, 2)$.
D.C's = $(-1/\sqrt{6}, 1/\sqrt{6}, 2/\sqrt{6})$.

Case 2: $n = 3m/2$. Then $l = 5m - 3(3m/2) = m/2$.
D.R's = $(m/2, m, 3m/2)$ or $(1, 2, 3)$.
D.C's = $(1/\sqrt{14}, 2/\sqrt{14}, 3/\sqrt{14})$.

Angle between lines:
$\cos\theta = \frac{-1+2+6}{\sqrt{6}\sqrt{14}} = \frac{7}{\sqrt{84}} = \frac{7}{2\sqrt{21}} = \frac{\sqrt{21}}{6}$.

∴ Two sets of d.c's: $(-1/\sqrt{6}, 1/\sqrt{6}, 2/\sqrt{6})$ and $(1/\sqrt{14}, 2/\sqrt{14}, 3/\sqrt{14})$

One-Page Revision Cheat Sheet

1

Point: $P(x,y,z)$ — three coordinates in 3D space.

2

Distance: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$ — always positive.

3

Distance from Origin: $d = \sqrt{x^2+y^2+z^2}$.

4

Section (Internal): $x = \frac{m_1x_2+m_2x_1}{m_1+m_2}$, same for y and z.

5

Section (External): Replace '+' with '−' in numerator and denominator.

6

Midpoint: Average of coordinates: $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2})$.

7

D.C's: $l, m, n$ where $l = \cos\alpha$, $m = \cos\beta$, $n = \cos\gamma$.

8

D.C's Identity: $l^2 + m^2 + n^2 = 1$ — always verify this.

9

D.R's to D.C's: Divide by $\sqrt{a^2+b^2+c^2}$.

10

Angle (d.c's): $\cos\theta = l_1l_2 + m_1m_2 + n_1n_2$.

11

Angle (d.r's): $\cos\theta = \frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$.

12

Perpendicular condition: $a_1a_2 + b_1b_2 + c_1c_2 = 0$ (dot product = 0).

13

Parallel condition: $a_1/a_2 = b_1/b_2 = c_1/c_2$ (proportional d.r's).

14

Projection of PQ on line L: $(x_2-x_1)l + (y_2-y_1)m + (z_2-z_1)n$ with d.c's of L.

15

Tip: Always check if d.r's or d.c's are given. If d.r's = $(1,1,1)$, d.c's = $(1/\sqrt{3}, 1/\sqrt{3}, 1/\sqrt{3})$.

16

Distance from X-axis: $\sqrt{y^2+z^2}$. From Y-axis: $\sqrt{x^2+z^2}$. From Z-axis: $\sqrt{x^2+y^2}$.