Points and coordinates in space

This chapter establishes the foundational principles for working with points and coordinates in three-dimensional space. It systematically introduces essential concepts such as distance calculation, the section formula, and midpoint determination. Mastery of these topics is crucial for solving a wide array of 3D geometry problems frequently encountered in examinations.

---

Chapter Contents

| Topic |

|---|-------| | 1 | Distance in 3D | | 2 | Section formula in 3D | | 3 | Midpoint problems |

---

We begin with Distance in 3D.

Part 1: Distance in 3D

Distance in 3D

Overview

Distance in 3D geometry is built from the same Pythagorean principle as in 2D, but with three coordinate differences instead of two. In exam problems, 3D distance appears not only in direct coordinate computation, but also in midpoint arguments, collinearity checks, sphere-style reasoning, and geometric interpretation of equations. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

Compute the distance between two points in 3D.

Find the distance of a point from the origin.

Use the distance formula to test equality of lengths.

Solve standard geometric questions using coordinate differences.

Interpret 3D distance problems confidently and accurately.

---

Core Formula

📐 Distance Between Two Points

If

$\qquad A(x_1,y_1,z_1)$
and
$\qquad B(x_2,y_2,z_2)$ ,

then the distance between them is

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

This is the fundamental formula for the topic. ---

Distance from the Origin

📐 Distance from the Origin

If a point is

$\qquad P(x,y,z)$ ,

then its distance from the origin is

$\qquad OP=\sqrt{x^2+y^2+z^2}$

This is just the special case of the distance formula with the origin

(0,0,0)

. ---

Squared Distance

💡 Use Squared Distance When Possible

In many problems, it is easier to compare squared distances:

$\qquad AB^2=(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2$

This avoids unnecessary square roots.

This is especially useful in equality-of-length questions. ---

Geometric Meaning

❗ Distance as Length of a Vector

If

$\qquad \overrightarrow{AB}=\langle x_2-x_1,\ y_2-y_1,\ z_2-z_1\rangle$

then

$\qquad AB=\left|\overrightarrow{AB}\right|$

So the distance formula is exactly the magnitude formula for the displacement vector from $A$ to $B$ .

---

Minimal Worked Examples

Example 1 Find the distance between

(1,2,3)

and

(4,6,3)

. We compute

\qquad AB=\sqrt{(4-1)^2+(6-2)^2+(3-3)^2}

\qquad =\sqrt{3^2+4^2+0}=\sqrt{25}=5

So the distance is

\boxed{5}

. --- Example 2 Find the distance of the point

(2,-1,2)

from the origin.

\qquad OP=\sqrt{2^2+(-1)^2+2^2}=\sqrt{4+1+4}=3

So the distance is

\boxed{3}

. ---

Equal Distances and Locus Thinking

📐 Equidistant Conditions

If a point $P(x,y,z)$ is equidistant from two fixed points $A$ and $B$ , then

$\qquad PA=PB$

or equivalently

$\qquad PA^2=PB^2$

Expanding the squared-distance form is usually the fastest method.

---

Common Uses in Problems

📐 Typical Uses

check whether three points form an isosceles triangle

test whether a point lies on a sphere centered at the origin

compare side lengths in a tetrahedron or triangle

determine whether two given lengths are equal

simplify geometric reasoning using coordinates

---

Common Mistakes

⚠️ Avoid These Errors

❌ Forgetting one of the three coordinate differences

❌ Dropping squares on negative differences

❌ Comparing raw expressions instead of squared distances

❌ Mixing distance formula with midpoint formula

---

CMI Strategy

💡 How to Attack Distance Problems

Write the two points clearly.

Subtract coordinates in the same order.

Use squared distances when the question is about equality or comparison.

Keep square roots to the final step only if a numerical distance is actually needed.

---

Practice Questions

:::question type="MCQ" question="The distance between

(1,2,3)

and

(4,6,3)

is" options=["

5

","

7

","

\sqrt{13}

","

25

"] answer="A" hint="Use the 3D distance formula." solution="We compute $\qquad \sqrt{(4-1)^2+(6-2)^2+(3-3)^2} =\sqrt{9+16} =\sqrt{25} =5$. Hence the correct option is

\boxed{A}

." ::: :::question type="NAT" question="Find the distance of the point

(2,-1,2)

from the origin." answer="3" hint="Use

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

." solution="The distance from the origin is

\qquad \sqrt{2^2+(-1)^2+2^2}=\sqrt{4+1+4}=\sqrt{9}=3

. Hence the answer is

\boxed{3}

." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["The distance formula in 3D uses three squared coordinate differences","The distance from

(x,y,z)

to the origin is

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

","If

PA=PB

, then necessarily

A=B

","Comparing squared distances is often useful"] answer="A,B,D" hint="Think about both formula and method." solution="1. True.

True.

False. A point can be equidistant from two distinct points.

True. Squared distances avoid square roots and are often cleaner.

Hence the correct answer is

\boxed{A,B,D}

." ::: :::question type="SUB" question="Show that the point

(1,2,2)

is equidistant from the origin and the point

(2,4,4)

." answer="True" hint="Compare the two distances." solution="Distance from

(1,2,2)

to the origin:

\qquad \sqrt{1^2+2^2+2^2}=\sqrt{1+4+4}=\sqrt{9}=3

. Distance from

(1,2,2)

(2,4,4)

: $\qquad \sqrt{(2-1)^2+(4-2)^2+(4-2)^2} =\sqrt{1+4+4} =\sqrt{9} =3$. Since both distances are equal to

3

, the point

(1,2,2)

is equidistant from the origin and

(2,4,4)

. Hence the answer is

\boxed{\text{True}}

." ::: ---

Summary

❗ Key Takeaways for CMI

The 3D distance formula is the basic metric tool in space geometry.

Distance from the origin is the special case $\sqrt{x^2+y^2+z^2}$ .
- Squared-distance comparison is often smarter than direct distance comparison.
- Many geometric relations can be translated into simple coordinate equalities using distance.
- Accuracy in coordinate subtraction is essential.

---

💡 Next Up

Proceeding to Section formula in 3D.

---

Part 2: Section formula in 3D

Section Formula in 3D

Overview

The section formula in 3D is the coordinate-geometry tool used to locate a point dividing a segment in a given ratio. In exam problems, the geometry is simple but sign control is crucial, especially in external division. The topic is best understood as weighted averaging of coordinates. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

Find the midpoint of a segment in 3D.

Use the internal section formula correctly.

Use the external section formula with the correct sign convention.

Recover an unknown endpoint or ratio from a given division point.

Interpret section formula geometrically as coordinate-wise weighted averaging.

---

Basic Setting

📖 Points in 3D

A point in 3D space is written as

$\qquad A(x_1,y_1,z_1)$

A segment joining two points

$\qquad A(x_1,y_1,z_1),\quad B(x_2,y_2,z_2)$

can be divided by another point $P$ internally or externally.

---

Midpoint Formula

📐 Midpoint in 3D

The midpoint of the segment joining

$\qquad A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$

is

$\qquad \left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2},\dfrac{z_1+z_2}{2}\right)$

This is just averaging the coordinates. ---

Internal Section Formula

📐 Internal Division

If the point $P$ divides the segment joining
$\qquad A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$
internally in the ratio

$\qquad AP:PB = m:n$

then

$\qquad P\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n},\dfrac{mz_2+nz_1}{m+n}\right)$

The point is closer to the endpoint with the larger attached coefficient in the numerator. ---

External Section Formula

📐 External Division

If $P$ divides the line joining
$\qquad A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$
externally in the ratio

$\qquad AP:PB = m:n,\quad m\ne n$

then

$\qquad P\left(\dfrac{mx_2-nx_1}{m-n},\dfrac{my_2-ny_1}{m-n},\dfrac{mz_2-nz_1}{m-n}\right)$

This is the same weighted idea, but with subtraction instead of addition. ::: ---

Why the Formula Is Coordinate-Wise

❗ Weighted Averaging

The section formula works separately in each coordinate:

first coordinate depends only on $x_1,x_2$

second coordinate depends only on $y_1,y_2$

third coordinate depends only on $z_1,z_2$

So the 3D formula is just the 2D section formula extended coordinate by coordinate.

---

Standard Uses

💡 Most Common Uses

finding a point dividing a segment in a given ratio

finding an unknown endpoint

checking collinearity in 3D

finding centroid-type points by repeated midpoint or weighted-average ideas

recovering the ratio from coordinates

---

Minimal Worked Examples

Example 1 Find the midpoint of

\qquad A(1,2,3)

and

B(5,6,7)

. Midpoint:

\qquad \left(\dfrac{1+5}{2},\dfrac{2+6}{2},\dfrac{3+7}{2}\right)=(3,4,5)

--- Example 2 Find the point dividing the segment joining

\qquad A(0,0,0)

and

B(6,9,12)

internally in the ratio

\qquad 1:2

. Using the internal section formula,

\qquad P=\left(\dfrac{1\cdot 6+2\cdot 0}{3},\dfrac{1\cdot 9+2\cdot 0}{3},\dfrac{1\cdot 12+2\cdot 0}{3}\right)=(2,3,4)

---

Standard Patterns

📐 High-Value Patterns

midpoint:

coordinate average

internal division:

weighted average with denominator

m+n

external division:

weighted difference with denominator

m-n

if ratio is $1:1$ , internal division gives midpoint

a valid internal division point lies between the two endpoints coordinate-wise only when the segment itself is aligned that way

---

Common Mistakes

⚠️ Avoid These Errors

❌ Reversing the weights in the section formula without checking the ratio convention

✅ If

AP:PB=m:n

, use the point formula exactly as given

❌ Using $m+n$ for external division

✅ External division uses

m-n

❌ Forgetting that external division requires $m\ne n$

❌ Treating the 3D case as different in principle

✅ It is the same formula applied coordinate-wise

---

CMI Strategy

💡 How to Attack These Questions

First decide whether the division is internal or external.

Write the ratio clearly in the form $AP:PB=m:n$ .

Apply the formula coordinate by coordinate.

Simplify only at the end.

In inverse problems, set the coordinates equal and solve systematically.

---

Practice Questions

:::question type="MCQ" question="The midpoint of the points

(1,2,3)

and

(5,6,7)

is" options=["

(3,4,5)

","

(2,3,4)

","

(6,8,10)

","

(4,4,4)

"] answer="A" hint="Average each coordinate." solution="The midpoint is

\qquad \left(\dfrac{1+5}{2},\dfrac{2+6}{2},\dfrac{3+7}{2}\right)=(3,4,5)

Hence the correct option is

\boxed{A}

." ::: :::question type="NAT" question="If the point

P

divides the segment joining

(0,0,0)

and

(6,9,12)

internally in the ratio

1:2

, find the

y

-coordinate of

P

." answer="3" hint="Use the internal section formula." solution="Using the internal section formula,

\qquad y=\dfrac{1\cdot 9+2\cdot 0}{1+2}=\dfrac{9}{3}=3

Hence the answer is

\boxed{3}

." ::: :::question type="MSQ" question="Which of the following are true?" options=["The midpoint is obtained by averaging coordinates","Internal division uses denominator

m+n

","External division uses denominator

m+n

","The section formula in 3D is applied coordinate-wise"] answer="A,B,D" hint="Check internal and external division carefully." solution="1. True.

True.

False. External division uses

m-n

True.

Hence the correct answer is

\boxed{A,B,D}

." ::: :::question type="SUB" question="Find the point dividing externally in the ratio

2:1

the segment joining

A(1,-2,3)

and

B(4,1,9)

." answer="

(7,4,15)

" hint="Use the external section formula." solution="Using the external section formula with

m=2

and

n=1

\qquad P=\left(\dfrac{2\cdot 4-1\cdot 1}{2-1},\dfrac{2\cdot 1-1\cdot(-2)}{2-1},\dfrac{2\cdot 9-1\cdot 3}{2-1}\right)

\qquad P=(7,4,15)

Hence the answer is

\boxed{(7,4,15)}

." ::: ---

Summary

❗ Key Takeaways for CMI

The midpoint is the average of coordinates.

Internal section uses denominator $m+n$ .

External section uses denominator $m-n$ .

The 3D formula is just the coordinate-wise version of the 2D formula.

Most mistakes come from ratio order and sign errors.

---

💡 Next Up

Proceeding to Midpoint problems.

---

Part 3: Midpoint problems

Midpoint Problems

Overview

Midpoint problems in 3D geometry are simple in formula but powerful in application. They appear in coordinate geometry, section arguments, line equations, symmetry, and distance-based constructions. In CMI-style problems, midpoint ideas are often combined with distance, collinearity, or algebraic constraints on coordinates. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

Find the midpoint of a segment in 3D.

Use midpoint coordinates to solve inverse problems.

Apply midpoint ideas in symmetry and line problems.

Translate midpoint conditions into equations.

Combine midpoint reasoning with distance and coordinate methods.

---

Core Formula

📐 Midpoint Formula in 3D

If

$\qquad A(x_1,y_1,z_1)$
and
$\qquad B(x_2,y_2,z_2)$ ,

then the midpoint of the segment $AB$ is

$\qquad M\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2},\ \dfrac{z_1+z_2}{2}\right)$

---

Why the Formula Makes Sense

📖 Equal Division

The midpoint divides a segment into two equal parts. So each coordinate of the midpoint is the average of the corresponding coordinates of the endpoints.

This is true in 1D, 2D, and 3D. ---

Inverse Midpoint Problems

📐 Recovering an Endpoint

If the midpoint $M$ of $AB$ is known along with one endpoint $A$ , then the other endpoint $B$ can be found by solving:

$\qquad x_B = 2x_M - x_A$

$\qquad y_B = 2y_M - y_A$

$\qquad z_B = 2z_M - z_A$

This comes directly from the midpoint formula. ---

Midpoint and Collinearity

❗ Common Use

If a point is claimed to be the midpoint of two given points, then:

its coordinates must equal the coordinate-wise averages, and

the three points must automatically be collinear because the midpoint lies on the segment joining the endpoints.

---

Minimal Worked Examples

Example 1 Find the midpoint of

(1,2,3)

and

(5,0,-1)

\qquad M=\left(\dfrac{1+5}{2},\ \dfrac{2+0}{2},\ \dfrac{3+(-1)}{2}\right)

\qquad = (3,1,1)

So the midpoint is

\boxed{(3,1,1)}

. --- Example 2 If the midpoint of

A(2,-1,4)

and

B(x,3,0)

(5,1,2)

, find

x

. Using the

x

-coordinate midpoint condition,

\qquad \dfrac{2+x}{2}=5

\qquad 2+x=10

\qquad x=8

Hence the answer is

\boxed{8}

. ---

Midpoint as an Average Vector

📐 Vector Form

If $\vec a$ and $\vec b$ are the position vectors of points $A$ and $B$ , then the midpoint has position vector

$\qquad \dfrac{\vec a+\vec b}{2}$

This is useful in vector-based solutions. ---

Common Uses in Problems

📐 Typical Midpoint Applications

finding the midpoint of a segment

determining an unknown coordinate

proving two segments have the same midpoint

checking whether diagonals bisect each other

using midpoint conditions to show a quadrilateral is a parallelogram

---

Common Mistakes

⚠️ Avoid These Errors

❌ Averaging incorrectly by forgetting one coordinate

❌ Mixing midpoint formula with distance formula

❌ Forgetting to divide by $2$

❌ Solving inverse midpoint problems without writing coordinate equations carefully

---

CMI Strategy

💡 How to Attack Midpoint Problems

Write the midpoint formula coordinate by coordinate.

For unknown coordinates, form one equation per coordinate.

Use midpoint equality to compare segments or diagonals.

In vector form, think of midpoint as an average.

Keep arithmetic clean; midpoint questions reward precision.

---

Practice Questions

:::question type="MCQ" question="The midpoint of

(1,2,3)

and

(5,0,-1)

is" options=["

(3,1,1)

","

(6,2,2)

","

(2,1,1)

","

(3,2,1)

"] answer="A" hint="Average each coordinate." solution="The midpoint is $\qquad \left(\dfrac{1+5}{2},\ \dfrac{2+0}{2},\ \dfrac{3+(-1)}{2}\right) =(3,1,1)$. Hence the correct option is

\boxed{A}

." ::: :::question type="NAT" question="If the midpoint of

(2,-1,4)

and

(x,3,0)

(5,1,2)

, find

x

." answer="8" hint="Use the midpoint formula on the first coordinate." solution="Using the midpoint condition on the first coordinate,

\qquad \dfrac{2+x}{2}=5

. So

\qquad 2+x=10

and therefore

\qquad x=8

. Hence the answer is

\boxed{8}

." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["The midpoint of two points is obtained by averaging corresponding coordinates","If two diagonals of a quadrilateral have the same midpoint, the quadrilateral is a parallelogram","The midpoint formula in 3D involves all three coordinates","The midpoint of

(x_1,y_1,z_1)

and

(x_2,y_2,z_2)

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

"] answer="A,B,C" hint="Check the exact formula carefully." solution="1. True.

True. Equal midpoints of diagonals imply that the diagonals bisect each other, which characterizes a parallelogram.

True.

False. Each sum must be divided by

2

Hence the correct answer is

\boxed{A,B,C}

." ::: :::question type="SUB" question="If the midpoint of the segment joining

(a,2,-1)

and

(3,b,5)

(4,1,2)

, find

a

and

b

." answer="

a=5,\ b=0

" hint="Equate the midpoint coordinates." solution="Using the midpoint formula,

\qquad \left(\dfrac{a+3}{2},\ \dfrac{2+b}{2},\ \dfrac{-1+5}{2}\right)=(4,1,2)

Now compare coordinates: From the first coordinate,

\qquad \dfrac{a+3}{2}=4

\qquad a+3=8

and

\qquad a=5

From the second coordinate,

\qquad \dfrac{2+b}{2}=1

\qquad 2+b=2

and

\qquad b=0

The third coordinate is already consistent:

\qquad \dfrac{-1+5}{2}=2

Hence the required values are

\boxed{a=5,\ b=0}

." ::: ---

Summary

❗ Key Takeaways for CMI

The midpoint of two points is obtained by averaging corresponding coordinates.

Midpoint problems often hide simple linear equations.

Inverse midpoint questions are solved coordinate by coordinate.

Midpoint equality is a strong tool in segment and diagonal problems.

Clean arithmetic is essential.

Chapter Summary

❗ Points and coordinates in space — Key Points

Distance Formula: The distance between two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ in 3D space is given by $PQ = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$ .
Internal Section Formula: A point $R$ dividing the line segment joining $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ internally in the ratio $m:n$ has coordinates:

R = \left(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}, \frac{mz_2+nz_1}{m+n}\right)

External Section Formula: A point $R$ dividing the line segment joining $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ externally in the ratio $m:n$ has coordinates:

R = \left(\frac{mx_2-nx_1}{m-n}, \frac{my_2-ny_1}{m-n}, \frac{mz_2-nz_1}{m-n}\right)

Midpoint Formula: The midpoint of the line segment joining $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ is a special case of the section formula (ratio $1:1$ ):

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

Centroid of a Triangle: The centroid of a triangle with vertices $A(x_1, y_1, z_1)$ , $B(x_2, y_2, z_2)$ , $C(x_3, y_3, z_3)$ is:

G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)

Collinearity: Three points $A, B, C$ are collinear if the sum of the distances of two segments equals the distance of the third segment (e.g., $AB+BC=AC$ ), or if one point divides the segment formed by the other two in some ratio.

Geometric Applications: These formulas are fundamental for solving problems involving properties of geometric figures in 3D, such as finding vertices of parallelograms, centers of spheres, or loci of points under specific conditions.

Chapter Review Questions

:::question type="MCQ" question="Find the coordinates of a point on the y-axis which is equidistant from the points $A(3, 1, 2)$ and $B(5, -1, 4)$ ." options=[" $(0, -7, 0)$ "," $(0, 7, 0)$ "," $(-7, 0, 0)$ "," $(0, 0, -7)$ "] answer=" $(0, -7, 0)$ " hint="A point on the y-axis has coordinates $(0, y, 0)$ . Use the distance formula and equate the squares of the distances from this point to A and B." solution="Let the point on the y-axis be $P(0, y, 0)$ .
The distance squared from $P$ to $A$ is $PA^2 = (3-0)^2 + (1-y)^2 + (2-0)^2 = 9 + (1-2y+y^2) + 4 = y^2 - 2y + 14$ .
The distance squared from $P$ to $B$ is $PB^2 = (5-0)^2 + (-1-y)^2 + (4-0)^2 = 25 + (1+2y+y^2) + 16 = y^2 + 2y + 42$ .
Since $PA^2 = PB^2$ :
$y^2 - 2y + 14 = y^2 + 2y + 42$
$-2y + 14 = 2y + 42$
$4y = -28$
$y = -7$ .
Thus, the point is $(0, -7, 0)$ ."
:::

:::question type="NAT" question="If the point $R(1, 0, -1)$ divides the line segment joining $P(x, 2, -3)$ and $Q(2, -1, z)$ in the ratio $2:1$ internally, find the value of $x+z$ ." answer="-1" hint="Apply the internal section formula for each coordinate of R. Solve for x and z, then find their sum." solution="Using the internal section formula for $R(1, 0, -1)$ dividing $P(x, 2, -3)$ and $Q(2, -1, z)$ in ratio $2:1$ :
For the x-coordinate:
$1 = \frac{2(2) + 1(x)}{2+1} \implies 1 = \frac{4+x}{3} \implies 3 = 4+x \implies x = -1$ .
For the y-coordinate:
$0 = \frac{2(-1) + 1(2)}{2+1} \implies 0 = \frac{-2+2}{3} \implies 0 = 0$ . (This confirms consistency).
For the z-coordinate:
$-1 = \frac{2(z) + 1(-3)}{2+1} \implies -1 = \frac{2z-3}{3} \implies -3 = 2z-3 \implies 2z = 0 \implies z = 0$ .
Therefore, $x = -1$ and $z = 0$ .
The value of $x+z = -1 + 0 = -1$ ."
:::

:::question type="MCQ" question="The vertices of a parallelogram are $A(1, 2, 3)$ , $B(-1, -2, -1)$ , $C(2, 1, -1)$ . Find the coordinates of the fourth vertex $D$ ." options=[" $(4, 5, 3)$ "," $(0, 1, -1)$ "," $(1, 0, 1)$ "," $(3, 2, 1)$ "] answer=" $(4, 5, 3)$ " hint="In a parallelogram, the diagonals bisect each other. This means the midpoint of $AC$ is the same as the midpoint of $BD$ ." solution="Let the fourth vertex be $D(x, y, z)$ .
In a parallelogram, the midpoint of the diagonal $AC$ must be the same as the midpoint of the diagonal $BD$ .
Midpoint of $AC = \left(\frac{1+2}{2}, \frac{2+1}{2}, \frac{3+(-1)}{2}\right) = \left(\frac{3}{2}, \frac{3}{2}, 1\right)$ .
Midpoint of $BD = \left(\frac{-1+x}{2}, \frac{-2+y}{2}, \frac{-1+z}{2}\right)$ .
Equating the coordinates:
$\frac{-1+x}{2} = \frac{3}{2} \implies -1+x = 3 \implies x = 4$ .
$\frac{-2+y}{2} = \frac{3}{2} \implies -2+y = 3 \implies y = 5$ .
$\frac{-1+z}{2} = 1 \implies -1+z = 2 \implies z = 3$ .
Thus, the coordinates of the fourth vertex $D$ are $(4, 5, 3)$ ."
:::

What's Next?

💡 Continue Your CMI Journey

This chapter has established the foundational understanding of points and their relative positions in three-dimensional space using coordinate geometry. The concepts of distance, section, and midpoint are critical building blocks for advanced topics.

In Vectors, you will discover an alternative and often more intuitive framework to represent points and perform geometric operations like finding distances, section points, and checking collinearity. Vectors provide powerful algebraic tools to solve complex 3D problems.

Moving forward into 3D Geometry, these foundational concepts will be extensively applied to define and analyze higher-level geometric entities such as lines, planes, and spheres. A solid grasp of points and coordinates is essential for understanding spatial relationships and deriving equations of these objects in 3D space.

Points and coordinates in space

Points and coordinates in space

Chapter Contents

| Topic |

Part 1: Distance in 3D

Distance in 3D

Overview

Learning Objectives

Core Formula

Distance from the Origin

Squared Distance

Geometric Meaning

Minimal Worked Examples

Equal Distances and Locus Thinking

Common Uses in Problems

Common Mistakes

CMI Strategy

Practice Questions

Summary

Part 2: Section formula in 3D

Section Formula in 3D

Overview

Learning Objectives

Basic Setting

Midpoint Formula

Internal Section Formula

External Section Formula

Why the Formula Is Coordinate-Wise

Standard Uses

Minimal Worked Examples

Standard Patterns

Common Mistakes

CMI Strategy

Practice Questions

Summary

Part 3: Midpoint problems

Midpoint Problems

Overview

Learning Objectives

Core Formula

Why the Formula Makes Sense

Inverse Midpoint Problems

Midpoint and Collinearity

Minimal Worked Examples

Midpoint as an Average Vector

Common Uses in Problems

Common Mistakes

CMI Strategy

Practice Questions

Summary

Chapter Summary

Chapter Review Questions

What's Next?

🎯 Key Points to Remember

Related Topics in Vectors, Matrices and 3D Geometry

Geometric use of vectors

Dot product

Systems of equations

Line and plane basics

More Resources

Study Notes

Short Notes

Test Series

Mock Tests

Previous Year Papers

Chapter-wise PYQs

Chapter Practice

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