Find the distance between the points (-3, 7, 2) and (2, 4, -1).

Question of 72

(-3, 7, 2)

Jharkhand JacJAC Intermediate Board (1st Year) 2022Subjective· 2mImportance★★★★★

0% · 0/72 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — 3D Coordinate Geometry

3D Coordinate Geometry

You already know 2D coordinate geometry — the $xy$ -plane where every point is described by two numbers $(x, y)$ . Now imagine lifting that plane into the air. That is three-dimensional geometry.

The Intuition: Three Numbers, One Point

In the real world you rarely locate something with just two numbers. To describe where a book sits on a shelf you might say: "third shelf up, fourth book from the left, and it is the one nearest the wall." That is three pieces of information — height, sideways position, and depth.

In 3D coordinate geometry we do exactly this. We keep the familiar $x$ and $y$ axes (which define a flat floor) and add a third axis — the $z$ -axis — pointing straight up. Every point in space now needs three numbers: $(x, y, z)$ .

Note

The three axes are mutually perpendicular. Picture the corner of a room: two floor edges give the $x$ - and $y$ -axes, and the vertical edge where the walls meet gives the $z$ -axis.

The Precise Statement

Definition: A rectangular 3D coordinate system consists of three mutually perpendicular number lines — the $x$ -axis, $y$ -axis and $z$ -axis — meeting at a common point, the origin $O(0, 0, 0)$ . Any point $P$ in space is uniquely represented by an ordered triple $(x, y, z)$ , where:

$x$ = signed distance from the $yz$ -plane,
$y$ = signed distance from the $zx$ -plane,
$z$ = signed distance from the $xy$ -plane.

P = (x,\, y,\, z)

How to Read a 3D Point

Take the point $A(2, -3, 4)$ . Start at the origin. Move 2 units along the $x$ -axis. From there move $-3$ units parallel to the $y$ -axis (backward, because it is negative). From that spot move 4 units parallel to the $z$ -axis (upward). You have reached $A$ .

Watch out

The order matters absolutely. $(2, -3, 4)$ is not the same point as $(2, 4, -3)$ . Always follow the sequence: $x$ first, then $y$ , then $z$ .

The Three Coordinate Planes

Each pair of axes determines a plane:

Plane	Equation	Description
$xy$ -plane	$z = 0$	the floor — all points with zero height
$yz$ -plane	$x = 0$	one wall — all points with zero $x$
$zx$ -plane	$y = 0$	the other wall — all points with zero $y$

These three planes cut space into 8 octants (the 3D analogue of the four quadrants of the plane). The first octant is where $x > 0$ , $y > 0$ and $z > 0$ .

Distance Between Two Points

This is the natural extension of the 2D distance formula. For $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ :

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

It is just the diagonal of a rectangular box whose edges are the differences in $x$ , $y$ and $z$ . The distance of $P$ from the origin is the special case $OP = \sqrt{x_1^2 + y_1^2 + z_1^2}$ .

Section Formula (Internal Division)

If $R$ divides the segment joining $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ internally in the ratio $m : n$ , then: …

Start your 14-day free trial to unlock the full solution →

Apply the 3D distance formula between $(-3,7,2)$ and $(2,4,-1)$ .

Distance formula in 3D: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$

$d = \sqrt{(2-(-3))^2 + (4-7)^2 + (-1-2)^2}$

…

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 26 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$P(1, -3, 4)$
(a) $5\sqrt{3}$ unit
(b) 7 unit
(c) $3\sqrt{5}$ unit
(d) 5 unit
›Reveal solutionSolution

Use the 3D distance formula $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 mark
$(1, 0, 0)$
›Reveal solutionSolution

Using the 3D distance formula, the distance between $(1,0,0)$ and $(0,1,0)$ works out to exactly $\sqrt{2}$ , confirming the statement.

Distance between $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 mark
$(4, 0, 0)$
›Reveal solutionSolution

Distance formula in 3D gives $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$ ; here it simplifies to 8, the direct separation along the x-axis.

Points: $(4,0,0)$ and $(-4,0,0)$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set 1A1 markMCQ
$(2, 3, 1)$
(1) $3 : 7$
(2) $2 : 7$
(3) $-3 : 7$
(4) $-2 : 4$
›Reveal solutionSolution

Setting the $y$ -coordinate to $0$ in the section formula gives ratio $-3:7$ .

Let the XOZ-plane (equation $y=0$ ) divide the join of $(2,3,1)$ and $(6,7,1)$ in the ratio $k:1$ . The $y$ -coordinate of the division point is …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. The coordinates of origin in three dimensional geometry are$
(a) $(0, 0)$
(b) $(0, 0, 0)$
(c) $(0, 0, 0, 0)$
(d) $(0, 0, 0, 0, 0)$
›Reveal solutionSolution

The origin in 3D geometry is $(0,0,0)$ .

In 3D space, every point needs three coordinates (along the $x$ , $y$ , $z$ axes) to be located. The origin $O$ is the common point where all three coor …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$(2, 3, 5)$
(a) $\sqrt{5}$
(b) $2\sqrt{5}$
(c) $3\sqrt{5}$
(d) 5
›Reveal solutionSolution

Distance between $(2,3,5)$ and $(4,3,1)$ is $2\sqrt5$ .

Using the distance formula: …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$(4, 6, 8)$
(a) $(0, 0, 0)$
(b) $(-2, 0, 2)$
(c) $(2, 0, -2)$
(d) $(2, 0, 2)$
›Reveal solutionSolution

Midpoint of $(4,6,8)$ and $(-8,-6,-4)$ is $(-2,0,2)$ .

Midpoint formula: $\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2},\dfrac{z_1+z_2}{2}\right)$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$A(-2, 1, -3)$
(a) 0 units
(b) 4 units
(c) 7 units
(d) 12 units
›Reveal solutionSolution

Apply the 3D distance formula directly to $A(-2,1,-3)$ and $B(4,3,-6)$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 mark
$(2, 3, 0)$
›Reveal solutionSolution

Any point of the form $(x, y, 0)$ , with $z$ -coordinate zero, lies in the $xy$ -plane.

In three-dimensional geometry, the $xy$ -plane consists of all points whose $z$ -coordinate is $0$ ; the $yz$ -plane consists of points whose $x$ -coordinate is $0$ ; the $xz$ -plane consists of points whose $y$ -coordinate is …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$(2, 4, 5)$
(a) 2
(b) 8
(c) $\dfrac{2}{3}$
(d) $\dfrac{2\tan x}{1 - \tan^2 x}$
(e) $\sin 2x$
(f) 10
(g) 20
(h) $\dfrac{2\tan x}{1 + \tan^2 x}$
(i) 4
›Reveal solutionSolution

The distance of any point $(x, y, z)$ from the $xz$ -plane equals $|y|$ ; here $y = 4$ .

The $xz$ -plane consists of all points with $y = 0$ . The perpendicular distance of a point $(x, y, z)$ from the $xz$ -plane is $|y|$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 mark
$xz$
›Reveal solutionSolution

The $xz$ -plane is defined as the set of all points with $y = 0$ .

In 3D coordinate geometry, the $xz$ -plane consists of all points of the form $(x, 0, z)$ — i.e. every point on this plane has $y$ -coor …

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$(1,-3,4)$
(a) 5
(b) 1
(c) $\sqrt{26}$
(d) None of these
›Reveal solutionSolution

The distance from a point to the $x$ -axis depends only on its $y$ and $z$ coordinates (the $x$ -coordinate doesn't matter, since the foot of the perpendicular slides along the axis).

For a point $(x, y, z)$ , its perpendicular distance from the $x$ -axis is:

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

…

Start your 14-day free trial to see the full solution →

←Previous chapterCh 10Conic Sections

Q.Find the distance between the points (−3,7,2)(-3, 7, 2)(−3,7,2) and (2,4,−1)(2, 4, -1)(2,4,−1).

🎓Unlock everything free for 14 days

Q.Find the distance between the points $(-3, 7, 2)$ and $(2, 4, -1)$ .

Unlock everything free for 14 days