Three vertices of a Parallelogram ABCD are A(1,2,3), B(-1,-2,-1) and C(2,3,2). Find the fourth vertex D.

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A(1,2,3)

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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: …

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The key property of a parallelogram is that its diagonals bisect each other, meaning their midpoints coincide. By equating the midpoint of diagonal AC with the midpoint of diagonal BD, we find the fourth vertex D(4,7,6).

To find the fourth vertex of a parallelogram when three are given, we rely on a fundamental property of parallelograms: their diagonals bisect each other. This means the midpoint of one diagonal is the same as the midpoint of the other diagonal.

Let the given vertices be $A(1,2,3)$ , $B(-1,-2,-1)$ , and $C(2,3,2)$ . Let the unknown fourth vertex be $D(x,y,z)$ . Since the vertices are usually given in cyclic order (A, B, C, D), the diagonals of the parallelogram ABCD are AC and BD.

Here's how we can use this property:

Identify the diagonals: In parallelogram ABCD, the diagonals are AC and BD.
Recall the Midpoint Formula: For any two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ , the midpoint $M$ is given by:
$M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right)$
Calculate the midpoint of diagonal AC:

Using $A(1,2,3)$ and $C(2,3,2)$ :

M_{AC} = \left(\frac{1+2}{2}, \frac{2+3}{2}, \frac{3+2}{2}\right)

M_{AC} = \left(\frac{3}{2}, \frac{5}{2}, \frac{5}{2}\right)

Calculate the midpoint of diagonal BD: Using $B(-1,-2,-1)$ and $D(x,y,z)$ :

M_{BD} = \left(\frac{-1+x}{2}, \frac{-2+y}{2}, \frac{-1+z}{2}\right)

Equate the midpoints: Since the diagonals bisect each other, $M_{AC}$ $M_{A C}$ must be equal to $M_{BD}$ $M_{B D}$ . We equate their corresponding coordinates:
- For the x-coordinate:

\frac{-1+x}{2} = \frac{3}{2}

-1+x = 3

x = 4

*   For the y-coordinate:

\frac{-2+y}{2} = \frac{5}{2}

-2+y = 5

y = 7

*   For the z-coordinate:
    $$\frac{-1+z}{2} = \frac{5}{2}$$ …

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Q.Three vertices of a Parallelogram ABCD are A(1,2,3)A(1,2,3)A(1,2,3), B(−1,−2,−1)B(-1,-2,-1)B(−1,−2,−1) and C(2,3,2)C(2,3,2)C(2,3,2). Find the fourth vertex D.

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Q.Three vertices of a Parallelogram ABCD are $A(1,2,3)$ , $B(-1,-2,-1)$ and $C(2,3,2)$ . Find the fourth vertex D.

Unlock everything free for 14 days