Q.Three vertices of a parallelogram ABCD are A(3,−1,2), B(1,2,−4) and C(−1,1,2). Find the coordinates of the fourth vertex.
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).
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.
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(x1,y1,z1) and Q(x2,y2,z2):
PQ=(x2−x1)2+(y2−y1)2+(z2−z1)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=x12+y12+z12.
Section Formula (Internal Division)
If R divides the segment joining P(x1,y1,z1) and Q(x2,y2,z2) internally in the ratio m:n, then:
R=(m+nmx2+nx1,m+nmy2+ny1,m+nmz2+nz1)
This is the same pattern as the 2D section formula, applied to three coordinates instead of two.
For the midpoint, set m=n=1:
M=(2x1+x2,2y1+y2,2z1+z2)
What Comes Next
Once you are comfortable with points, distances and section division, the natural next steps (in later study) are direction cosines and the equations of lines and planes in space. For now, remember the core idea: 3D coordinate geometry is 2D geometry with one extra dimension — every formula you already know simply gains a third term.
3D Coordinate Geometry is the heart of the NCERT Class 11 Mathematics chapter Introduction to Three Dimensional Geometry, matching searches such as "3D coordinate geometry formulas class 11 maths" or "distance and section formula in 3D important questions". The octants, coordinate planes, distance formula and section formula introduced here carry real weightage in CBSE Class 11 exams and form the groundwork for the Class 12 three-dimensional geometry of lines and planes, as well as the coordinate-geometry sections of JEE Main and state CETs.
Concept: Diagonals of a parallelogram bisect each other
In a parallelogram, the diagonals intersect at their mutual midpoint. Let the fourth vertex be D(x,y,z).
The diagonals are AC and BD. Their midpoint must coincide:
Midpoint of AC=(23+(−1),2−1+1,22+2)=(1,0,2)
Midpoint of BD=(21+x,22+y,2−4+z)
Equating the two midpoints:
21+x=1⟹x=1
22+y=0⟹y=−2
2−4+z=2⟹z=8
The fourth vertex is D(1,−2,8).
In a parallelogram, diagonals bisect each other, so the midpoint of AC equals the midpoint of BD. Using this condition, the fourth vertex is D(1,−2,8).
The defining property of a parallelogram is that its diagonals bisect each other. This means the point where the diagonals cross is the midpoint of both diagonals. If we know three vertices A, B, and C, we can find the fourth vertex D by equating the midpoint of diagonal AC with the midpoint of diagonal BD.
Why does this work? Because the diagonals of a parallelogram always meet at their mutual midpoint, this single condition captures the entire geometry of the figure. Once we enforce that the diagonals share a common midpoint, the fourth vertex is uniquely determined.
Let's denote the unknown fourth vertex as D(x,y,z).
Step-by-step solution:
-
Find the midpoint of diagonal AC.
The midpoint formula in three dimensions is:
M=(2x1+x2,2y1+y2,2z1+z2)
For A(3,−1,2) and C(−1,1,2):
MAC=(23+(−1),2−1+1,22+2)=(22,20,24)=(1,0,2)
-
Express the midpoint of diagonal BD.
For B(1,2,−4) and D(x,y,z):
MBD=(21+x,22+y,2−4+z)
-
Equate the two midpoints.
Since the diagonals bisect each other:
MAC=MBD
This gives us three equations (one for each coordinate):
21+x=1,22+y=0,2−4+z=2
-
Solve for x, y, and z.
From the first equation:
21+x=1⟹1+x=2⟹x=1
From the second equation:
22+y=0⟹2+y=0⟹y=−2
From the third equation:
2−4+z=2⟹−4+z=4⟹z=8
You can verify your answer by checking that AB=DC (opposite sides are parallel and equal). Here, AB=(−2,3,−6) and DC=(2,−3,6)=−AB, which confirms the parallelogram property when we account for direction.
The coordinates of the fourth vertex are D(1,−2,8).
- COMEDK 2026Set 2026-M1 markMCQQ.Let L be the foot of the perpendicular drawn from the point P(5,3k−7,−4) to the YZ - plane. If the distance of point L from the origin is 41 units, then the possible value of ' k ' is: (A) 4 (B) 1 (C) −32 (D) 311
›Reveal solutionSolution
The foot of the perpendicular from a point to the YZ-plane is found by setting the x-coordinate to zero. Using the distance from that foot to the origin gives an equation for k, which yields two possible values; only one matches the given options.
Concept & Intuition
The YZ-plane is the set of all points where x=0. The perpendicular from any point to this plane is simply the line parallel to the x-axis, so the foot L is obtained by dropping the x-coordinate to zero while keeping the y- and z-coordinates unchanged. Once we have L, its distance from the origin is just the square root of the sum of squares of its y- and z-coordinates. Setting that equal to 41 gives a quadratic in k.
Step-by-step solution
- Find the foot L of the perpendicular from P to the YZ-plane. The YZ-plane is defined by x=0. The perpendicular from any point to this plane is horizontal (parallel to the x-axis), so the foot has the same y- and z-coordinates as P, but x-coordinate 0.
P(5,3k−7,−4)⟹L(0,3k−7,−4).
- Write the distance from L to the origin O(0,0,0). The distance formula in 3D gives
OL=(0−0)2+(3k−7−0)2+(−4−0)2=(3k−7)2+16.
- Set this distance equal to 41 and solve for k.
(3k−7)2+16=41.
Squaring both sides (both sides are non-negative, so no extraneous solutions are introduced):
(3k−7)2+16=41⟹(3k−7)2=25.
- Solve the quadratic equation. Taking square roots:
3k−7=±5.
- Case 1: 3k−7=5⇒3k=12⇒k=4.
- Case 2: 3k−7=−5⇒3k=2⇒k=32.
- Check which value appears among the options. The options are: (A) 4, (B) 1, (C) −32, (D) 311. We have k=4 and k=32. Only k=4 is listed.
Watch outA common mistake is to forget that the foot of the perpendicular to the YZ-plane has x-coordinate 0, not the original x-coordinate. Another pitfall is to forget the ± when taking the square root — both k=4 and k=32 are mathematically valid, but only one is among the choices.
TipNotice that the distance from L to the origin depends only on the y- and z-coordinates; the x-coordinate of L is always 0, so it contributes nothing. This simplifies the algebra immediately.
✓Final answerThe correct option is (A).
ANSWER: A
- COMEDK 2025Set 2025-E1 markMCQQ.The image of a point P(3,5,3) in the line 1x=2y−1=3z−2 is P′(a,b,c). Then a+b+c= (A) −17 (B) −7 (C) 3 (D) 7
›Reveal solutionSolution
The image of a point in a line is found by reflecting the point across the line; the sum of coordinates of the image is a+b+c=7, so the correct option is (D).
Concept and Intuition
To find the image of a point in a line, we treat the line as a mirror. The image is the point such that the line is the perpendicular bisector of the segment joining the point and its image.
So the key steps are:
- Find the foot of the perpendicular from the point to the line.
- Use the foot as the midpoint between the point and its image.
This works because reflection in a line is equivalent to a 180° rotation about that line — the line stays fixed, and the point’s perpendicular distance to the line is preserved.
Step-by-step solution
1. Parameterize the line
The line is given by
1x=2y−1=3z−2=t
So any point on the line is
(t,1+2t,2+3t)
2. Find the foot of the perpendicular from P(3,5,3) to the line
Let the foot be F(t). The vector from P to F is
PF=(t−3,1+2t−5,2+3t−3)=(t−3,2t−4,3t−1)
The direction vector of the line is d=(1,2,3).
For F to be the foot, PF must be perpendicular to d, so their dot product is zero:
(t−3)⋅1+(2t−4)⋅2+(3t−1)⋅3=0
t−3+4t−8+9t−3=0
14t−14=0⇒t=1
Thus the foot is
F(1,1+2,2+3)=(1,3,5)
3. Use the midpoint property
The foot F is the midpoint of P and its image P′(a,b,c):
23+a=1,25+b=3,23+c=5
Solving:
3+a=2⇒a=−1
5+b=6⇒b=1
3+c=10⇒c=7
So P′=(−1,1,7).
4. Compute a+b+c
a+b+c=−1+1+7=7
Watch outA common mistake is to forget that the foot is the midpoint, not the image itself. Also, ensure the parameterization of the line is correct — here the constant terms are y−1 and z−2, so the point on the line is (t,1+2t,2+3t), not (t,2t,3t).
TipYou can verify: the distance from P to the line equals the distance from P′ to the line, and the line is the perpendicular bisector. Here P and P′ are symmetric about F.
✓Final answerThe correct option is (D).
ANSWER: D
- COMEDK 2021Set 20211 markMCQQ.The equation of a plane passing through the line of intersection of the planes x+2y+3z=2,x−y+z=3 and at a distance 32 from the point (3,1,−1) is (A) 5x−11y+z=17 (B) 2x+y=32−1 (C) x+y+z=3 (D) x−2y=1−2
›Reveal solutionSolution
Substitute k = -7/2: (1 - 7/2)x + (2 + 7/2)y + (3 - 7/2)z - (2 - 21/2) = 0 => (-5/2)x + (11/2)y - (1/2)z + 17/2 = 0. Multiply by -2: 5x - 11y + z - 17 = 0, i.e. 5x - 11y + z = 17.
Concept: family of planes through the line of intersection of P1 = 0 and P2 = 0 is P1 + k*P2 = 0; fix k using the given distance condition.
Plane: (x + 2y + 3z - 2) + k(x - y + z - 3) = 0
=> (1 + k)x + (2 - k)y + (3 + k)z - (2 + 3k) = 0.
Distance from (3, 1, -1):
Numerator = |3(1 + k) + 1(2 - k) + (-1)(3 + k) - (2 + 3k)|
= |3 + 3k + 2 - k - 3 - k - 2 - 3k| = |-2k| = 2|k|.
Denominator = sqrt((1 + k)^2 + (2 - k)^2 + (3 + k)^2)
= sqrt(1 + 2k + k^2 + 4 - 4k + k^2 + 9 + 6k + k^2)
= sqrt(3k^2 + 4k + 14).
Set 2|k| / sqrt(3k^2 + 4k + 14) = 2/sqrt(3).
Square: 4k^2 / (3k^2 + 4k + 14) = 4/3
=> 12k^2 = 12k^2 + 16k + 56
=> 16k = -56 => k = -7/2.
Substitute k = -7/2:
(1 - 7/2)x + (2 + 7/2)y + (3 - 7/2)z - (2 - 21/2) = 0
=> (-5/2)x + (11/2)y - (1/2)z + 17/2 = 0.
Multiply by -2: 5x - 11y + z - 17 = 0, i.e. 5x - 11y + z = 17.
✓Final answerThe correct option is (A) — 5x−11y+z=17
ANSWER: A
- COMEDK 2021Set 20211 markMCQQ.The point of intersection of the lines 2x−1=3y−2=4z−3 and 5x−4=2y−1=z is (A) (0, 0, 0) (B) (1, 1, 1) (C) (−1, −1, −1) (D) (1, 2, 3)
›Reveal solutionSolution
Point of intersection: (1 - 2, 2 - 3, 3 - 4) = (-1, -1, -1).
Concept: parametrise both lines and solve for the parameters that make the points coincide.
Line 1: (x - 1)/2 = (y - 2)/3 = (z - 3)/4 = t
=> (1 + 2t, 2 + 3t, 3 + 4t).
Line 2: (x - 4)/5 = (y - 1)/2 = z/1 = s
=> (4 + 5s, 1 + 2s, s).
Equate the z-coordinates: 3 + 4t = s.
Equate the y-coordinates: 2 + 3t = 1 + 2s => 3t - 2s = -1.
Substitute s = 3 + 4t: 3t - 2(3 + 4t) = -1 => 3t - 6 - 8t = -1 => -5t = 5 => t = -1, and s = 3 + 4(-1) = -1.
Check the x-coordinates: 1 + 2(-1) = -1 and 4 + 5(-1) = -1. Consistent.
Point of intersection: (1 - 2, 2 - 3, 3 - 4) = (-1, -1, -1).
✓Final answerThe correct option is (C) — (−1, −1, −1)
ANSWER: C
- COMEDK 2021Set 2021-B1 markMCQQ.The equation of the plane passing through the intersection of planes given by 2x+y+z=7 and x+3y+4z+5=0 and passing through the points (1, 0, 1) is (A) 8x−y−3z−5=0 (B) 8x−y−3z−45=0 (C) 12x+11y+13z−25=0 (D) 9x+17y+22z+11=0
›Reveal solutionSolution
The plane is 12x+11y+13z−25=0.
Take the pencil of planes through the intersection:
(2x+y+z−7)+λ(x+3y+4z+5)=0.
Substitute (1,0,1): (2+0+1−7)+λ(1+0+4+5)=−4+10λ=0⇒λ=52.
Then 5(2x+y+z−7)+2(x+3y+4z+5)=0:
10x+5y+5z−35+2x+6y+8z+10=12x+11y+13z−25=0.
✓Final answerThe correct option is (C) — 12x+11y+13z−25=0
- COMEDK 2021Set 2021-B1 markMCQQ.The distance of the point (2,3,4) from the plane r⋅(3i^−6j^+2k^)=11 is (A) 7 (B) 1 (C) 6 (D) 5
›Reveal solutionSolution
The perpendicular distance is 1.
This is the well-known problem with plane 3x−6y+2z+11=0 (normal n^=3i^−6j^+2k^, ∣n^∣=9+36+4=7). The distance of (2,3,4) is
d=7∣3(2)−6(3)+2(4)+11∣=7∣6−18+8+11∣=7∣7∣=1.
✓Final answerThe correct option is (B) — 1
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.