MHT-CET Maths · Line and Plane

Tetrahedron Geometry — Centroid, Volume, and Vertices

Average the four vertices to get a tetrahedron's centroid (or the three vertices for a triangle), reverse that average to recover a missing vertex or coordinate, and use one-sixth the scalar triple product to get a tetrahedron's volume — including the plane-cuts-the-axes volume of OABC.

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Why this matters

This thin subtopic is pure plug-in: across the 9 PYQs here, only THREE shapes appear and every one is a one-line formula. The centroid shape (direct average, or the inverse: solve for a missing vertex/coordinate) is the most frequent and runs EASY-to-MODERATE. The volume shape is a single scalar triple product set equal to a given value and solved for one unknown coordinate. The third shape — a plane parallel to two given lines, cutting the axes — chains a cross product, the intercept form, and the product-of-intercepts volume of OABC. There are no proofs and no tricks: memorise the three formulas, watch the factor of 1/4 vs 1/3 and the 1/6 on the volumes, and the subtopic is yours.

Concept 1 of 3: Centroid of a tetrahedron and a triangle

The centroid is just the average of the corner points: add up the vertices and divide by how many there are — 4 for a tetrahedron, 3 for a triangle. The same averaging, read backwards, lets you recover a missing vertex (or a missing coordinate) when the centroid is given.

Definition

For a tetrahedron with vertices $A, B, C, D$ , the centroid is the average of all four:

$G = \dfrac{A + B + C + D}{4}$ , i.e. $\left(\dfrac{x_A+x_B+x_C+x_D}{4},\ \dfrac{y_A+\cdots}{4},\ \dfrac{z_A+\cdots}{4}\right)$ .

For a triangle with vertices $A, B, C$ , divide by 3 instead:

$G = \dfrac{A + B + C}{3}$ .

Inverse use: if $G$ and all-but-one vertex are known, isolate the missing one — e.g. for a triangle $x_C = 3x_G - x_A - x_B$ (and likewise for a tetrahedron with the factor 4).

Centroid (tetrahedron and triangle)

G_{\text{tet}} = \dfrac{A + B + C + D}{4} \qquad G_{\triangle} = \dfrac{A + B + C}{3}

$A, B, C, D$ position vectors (or coordinate triples) of the vertices
$G$ centroid — the component-wise average of the vertices

Diagram · coordinate planes & octants (drag to rotate)

Three planes (XY, YZ, ZX), each splitting space in two → 2 × 2 × 2 = 8 octants. P sits in the first octant (all coordinates positive).

Worked example

Find the centroid of the tetrahedron with vertices

A(2,0,1)

B(0,4,-1)

C(2,2,2)

D(0,2,2)

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 12th May Shift 1 · Q102Easy

Example 1 · Line and Plane · Tetrahedron Geometry — Centroid, Volume, and Vertices

The centroid of tetrahedron with vertices at

A(-1,2,3)

B(3,-2,1)

C(2,1,3)

and

D(-1,-2,4)

Divide by 4 for a tetrahedron, by 3 for a triangle

The single most common slip: a four-vertex solid averages over 4, a three-vertex triangle over 3. Read the figure named in the stem — "tetrahedron" means

/4

, "triangle" means

/3

. Using the wrong divisor lands you on a tempting distractor every time.

Inverse problems: rearrange, don't re-guess

When the centroid is given and a vertex is unknown, multiply through by the divisor first:

x_A + x_B + x_C = 3x_G

, then subtract the knowns. Skipping the multiply-through step is where sign errors creep in.

Watch which coordinate the puzzle reuses

Some stems reuse a letter across the centroid AND a vertex (e.g. centroid

(r, q, 1)

with a vertex coordinate

q

). Match coordinates by POSITION, solve the chain in order, and don't conflate a centroid component with a same-named vertex component.

Drill 3 more on centroid of a tetrahedron and a triangle

Concept 2 of 3: Volume of a tetrahedron via the scalar triple product

Pick any vertex as a base point and form the three edge vectors leaving it. The scalar triple product of those edges measures the volume of the parallelepiped they span; a tetrahedron is exactly one-sixth of that box. When one coordinate is unknown, set the volume equal to the given value and solve the resulting linear equation.

Definition

For a tetrahedron with vertices $A, B, C, D$ , build the three edges from $A$ : $\overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD}$ . Then:

Volume $V = \dfrac{1}{6}\,\big|\,[\,\overrightarrow{AB}\ \ \overrightarrow{AC}\ \ \overrightarrow{AD}\,]\,\big|$ , where the bracket is the scalar triple product — equal to the $3\times 3$ determinant of the three edge rows.
The absolute value is taken because volume is non-negative (the determinant's sign only records orientation).

Solve-for-a-coordinate variant: with one vertex coordinate as $x$ , the determinant becomes linear in $x$ ; set $\dfrac{1}{6}|\det| = V$ (so $|\det| = 6V$ ) and solve.

Tetrahedron volume

V = \dfrac{1}{6}\left|\det\begin{bmatrix} \overrightarrow{AB} \\ \overrightarrow{AC} \\ \overrightarrow{AD} \end{bmatrix}\right| = \dfrac{1}{6}\big|\,\overrightarrow{AB}\cdot(\overrightarrow{AC}\times\overrightarrow{AD})\,\big|

$\overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD}$ the three edge vectors from a common vertex $A$
$[\,\cdot\ \cdot\ \cdot\,]$ scalar triple product = determinant of the edge rows
$\tfrac{1}{6}$ a tetrahedron is one-sixth of the spanning parallelepiped

Diagram · triple product = box volume (SVG, drag to rotate)

[a b c] = a · (b × c) = 13.22 (signed volume of the box)

The box spanned by a, b, c has volume |[a b c]|. Painter's-ordered faces fake the solidity — edges don't truly hide behind nearer faces, which is the SVG limit this comparison is testing.

Worked example

Find the volume of the tetrahedron with vertices

A(0,0,0)

B(2,0,0)

C(0,3,0)

D(0,0,4)

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 12th May Shift 2 · Q135Moderate

Example 2 · Line and Plane · Tetrahedron Geometry — Centroid, Volume, and Vertices

If the volume of tetrahedron, whose vertices are

A(1,2,3),\, B(-3,-1,1),\, C(2,1,3)

and

D(-1,2,x)

\frac{11}{6}

cubic units, then the value of

x

It's $\tfrac{1}{6}$ for a tetrahedron, not $\tfrac{1}{3}$ or 1

The parallelepiped volume is

|\det|

; the tetrahedron is one-SIXTH of it, not one-third. (

\tfrac{1}{3}

belongs to the pyramid-volume formula

\tfrac{1}{3}\times\text{base}\times\text{height}

, a different setup.) Forgetting the

\tfrac{1}{6}

gives an answer six times too big.

Build edges from ONE common vertex

Use

\overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD}

— all three leaving the same vertex

A

(head minus the SAME tail). Mixing tails (e.g.

\overrightarrow{AB}, \overrightarrow{BC}, \overrightarrow{AD}

) breaks the triple-product meaning and the volume is wrong.

Set $|\det| = 6V$ , not $|\det| = V$

In a solve-for-

x

problem, multiply the target volume by 6 BEFORE equating to the determinant:

\dfrac{1}{6}|\det| = \tfrac{11}{6}

means

|\det| = 11

, not

\tfrac{11}{6}

. Dropping the

\times 6

corrupts the linear equation for

x

Concept 3 of 3: Volume of OABC from a plane cutting the axes

A favourite MHT-CET chain: a plane is parallel to two given lines, so its normal is the cross product of their direction ratios. Write the plane in the form

x+y+z=k

(or the matching intercept form), read off where it crosses each axis, and the tetrahedron

OABC

— origin plus the three intercepts — has volume one-sixth the product of the intercepts.

Definition

Step by step, for a plane parallel to two lines with direction ratios $\vec{d_1}$ and $\vec{d_2}$ , passing through a point $P$ :

Normal $\vec{n} = \vec{d_1}\times\vec{d_2}$ (the plane is parallel to both lines, so $\vec{n}$ is perpendicular to both).
Plane: $\vec{n}\cdot(\vec{r}-\vec{P}) = 0$ ; when $\vec{n}\propto(1,1,1)$ this is $x+y+z = k$ with $k$ fixed by $P$ .
Intercepts: set two coordinates to 0 — $A(a,0,0), B(0,b,0), C(0,0,c)$ . For $x+y+z=k$ each intercept is $k$ .
Volume of $OABC$ : the edges from $O$ are along the axes, so $V = \dfrac{1}{6}\,|abc|$ (for $x+y+z=k$ , $V = \dfrac{k^3}{6}$ ).

Plane normal, intercepts, and OABC volume

\vec{n} = \vec{d_1}\times\vec{d_2} \qquad V_{OABC} = \dfrac{1}{6}\,|a\,b\,c| = \dfrac{k^3}{6}\ \ (\text{for } x+y+z=k)

$\vec{d_1}, \vec{d_2}$ direction ratios of the two parallel lines
$a, b, c$ the $x$ -, $y$ -, $z$ -intercepts of the plane
$k$ the constant in $x+y+z=k$ ; each intercept when $\vec{n}\propto(1,1,1)$

Worked example

A plane is parallel to the lines with direction ratios

(1,0,-1)

and

(0,1,-1)

and passes through

(1,1,2)

. It cuts the axes at

A, B, C

. Find the volume of tetrahedron

OABC

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 13th May Shift 2 · Q138Moderate

Example 3 · Line and Plane · Tetrahedron Geometry — Centroid, Volume, and Vertices

A plane is parallel to two lines whose direction ratios are

2,0,-2

and

-2,2,0

and it contains the point

(2,2,2)

. If it cuts co-ordinate axes at A, B, C, then the volume of the tetrahedron OABC (in cubic units) is

Normal = cross product, then the constant comes from the POINT

Two steps, in order: first

\vec{n} = \vec{d_1}\times\vec{d_2}

gives the coefficients, THEN plug the given point into

\vec{n}\cdot\vec{r}=k

to fix

k

. Skipping the point and assuming

k

is wrong; the point is what pins the plane down.

Volume of OABC is $\tfrac{1}{6}\,abc$ , not $\tfrac{1}{6}k$ or $abc$

It's one-sixth the PRODUCT of the three intercepts. For

x+y+z=k

that's

\tfrac{1}{6}k^3

(since all three intercepts equal

k

) — e.g.

k=3

gives

\tfrac{27}{6}=\tfrac{9}{2}

, not

\tfrac{3}{6}

. Don't forget to cube

k

before dividing by 6.

Simplify the cross-product normal before reading intercepts

A normal like

(4,4,4)

is parallel to

(1,1,1)

— reduce it so the plane reads cleanly as

x+y+z=k

. Carrying the un-reduced

(4,4,4)

into the intercept step muddies the constant and the intercepts.

Drill 2 more on volume of oabc from a plane cutting the axes

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Centroid of a tetrahedron and a triangle
Centroid (tetrahedron and triangle)
$G_{\text{tet}} = \dfrac{A + B + C + D}{4} \qquad G_{\triangle} = \dfrac{A + B + C}{3}$
Volume of a tetrahedron via the scalar triple product
Tetrahedron volume
$V = \dfrac{1}{6}\left|\det\begin{bmatrix} \overrightarrow{AB} \\ \overrightarrow{AC} \\ \overrightarrow{AD} \end{bmatrix}\right| = \dfrac{1}{6}\big|\,\overrightarrow{AB}\cdot(\overrightarrow{AC}\times\overrightarrow{AD})\,\big|$
Volume of OABC from a plane cutting the axes
Plane normal, intercepts, and OABC volume
$\vec{n} = \vec{d_1}\times\vec{d_2} \qquad V_{OABC} = \dfrac{1}{6}\,|a\,b\,c| = \dfrac{k^3}{6}\ \ (\text{for } x+y+z=k)$

Watch out for (9)

Divide by 4 for a tetrahedron, by 3 for a triangle
→ Centroid of a tetrahedron and a triangle
Inverse problems: rearrange, don't re-guess
→ Centroid of a tetrahedron and a triangle
Watch which coordinate the puzzle reuses
→ Centroid of a tetrahedron and a triangle
It's $\tfrac{1}{6}$ for a tetrahedron, not $\tfrac{1}{3}$ or 1
→ Volume of a tetrahedron via the scalar triple product
Build edges from ONE common vertex
→ Volume of a tetrahedron via the scalar triple product
Set $|\det| = 6V$ , not $|\det| = V$
→ Volume of a tetrahedron via the scalar triple product
Normal = cross product, then the constant comes from the POINT
→ Volume of OABC from a plane cutting the axes
Volume of OABC is $\tfrac{1}{6}\,abc$ , not $\tfrac{1}{6}k$ or $abc$
→ Volume of OABC from a plane cutting the axes
Simplify the cross-product normal before reading intercepts
→ Volume of OABC from a plane cutting the axes

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