The point (-2,-3,-4) lies in the (A) First octant (B) Seventh octant (C) Second octant (D) Eighth octant

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(-2,-3,-4)

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Concept understanding — 3D Coordinate Octants

3D Coordinate Octants

Stand at the corner of a room where two walls meet the floor. That corner is the origin $O$ , and the three edges meeting there are the three coordinate axes:

the edge where the floor meets one wall is the x-axis,
the edge where the floor meets the other wall is the y-axis,
the vertical edge where the two walls meet is the z-axis.

These three axes are mutually perpendicular. Taken in pairs they determine three coordinate planes — the $xy$ -plane, the $yz$ -plane and the $zx$ -plane — and these planes slice all of space into eight regions. Each region is called an octant (from octo, meaning eight).

Why exactly eight

In a plane, the two axes make $4$ quadrants. Adding a third axis doubles this: each of the three coordinate planes splits space into two halves, so there are $2 \times 2 \times 2 = 2^3 = 8$ regions. Equivalently, an octant is fixed by choosing a sign — positive or negative — for each of $x$ , $y$ and $z$ , and there are $2^3 = 8$ such sign-triples $(\pm, \pm, \pm)$ .

The eight octants and their signs

(x, y, z)

The standard NCERT labelling of the signs is:

Octant	Sign of $x$	Sign of $y$	Sign of $z$	Example point
I	$+$	$+$	$+$	$(1, 2, 3)$
II	$-$	$+$	$+$	$(-1, 2, 3)$
III	$-$	$-$	$+$	$(-1, -2, 3)$
IV	$+$	$-$	$+$	$(1, -2, 3)$
V	$+$	$+$	$-$	$(1, 2, -3)$
VI	$-$	$+$	$-$	$(-1, 2, -3)$
VII	$-$	$-$	$-$	$(-1, -2, -3)$
VIII	$+$	$-$	$-$	$(1, -2, -3)$

Notice the pattern: octants I–IV all have $z > 0$ (above the $xy$ -plane) and V–VIII all have $z < 0$ (below it).

How to read a point's octant

Just look at the signs of its coordinates. For example, $(-3, 1, 2)$ has $x < 0$ , $y > 0$ , $z > 0$ , which is the sign pattern of octant II. The point $(-3, 1, -2)$ has $x < 0$ , $y > 0$ , $z < 0$ , placing it in octant VI. …

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The sign pattern $(-,-,-)$ places the point in the seventh octant, which is the only octant where all three coordinates are negative.

In 3D coordinate geometry, the three axes — $x$ , $y$ , and $z$ — divide space into eight regions called octants. Each octant is defined by the signs of the three coordinates. The naming convention is standardised: the first octant is where all three coordinates are positive $(+,+,+)$ . From there, the octants are numbered by moving around the $xy$ -plane and then considering the $z$ -sign.

The key idea: you don't need to memorise all eight sign patterns. Just remember that the first octant is $(+,+,+)$ , and then the octant number increases as you change signs in a systematic way. But for a quick check, the seventh octant is the one where all three coordinates are negative — that is, $(-,-,-)$ .

Let’s verify step by step.

Identify the signs of each coordinate.

The point is $(-2, -3, -4)$ .
- $x = -2$ → negative
- $y = -3$ → negative
- $z = -4$ → negative
So the sign pattern is $(-,-,-)$ .
Recall the octant numbering convention.

The standard convention (used in most Indian textbooks and exams) is:
- First octant: $(+,+,+)$
- Second octant: $(-,+,+)$
- Third octant: $(-,-,+)$
- Fourth octant: $(+,-,+)$
- Fifth octant: $(+,+,-)$
- Sixth octant: $(-,+,-)$
- Seventh octant: $(-,-,-)$
- Eighth octant: $(+,-,-)$
This pattern is derived by first fixing the $z$ -sign: octants 1–4 have $z>0$ , octants 5–8 have $z<0$ . Then within each group, the $x$ and $y$ signs cycle through the four quadrants of the $xy$ -plane.
Match the pattern.

Our pattern $(-,-,-)$ matches exactly the seventh octant. …

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

KCET 2022Set C-41 markMCQ
$2x-3y+4z=29$
›Reveal solutionSolution

The line from the origin perpendicular to the plane is parallel to the plane's normal vector, so the foot is $t\vec n$ ; plug it into the plane equation and solve the single unknown $t$ .

Step 1 — Identify the normal.

For a plane $ax + by + cz = d$ , the coefficients form the normal vector
$\vec n = a\hat i + b\hat j + c\hat k .$
Here $2x - 3y + 4z = 29$ , so
$\vec n = 2\hat i - 3\hat j + 4\hat k .$
Step 2 — Why the foot must be a multiple of $\vec n$ .

The perpendicular dropped from the origin to the plane is, by definition, a line parallel to $\vec n$ passing through $O(0,0,0)$ . Any point on it is
$P = t\,\vec n = (2t,\; -3t,\; 4t), \qquad t \in \mathbb{R}.$
So the foot of the perpendicular is that one value of $t$ for which $P$ actually lies on the plane. This reduces a 3-variable problem to one unknown — that is the whole trick.

Step 3 — Impose the plane condition.

Substitute $x = 2t,\; y = -3t,\; z = 4t$ into $2x - 3y + 4z = 29$ :
$2(2t) - 3(-3t) + 4(4t) = 29$ $4t + 9t + 16t = 29$
$29t = 29 \quad\Longrightarrow\quad t = 1 .$ …
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KCET 2021Set A-11 markMCQ
$A$
›Reveal solutionSolution

Use the intercept form of a plane; the centroid of the triangle cut off on the axes is just one-third of each intercept.

Step 1 — Why the intercept form is the right tool.

A plane that meets the three coordinate axes can always be written as
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 ,$
because putting $y=z=0$ gives $x=a$ , and so on. So its axis intercepts are exactly
$A(a,0,0),\qquad B(0,b,0),\qquad C(0,0,c).$
This form turns the geometric condition into simple arithmetic on $a,b,c$ .

Step 2 — Centroid of the triangle $ABC$ .

The centroid of a triangle is the average of its three vertices:
$G=\left(\frac{a+0+0}{3},\ \frac{0+b+0}{3},\ \frac{0+0+c}{3}\right)=\left(\frac{a}{3},\frac{b}{3},\frac{c}{3}\right).$
Step 3 — Impose $G=(1,2,3)$ .
$\frac{a}{3}=1\Rightarrow a=3,\qquad \frac{b}{3}=2\Rightarrow b=6,\qquad \frac{c}{3}=3\Rightarrow c=9 .$
Step 4 — Write the plane. …
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Q.The point (−2,−3,−4)(-2,-3,-4)(−2,−3,−4) lies in the (A) First octant (B) Seventh octant (C) Second octant (D) Eighth octant

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Q.The point $(-2,-3,-4)$ lies in the
(A) First octant
(B) Seventh octant
(C) Second octant
(D) Eighth octant

Unlock everything free for 14 days