Name the octants in which the following points lie: \[(1,2,3)\], \[(4,-2,3)\], \[(4,-2,-5)\], \[(4,2,-5)\], \[(-4,2,-5)\], \[(-4,2,5)\], \[(-3,-1,6)\] ,(2, – 4, –7). - Noon Academy

Name the octants in which the following points lie:

$(1,2,3)$

,

$(4,-2,3)$

,

$(4,-2,-5)$

,

$(4,2,-5)$

,

$(-4,2,-5)$

,

$(-4,2,5)$

,

$(-3,-1,6)$

,(2, – 4, –7).

Class 11 | Exercise 12.1 | Maths | NCERT | Three Dimensional Geometry

Name the octants in which the following points lie:

$(1,2,3)$

,

$(4,-2,3)$

,

$(4,-2,-5)$

,

$(4,2,-5)$

,

$(-4,2,-5)$

,

$(-4,2,5)$

,

$(-3,-1,6)$

,(2, – 4, –7).

Solution:

The below table which represents the octants:

(i)

$(1,2,3)$

Here x, y and z are positive.

So it lies in I octant.

(ii)

$(4,-2,3)$

Here x and z are positive, y is negative

So it lies in IV octant.

(iii)

$(4,-2,-5)$

Here x is positive, y and z are negative

So it lies in VIII octant.

(iv)

$(4,2,-5)$

Here x and y are positive and z is negative.

So it lies in V octant.

(v)

$(-4,2,-5)$

Here x and z are negative and y is positive

So it lies in VI octant.

(vi)

$(-4,2,5)$

Here x is negative, y is positive and z is positive.

So it lies in II octant.

(vii)

$(-3,-1,6)$

Here x and y are negative and z is positive.

So it lies in III octant.

(viii)

$(2,-4,-7)$

Here x is positive and y and z are negative.

So it lies in VIII octant.

i

More Material

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Evaluate:

Evaluate:

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If $f(x) = |x| - 3,find\mathop {\lim }\limits_{x \to 3} f(x)$

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Let $\begin{array}{l} f(x) = \left\{ \begin{array}{l} \frac{|x-3|}{{(x-3)'}},x \ne 3\\ 0,x = 3 \end{array} \right.\\ \end{array}$ . Show that $\begin{array}{l} \mathop {\lim }\limits_{x \to 3} f(x) \end{array}$ does not exist.

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