Tag: introduction to three dimensional geometry

Questions Related to introduction to three dimensional geometry

The points (-5,12), (-2,-3),(9,-10),(6,5) taken in order, form

maths introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

  1. Parallelogram

  2. rectangle

  3. rhombus

  4. square

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Correct Option: A
Explanation:

Given points $A(-5, 12)\quad B(-2, -3), C(9, -10), D(6, 5)$


Distance $AB=\sqrt{(-5+2)^2+(12+3)^2}=\sqrt{9+225}=\sqrt{234}$
Distance $BC=\sqrt{(-2-9)^2+(-3+10)^2}=\sqrt{121+49}=\sqrt{170}$

Distance $CD=\sqrt{(9-6)^2+(-10-5)^2}=\sqrt{9+225}=\sqrt{234}$
Distance $AD=\sqrt{(-5-6)^2+(12-5)^2}=\sqrt{121+49}=\sqrt{170}$

Distance $AC=\sqrt{(-5-9)^2+(12+10)^2}=\sqrt{196+484}=\sqrt{680}$
Distance $BD=\sqrt{(-2-6)^2+(-3-5)^2}=\sqrt{64+64}=\sqrt{128}$

These points forms a parallelogram, opposite pair of sides are equal and adjacent sides do not form right angles.

If G is the centroid of $\triangle ABC$ and BC = 3, CA = 4, AB = 5 then BG =

maths introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

  1. $\dfrac { \sqrt { 73 } }{ 3 } $

  2. $\dfrac { \sqrt { 13 } }{ 3 } $

  3. $\dfrac { \sqrt { 52 } }{ 3 } $

  4. $\dfrac { \sqrt { 26 } }{ 3 } $

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Correct Option: C

If $( 3,4 )$ and $( 6,5 )$ are the extremities of a diagonal of a parallelogram and $( 2,1 )$ is is third vertex, then its fourth vertex is _______.

maths introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

  1. $( - 1,0 )$

  2. $( - 1,1 )$

  3. $( 0,-1 )$

  4. $( 7,8 )$

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Correct Option: D

The foot of the perpendicular from the point $A(7, 14, 5)$ to the plane $2x+4y-z=2$ is?

maths introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

  1. $(3, 1, 8)$

  2. $(1, 2, 8)$

  3. $(3, -3, 5)$

  4. $(5, -3, -4)$

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Correct Option: B
Explanation:

Let N be the foot of the perpendicular drawn from the point $A(7, 14, 5)$ and perpendicular to the plane $2x+4y-z=2$

Then, the equation of the line PN is $\dfrac{x-7}{2}=\dfrac{y-14}{4}=\dfrac{z-5}{-1}=\lambda$ (say)
Let the coordinates of N be $N(2\lambda +7, 4\lambda +14, -\lambda +5)$
Since N lies on the plane $2x+4y-z=2$, so
$2(2\lambda +7)+4(4\lambda +14)-(-\lambda +5)=2$
$\Rightarrow 21\lambda =-63$
$\Rightarrow \lambda =-3$
$\therefore$ required foot of the perpendicular is
$N(-6+7, -12+14, 3+5)$, i.e., $N(1, 2, 8)$.

In geometry, we take a point, a line and a plane as undefined terms.

maths introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

  1. True

  2. False

  3. Ambiguous

  4. Data Insufficient

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Correct Option: A
Explanation:

 In Geometry, we define a point as a location and no size. A line is defined as something that extends infinitely in either direction but has no width and is one dimensional while a plane extends infinitely in two dimensions.

Arrange the points: $\mathrm{A}(1,2-3), \mathrm{B}(-1,2,-3), \mathrm{C}(-1,-2-3)$ and $\mathrm{D}(1,-2, -3)$ in the increasing order of their octant numbers:

maths introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

  1. $A,B,C,D$

  2. $B,C,D,A$

  3. $C,D,A,B$

  4. $D,C,B,A$

Show answer
Correct Option: A
Explanation:
 Octant $I$  $II$ $III$  $IV$  $V$  $VI$  $VII$ $VIII$
 Signs: $+,+,+$  $-,+,+$ $-,-,+$  $+,-,+$  $+,+,-$ $-,+,-$ $-,-,-$  $+,-,-$ 

Based on this, increasing order is

$ A,B ,C,D$

Graph $x^2+y^2=4$ in 3D looks like

maths introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

  1. Circle

  2. Cylinder

  3. Hemisphere

  4. Sphere

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Correct Option: B
Explanation:

The given curve is $x^2+y^2=4$ 

So $x$ coordinate and y-coordinate are connected by $x^2+y^2=4$
which is locus of a circle with radius $2$
But z-coordinate can be anything, so in three dimension the circle $x^2+y^2=4$ will be 
stretched which will be a cylinder with radius same as the radius of the circle .

The point $(0 , -2 , 5)$ lies on the

maths introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

  1. z axis

  2. x axis

  3. xy plane

  4. yz plane

  5. xz plane

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Correct Option: D
Explanation:

Given point is $(0,-2,5)$
The $X$-coordinate in the given point is zero. 

So, the point lies on $yz$ plane.

The coordinates of any point, which lies in $yz$ plane, are

maths introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

  1. $(x,y,y)$

  2. $(0,y,y)$

  3. $(0,y,x)$

  4. $(x,y,z)$

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Correct Option: B,C
Explanation:

In a $y-z$ plane, x co-ordinate is always $0$

So $(0,y,y)$ and $(0,y,x)$ are point in a $y-z$ plane.

An equation of sphere with centre at origin and radius $r$ can be represented as

maths introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

  1. $x^2+y^2+z^2=r$

  2. $x^2+y^2+z^2=r^2$

  3. $x^2+y^2+z^2=2r^2$

  4. None of the above

Show answer
Correct Option: B
Explanation:

Sphere is locus of a point in 3D whose distance from a fixed point(center) is constant (radius)

$\Rightarrow \sqrt{(x-0)^2+(y-0)^2+(z-0)^2}=|r|$
$\Rightarrow x^2+y^2+z^2=r^2$, square both sides