0

Coordinate axes and coordinate planes in 3d space - class-XII

Description: coordinate axes and coordinate planes in 3d space
Number of Questions: 32
Created by:
Tags: three dimensional coordinates three dimensional geometry introduction to three-dimensional geometry maths introduction to three dimensional geometry
Attempted 0/32 Correct 0 Score 0

Examine if the following is true statement.
The cube can cast a shadow in the shape of a rectangle.

  1. True

  2. False


Correct Option: A
Explanation:

The formation of shadow is equivalent to cutting of cube . We know that cutting cube always gives a square or a rectangle . so, it can cast a shadow of a rectangle.

Examine if the following is true statement.
The cube can cast a shadow in the shape of a hexagon.

  1. True

  2. False


Correct Option: B
Explanation:
False
The cube cannot cast shadow in shape of hexaxon because of following reason:
no matter whatever the position of light source focussing on cube it always cast a shadow of either square or rectangle .

State the following statement is True or False
Two distinct lines intersecting each other in a plane can have two points in common

  1. True

  2. False


Correct Option: B
Explanation:

Answer is option B (False)
Two distinct lines intersecting each other  in a plane cannot have more than one point in common

Name three undefined terms.

  1. Point

  2. Line

  3. Plane

  4. All of the above


Correct Option: D
Explanation:

The basic undefined term is point.
Line is formed from points and plane is formed from many lines.
Undefined terms are point, line and plane.

State the following statement is True or False
If two distinct lines are intersecting each other in a plane then they cannot have more than one point in common.

  1. True

  2. False


Correct Option: A
Explanation:
suppose lines $l1$ and $l2$ intersect at two disticnt points say P and Q.Then $l1$ contains points P and Q.
Also, $l2$ contains points P and Q.
So two lines $l1$ and $l2$ pass through two distinct points P and Q.
But only one line can pass through two different points.  (axiom 3)
Hence, two lines cannot have more than one point in common.
$\therefore$ Option A  is correct.

Platonic solids are regular  Polyhedra.

  1. True

  2. False


Correct Option: A
Explanation:

Platonic solids are regular  Polyhedra.

The coordinate of any point, which lies in $xy$ plane , is

  1. $(x,0,y)$

  2. $(x,x,0)$

  3. $(x, 0, x)$

  4. $(y,0,x)$


Correct Option: B
Explanation:

Given that the point lies in $xy$ plane

In $xy$ plane , the coordinate of $z$ will be zero
So $(x,x,0)$ represents a point which lies in $xy$ plane
Therefore option $B$ is correct

In three dimensions, the coordinate axes of a rectangular cartesian coordinate system are

  1. three mutually parallel lines

  2. three mutually perpendicular lines

  3. two mutually perpendicular lines and any two parallel

  4. None of these


Correct Option: B
Explanation:

In three dimensions, the coordinate axes, i.e. x, y and z axes of a rectangular cartesian coordinate system are three mutually perpendicular lines.

The word rectangular is used to indicate perpendicularity among the axes.

Who gave the systematic development of analytical geometry for the first time?

  1. Leonard Euler

  2. J. Bernoulli

  3. Rene' Descartes

  4. Pierre Fermat


Correct Option: B
Explanation:

J. bernoulli in a letter of $1715 A.D$ to Leibnitz introduced the three coordinate planes which we are using today

He is the fist one to give the systematic development of analytical geometry
Therefore the correct option is $B$

An ordered triplet corresponds to ___________ in three dimensional space.

  1. three points

  2. a unique point

  3. a point in each octant

  4. infinite number of points


Correct Option: B
Explanation:

It is fundamental fact that:
The ordered triplet $(x,y,z)$ represents an unique point in three dimensional space

$(-1,-5,-7)$ lies in Octant

  1. I

  2. VII

  3. V

  4. III


Correct Option: B
Explanation:

Here all the three $x,y,z$ coordinate are negative of the given point.

Therefore, it will lie in the seventh Octant.

The number of dimension, a point has :

  1. 0

  2. 1

  3. 2

  4. 3


Correct Option: A
Explanation:

A point has no length, width and depth. 
Hence, it has no dimensions.

A cube of side 5 has one vertex at the point (1,0,-1), and the three edges from this vertex are, respectively, parallel to the negative x and y axes and positive  z-axis. Find the coordinates of the other vertices of the cube.

  1.  (1,0,1)

  2.  (0,-1,0)

  3.  (0,0,-1)

  4. (1,0,0)


Correct Option: B
Explanation:

Consider the problem 

Below, are four complete cube face on $XZ-plane,\,(y=0)$  

Given point
$(1,0,-1)$

End of the edge parallel to negative $x-axis$ 
$(0,0-1)$

Origin 
$(0,0,0)$

End of the edge parallel to positive $z-axis $
$(1,0,0)$

And, below 
Four point complete the opposite face of cube. 

consider $P$, end of edge parallel to negative  $y-axis $
$(1,-1,-1)$

Edge from $P$ parallel to positive $z-axis $
$(0,-1,0)$

Edge from $P$ parallel to negative $x-axis $
$(0,-1,-1)$

And 
$(0,-1,0)$

The graph of the equation $y^{2}+z^{2}=0$ in three dimensional space is

  1. x- axis

  2. y- axis

  3. z- axis

  4. yz-plane


Correct Option: A
Explanation:

Consider the problem 


${y^2} + {z^2} = 0$

$x=0$ and $z=0$

Therefore, 
The graph of the equation 

${y^2} + {z^2} = 0$ is $x-axis$.

Hence, the correct option is $x-axis$.

The points $(3,\ 2,\ 0),\ (5,\ 3,\ 2)$ and $(-9,\ 6,\ -3)$, are the vertices of a triangle $ABC.AD$ is the internal bisector of $\angle\ BAC$ which meets $BC$ at $D$. Then the co-ordinates of $D$, are

  1. $\left[ {\dfrac{{17}}{{16}},\ \dfrac{{57}}{{16}},\ \dfrac{{19}}{8}} \right]$

  2. $\left[ {\dfrac{{19}}{{8}},\ \dfrac{{57}}{{16}},\ \dfrac{{17}}{16}} \right]$

  3. $\left[0,\ 0,\ {\dfrac{{17}}{{16}}}\right]$

  4. $\left[{\dfrac{{17}}{{16}}},\ 0,\ 0\right]$


Correct Option: B
Explanation:
$AD$ is bisector of $ \angle BAC $ 

$ \Rightarrow $ Ratio at $D$ is $ c:b$ where 

$ c = AB = \sqrt{(3-5)^{2}+(2-3)+(0-2)^{2}} $

$ = \sqrt{4+1+4} = \sqrt{9} = 3 $

$ b = AC = \sqrt{(3+9)^{2}+(2-6)^{2}+(0+3)^{2}} $

$ = \sqrt{144+16+9} = \sqrt{162} = 13 $

for point $D$ 

$ x = \dfrac{c(-9)+b(6)}{c+b} = \dfrac{3(-9)+13(5)}{3+13} = \dfrac{38}{16} = \dfrac{19}{8} $

$ y = \dfrac{c(6)+b(3)}{c+b} = \dfrac{3(6)+13(3)}{3+13} = \dfrac{57}{16} $

$ z = \dfrac{c(-3)+b(2)}{c+b} = \dfrac{3(-3)+13(2)}{3+13} = \dfrac{17}{16} $

Hence, point $D$ is $ [\dfrac{19}{8},\dfrac{57}{16},\dfrac{17}{6}] $ 

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

  1. Parallelogram

  2. rectangle

  3. rhombus

  4. square


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 =

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

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

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

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


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 _______.

  1. $( - 1,0 )$

  2. $( - 1,1 )$

  3. $( 0,-1 )$

  4. $( 7,8 )$


Correct Option: D

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

  1. $(3, 1, 8)$

  2. $(1, 2, 8)$

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

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


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.

  1. True

  2. False

  3. Ambiguous

  4. Data Insufficient


Correct Option: A
Explanation:

 In Geometrywe 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:

  1. $A,B,C,D$

  2. $B,C,D,A$

  3. $C,D,A,B$

  4. $D,C,B,A$


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

  1. Circle

  2. Cylinder

  3. Hemisphere

  4. Sphere


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

  1. z axis

  2. x axis

  3. xy plane

  4. yz plane

  5. xz plane


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

  1. $(x,y,y)$

  2. $(0,y,y)$

  3. $(0,y,x)$

  4. $(x,y,z)$


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

  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


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 

The equation of plane passing through $(-1,0,-1)$ parallel to $xz$ plane is

  1. $y=-2$

  2. $y=0$

  3. $-x-z=0$

  4. None of the above


Correct Option: B
Explanation:

Given that the plane is parallel to $xz$ plane and the plane passes through $(-1,0,-1)$

Since the plane is parallel to $xz$ plane , the $y$ coordinate should be constant
Given that it passes through point $(-1,0,-1)$ , therefore the plane lies on $xz$ plane
Therefore the equation of plane is $y=0$
The correct options are $B$

The planes $2x-y+4z=5$ and $5x-2.5y+10z=6$ are

  1. Parallel

  2. Perpendicular

  3. Intersect

  4. intersect $x$ axis


Correct Option: A
Explanation:

Planes are $2x-y+4z=5$ 

and $5x-2.5y+10z=6$
Multiply both sides by 2 to the second equation
$\Rightarrow 10x-5y+20=12$
Now divide both sides by $2$
$\Rightarrow 2x-y+4z=\dfrac{12}{5}$

Clearly both planes are parallel 

In a three-dimensional space, the equation $3x - 4y = 0$ represents.

  1. A plane containing $Z-axis$

  2. A plane containing $X-axis$

  3. A plane containing $Y-axis$

  4. Passing through $(0, 0)$


Correct Option: A
Explanation:

If we consider the z-part also, then the equation is $3x-4y+0z=0$

Which means on putting any value of z equation will have no effect as $0\times z=0$
$\therefore$ it will be a plane containing $Z-axis$.($\because$ it will pass through all points z if it satisfy condition for$ (x,y) $)
(D) is not right because its plane and not a line so it will pass through $(0,0,0)$ not $(0,0)$.
Hence, $(A)$


The point $(3, 0, -4)$ lies on the

  1. Y-axis

  2. Z-axis

  3. XY-plane

  4. XZ-plane

  5. YZ-plane


Correct Option: D
Explanation:
$(3, 0, -4)$   $\rightarrow$   Given point
Clearly, $y = 0$ and $ x$ and $z$ have non-zero value.
If the point lies on $x-z$ plane, this condition is possible.
Hence, the answer is $XZ$- plane.

Which of the following is true for a plane?

  1. A locus is called a plane if the line joining any two arbitrary points on the locus is also a part of the locus.

  2. Value of $y$ in a $zx$ plane is non-zero.

  3. Value of $z$ in a $xy$ plane is zero.

  4. None of the above


Correct Option: A,C
Explanation:

Option A and C are correct 

A locus is called a plane if the line joining any two arbitrary points on the locus is also a part of the locus. and also Value of z in a xy plane is zero.

There are three points with position vectors $ -2a+3b+5c, a+2b+3c $ and$ 7a-c$. What is the relation between the three points?

  1. Collinear

  2. Forms a triangle

  3. In different plane

  4. None of the above


Correct Option: A
Explanation:

The relation between the three points are collinear

Thus option A is correct answer 

The coordinates of the point where the line through $(3, -4, -5)$ and $(2, -3, 1)$ crosses the plane passing through three points $(2, 2, 1),(3, 0, 1)$ and $(4, -1, 0)$ is

  1. $(1, 2, 7)$

  2. $(-1, 2, -7)$

  3. $(1, -2, 7)$

  4. None of these


Correct Option: A
- Hide questions