Tag: coordinates in 3d

Questions Related to coordinates in 3d

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

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

  1. $(x,0,y)$

  2. $(x,x,0)$

  3. $(x, 0, x)$

  4. $(y,0,x)$

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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

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

  1. three mutually parallel lines

  2. three mutually perpendicular lines

  3. two mutually perpendicular lines and any two parallel

  4. None of these

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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?

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

  1. Leonard Euler

  2. J. Bernoulli

  3. Rene' Descartes

  4. Pierre Fermat

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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.

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

  1. three points

  2. a unique point

  3. a point in each octant

  4. infinite number of points

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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

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

  1. I

  2. VII

  3. V

  4. III

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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 :

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

  1. 0

  2. 1

  3. 2

  4. 3

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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.

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

  1.  (1,0,1), 

  2.  (0,-1,0), 

  3.  (0,0,-1), 

  4. (1,0,0)

Show answer
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

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

  1. x- axis

  2. y- axis

  3. z- axis

  4. yz-plane

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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

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

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

Show answer
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

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.