Mathematics

3D Geometry and Coordinate Distance

137 Questions

Three dimensional geometry and coordinate distance problems involve calculating spatial measurements between points and lines. Questions feature vector distances, perpendicular distances, and coordinate geometry theorems. This advanced topic is typically found in mathematics examinations.

Point distance calculationsPerpendicular distancesVector coordinatesLine ratiosAxis distances

3D Geometry and Coordinate Distance Questions

Multiple choice using trigonometric tables trigonometric ratios of some specific angles trigonometric identities trigonometry maths

The distance between $A ( \cos \theta , \sin \theta )$ and $B ( - \sin \theta , \cos \theta )$ is

  1. 1
  2. $2 + 2 \sin \theta$
  3. $1 + \sin \theta$
  4. $\sqrt { 2 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given $A(\cos \theta,\sin \theta)$ and $B(-\sin \theta,\cos \theta)$

Distance between $AB=\sqrt{(\cos \theta+\sin \theta)^2+(\sin \theta-\cos \theta)^2}=\sqrt{\cos^2 \theta+\sin ^2 \theta+2\sin \theta\cos \theta+\cos^2\theta+\sin ^2 \theta-2\sin \theta\cos \theta}=\sqrt{1+1}=\sqrt{2}$

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

For $a> b> c> 0$, the distance between $(1,1)$ and the point of intersection of the lines $ax+by+c=0$ and $bx+ay+c=0$ is less then $2\sqrt{2}$. Then

  1. $a+b-c> 0$
  2. $a-b+c< 0$
  3. $a-b+c> 0$
  4. $a+b-c< 0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The intersection of ax+by+c=0 and bx+ay+c=0 is (-c/(a+b), -c/(a+b)). The distance from (1,1) to this point is sqrt((1+c/(a+b))^2 + (1+c/(a+b))^2) = sqrt(2)|1+c/(a+b)|. Given this is < 2sqrt(2), we find |1+c/(a+b)| < 2, leading to a+b-c > 0.

Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

lf the distance between two given points is $2$ units and the points are transferred by shifting the origin to $(2, 2)$, then the distance between the points in their new position is.

  1. $2$
  2. $5$
  3. $6$
  4. $7$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Shifting the origin to $(2 , 2)$ transforms the coordinates to
$(x , y)$ to $(x - 2, y - 2)$.

$\therefore$ distance between two points is

$\sqrt{(x _{1}-x _{2})^{2}+(y _{1}-y _{2})^{2}}$

$=\sqrt{((x _{1}-2)-(x _{2}-2))^{2}+((y _{1}-2)-(y _{2}-2))}^{2}$

$=\sqrt{(x _{1}-x _{2})^{2}+(y _{1}-y _{2})}^{2}$

$\therefore$ distance is unaltered $ = 2$ units.

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

The distance of origin from the image of (1, 2, 3) in plane x - y + z = 5 is 

  1. $\sqrt{17}$
  2. $\sqrt{29}$
  3. $\sqrt{34}$
  4. $\sqrt{41}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$P(1,2,3),$ Plane :$x-y+z=5$

$F$ is foot of perpendicular form $P$ to plane and $I$ is image,then $PF=FI$
$\therefore$ If $(x,y,z)=(r+1,-r+2,r+3)$ are foot of perpendicular.
$ \Rightarrow (r+1)-(-r+2)+r+3=5\quad \quad \Rightarrow r=1\ \therefore F=(2,1,4)\ \therefore I=(3,0,5)$
$ \therefore$ distance of $I$ from origin  $=\sqrt { { 3 }^{ 2 }+{ 0 }^{ 2 }+{ 5 }^{ 2 } } =\sqrt { 34 } $

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

If the distance between a point $P$ and the point $(1, 1, 1)$ on the line $\dfrac {x - 1}{3} = \dfrac {y - 1}{4} = \dfrac {z - 1}{12}$ is $13$, then the coordinates of $P$ are

  1. $(3, 4, 12)$
  2. $\left (\dfrac {3}{13}, \dfrac {4}{13}, \dfrac {12}{13}\right )$
  3. $(4, 5, 13)$
  4. $(40, 53, 157)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the given points be $A(1,1,1)$

Consider,
$\dfrac{{x - 1}}{3} = \dfrac{{y - 1}}{4} = \dfrac{{z - 1}}{{12}} = \lambda $

$\begin{array}{l} x=3\lambda +1 \  \ y=4\lambda + 1\  \ z=12\lambda +1 \end{array}$

General point on the line is 
$3\lambda  + 1,\,4\lambda  + 1,\,12\lambda  + 1$

Given that,
$AP=13$ 

$\sqrt {{{\left( {3\lambda  + 1 - 1} \right)}^2} + {{\left( {\,4\lambda  + 1 - 1} \right)}^2} + {{\left( {12\lambda  + 1 - 1} \right)}^2}}  = 13$

$13\lambda =13$

$\lambda =1$

$\begin{array}{l} 3\lambda +1=4 \  \ 4\lambda +1=5 \  \ 12\lambda +1=13 \end{array}$

Therefore, required point $P$ is $(4,5,13)$.
Hence the correct option is $C$.

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

The distance between (5,1,3) and the line x=3, y=7+t, z=1+t is

  1. 4

  2. 2

  3. 6

  4. 8
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\begin{array}{l} x=3\, \, \, ,y=7+t\, \, ,z=1+t \ A\left( { 3,7+t,1+t } \right)  \ 0.\left( { 3-5 } \right) +1\left( { 7+t-1 } \right) +1\left( { 1+t-3 } \right) =0 \ 6+t+t-2=0 \ 2t=-4 \ t=-2 \ A:\left( { 3,5,-1 } \right)  \ dis\tan  ce=\sqrt { 4+16+16 }  \ =6 \ Option\, \, C\, \, is\, \, correct\, \, answer. \end{array}$

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

Distance between the points $(12,4,7)$ and $(10,5,3)$ is

  1. $\sqrt{21}$
  2. $\sqrt{5}$
  3. $\sqrt{17}$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Consider the problem,

Let the given points 
$A(12,4,7)$ and $B(10,5,3)$
So, distance between $A$ and $B$ by distance formula.
$AB=\sqrt{(10-12)^2+(5-4)^2+(3-7)^2}=\sqrt{(-2)^2+1^2+(-4)^2}$ 
$=\sqrt{4+1+16}=\sqrt{21}$
So, distance between the points $(12,4,7)$ and $(10,5,3)$ is $\sqrt{21}$ sq. units.

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

Find the distance between $(12,3,4)$ and $(4,5,2)$

  1. $\sqrt {72}$
  2. $\sqrt {62}$
  3. $\sqrt {64}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Consider the problem,

Let the given points 
$A(12,3,4)$ and $B(4,5,2)$
So, distance between $A$ and $B$ by distance formula.
$AB=\sqrt{(4-12)^2+(5-3)^2+(2-4)^2}=\sqrt{(-8)^2+2^2+(-2)^2}$ 
$=\sqrt{64+4+4}=\sqrt{72}$
So, distance between the points $(12,3,4)$ and $(4,5,2)$ is $\sqrt{72}$ sq. units.

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

Find the co-ordinates of a point lying on the line $\dfrac{x -2}{3} = \dfrac{y + 3}{4} = \dfrac{z - 1}{7}$ which is at a distance $10$ units from $(2, -3, 1)$.

  1. $(32,37,71)$
  2. $(-28,-43,-69)$
  3. $(-32,-37,-71)$

  4. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given that the required point lies on the line $\dfrac{x-2}{3}=\dfrac{y+3}{4}=\dfrac{z-1}{7}$.

The required point is at a distance of $10$ units from $(2,-3,1)$

By option verification,

(A) Substituting the point $(32,37,71)$ 

$\implies \dfrac{32-2}{3}=\dfrac{37+3}{4}=\dfrac{71-1}{7}$

$\implies 10=10=10$

Therefore, distance is $\sqrt{(32-2)^2+(37+3)^2+(71-1)^2} =86.02$ units

Hence, $(32,37,71)$ is not the required point.

(B) Substituting the point $(-28,-43,-69)$ 

$\implies \dfrac{-28-2}{3}=\dfrac{-43+3}{4}=\dfrac{-69-1}{7}$

$\implies -10=-10=-10$

Therefore, distance is $\sqrt{(-28-2)^2+(-43+3)^2+(-69-1)^2} =86.02$ units

Hence, $(-28,-43,-69)$ is not the required point.


(C) Substituting the point $(-32,-37,-71)$ 

$\implies \dfrac{-32-2}{3}=\dfrac{-37+3}{4}=\dfrac{-71-1}{7}$

$\implies 11.3=11.3=10.28$

Therefore, distance is $\sqrt{(-32-2)^2+(-37+3)^2+(-71-1)^2} =86.57$ units

Hence, $(-32,-37,-71)$ is not the required point.

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

The distance between the points $(\cos \, \theta , \, \sin \, \theta) $ and $ (\sin \, \theta - \cos \, \theta)$ is 

  1. $\sqrt{3}$
  2. $\sqrt{2}$
  3. $2$
  4. $1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Distance between the point $(\cos\theta, \sin\theta)$ and $(\sin\theta, -\cos\theta)$ is :
$ = \sqrt{(\sin\theta - \cos\theta)^2 + (-\cos\theta - \sin\theta)^2}$
$ = \sqrt{1 - 2 \sin\theta \cos\theta + 1 + 2 \sin\theta \cos\theta}$
$(\cos^2\theta + \sin^2\theta = 1)$
$= \sqrt{2}$
Option B is the correct answer
Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

If the distance between a point P and the point (1, 1, 1) on the line $\frac{{x\, - \,1}}{3}\, = \,\frac{{y - \,1}}{4}\, = \,\frac{{z\, - 1}}{{12}}$ is 13, then the coordinates of P are

  1. (3, 4, 12)
  2. $\left( {\frac{3}{{13}},\,\frac{4}{{13}},\,\frac{{12}}{{13}}} \right)$

  3. (4, 5, 12)

  4. (40, 53, 157)
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\begin{array}{l} According\, to\, question: \ we\, have\, the\, line\, equ=\frac { { x-1 } }{ 3 } =\frac { { y-1 } }{ 4 } =\frac { { z-1 } }{ { 12 } } \, and\, point\, (P)\, is\, 13. \ Now, \ line\, of\, equ: \ \Rightarrow \frac { { x-1 } }{ 3 } =\frac { { y-1 } }{ 4 } =\frac { { z-1 } }{ { 12 } } =13 \ Now\, find\, x,y\, &amp; \, z\, \, coordinates: \ \, \Rightarrow \frac { { x-1 } }{ 3 } =13 \ \, \, \, \, \, \, \, \therefore \, \, \, \, x=39+1=40 \ \Rightarrow \frac { { y-1 } }{ 4 } =13 \ \, \, \, \, \, \, \, \therefore \, \, y=\, 52+1=53 \ \Rightarrow \frac { { z-1 } }{ { 12 } } =13 \ \, \, \, \, \, \, \, \, \therefore \, \, z=156+1=157 \ Now,\, we\, get\, \, new\, coordinates\, of\, P:\, \, \, \left( { 40,53,157 } \right) \, \,  \ so\, that\, \, the\, correct\, option\, is\, D. \end{array}$

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

The area of triangle whose vertices are $(1, 2, 3), (2, 5, -1)$ and $(-1, 1, 2)$ is

  1. $150\ sq. units$
  2. $145\ sq. units$
  3. $\sqrt {155}/2\ sq. units$
  4. $155/2\ sq. units$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the vertices of triangle are 


$A(1,2,3),B(2,5,-1)$ and $C(-1,1,2)$ 

Then, 

$\begin{array}{l} \overrightarrow { AB } =\overrightarrow { OB } -\overrightarrow { OA } =\hat { i } +3\hat { j } -4\hat { k }  \  \ \overrightarrow { AC } =\overrightarrow { OC } -\overrightarrow { OA } =-2\hat { i } -\hat { j } -\hat { k }  \end{array}$

Then, 

$\begin{array}{l} \overrightarrow { AB } \times \overrightarrow { AC } =\left| { \begin{array} { *{ 20 }{ c } }{ \hat { i }  } & { \hat { j }  } & { \hat { k }  } \ 1 & 3 & { -4 } \ { -2 } & { -1 } & { -1 } \end{array} } \right|  \  \ =-7\hat { i } +9\hat { j } +5\hat { k }  \  \ \left| { \overrightarrow { AB } \times \overrightarrow { AC }  } \right| =\sqrt { { { \left( { -7 } \right)  }^{ 2 } }+{ { \left( 9 \right)  }^{ 2 } }+{ { \left( 5 \right)  }^{ 2 } } }  \  \ =\sqrt { 49+8+25 }  \  \ =\sqrt { 155 }  \end{array}$

Area of triangle $ABC=\frac{1}{2}|\vec {AB} \times \vec {AC}|$

$=\frac{1}{2} \times \sqrt {155}$

$=\frac{\sqrt {155}}{2}$ sq. units 

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

The shortest distance of the point $(1,2,3)$ from ${x}^{2}+{y}^{2}=0$ is 

  1. $5$
  2. $\sqrt{5}$
  3. $2$
  4. $\sqrt{14}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

$\therefore$ $(x,y)\equiv (0,0)$
Let  point be $\equiv (0,0,z)$
$d\equiv \sqrt { { (1-0) }^{ 2 }+{ (2-0) }^{ 2 }+{ (z-3) }^{ 2 } } =\sqrt { 1+4+{ (z-3) }^{ 2 } } \Rightarrow { (z-3) }^{ 2 }\ge 0$
${ d } _{ min }=\sqrt { 5 } $

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

The distance between two points $(1,1)$ and $\left( {\dfrac{{2{t^2}}}{{1 + {t^2}}},\dfrac{{{{\left( {1 - t} \right)}^2}}}{{1 + {t^2}}}} \right)$ is 

  1. 4t

  2. 3t

  3. 1

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Two point are $A(1,1)$ and $B\left(\dfrac{2t^{2}}{1+t^{2}}, \dfrac{(1+t)^{2}}{1+t^{1}}\right)$

$AB=\sqrt{\left(1-\dfrac{2t^{2}}{1+t^{2}}\right)^{2}+\left(1-\dfrac{(1-t^{2})}{1+t^{2}}\right)^{2}}$

$=\sqrt{\left(\dfrac{1+t^{2}-2t^{2}}{1+t^{2}}\right)^{2}+\left(\dfrac{1+t^{2}-1-t^{2}+2t}{1+t^{2}}\right)^{2}}$

$=\sqrt{\left(\dfrac{1-t^{2}}{1+t^{2}}\right)^{2}+\left(\dfrac{RT}{1+t^{2}}\right)^{2}}$

$=\sqrt{\dfrac{1+t^{4}-2t^{2}+4t^{2}}{(1+t^{2})^{2}}}$

$=\sqrt{\dfrac{(1+t^{2})^{2}}{(1+t^{2})^{2}}}$

$=1$