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

The shortest distance between line $y-x=1$ and curve $x=y^{2}$ is :-

  1. $\dfrac{8}{3\sqrt{2}}$
  2. $\dfrac{4}{\sqrt{3}}$
  3. $\dfrac{\sqrt{3}}{4}$
  4. $\dfrac{3\sqrt{2}}{8}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The shortest distance between a curve and a line occurs where the tangent to the curve is parallel to the given line. By setting the derivative of the parabola x = y^2 equal to the slope of the line y - x = 1, we find the closest point and compute the perpendicular distance.

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 the point $\left( 1,-2,3 \right) $ from the plane $x-y+z=5$ measured parallel to the line $\displaystyle \frac { x }{ 2 } =\frac { y }{ 3 } =\frac { z-1 }{ -6 } $ is

  1. $1$
  2. $2$
  3. $4$

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

Equation of the line through $\left( 1,-2,3 \right) $ parallel to the line $\displaystyle \dfrac { x }{ 2 } =\dfrac { y }{ 3 } =\dfrac { z-1 }{ -6 } $ is


$\displaystyle \dfrac { x-1 }{ 2 } =\dfrac { y+2 }{ 3 } =\dfrac { z-1 }{ -6 } =r$ (say)   ...$(1)$


Then any point on $(1)$ is $\left( 2r+1,3r-2,-6r+3 \right) $.


If this point lies on the plane $x-y+z=5$, then 


$\displaystyle \left( 2r+1 \right) -\left( 3r-2 \right) +\left( -6r+3 \right) =5\Rightarrow -7r+6=5\Rightarrow r=\dfrac { 1 }{ 7 } $


Hence, the point is $\displaystyle \left( \dfrac { 9 }{ 7 } ,-\dfrac { 11 }{ 7 } ,\dfrac { 15 }{ 7 }  \right) $


Distance between $\left( 1,-2,3 \right) $ and $\displaystyle \left( \dfrac { 9 }{ 7 } ,-\dfrac { 11 }{ 7 } ,\dfrac { 15 }{ 7 }  \right) $


$\displaystyle =\sqrt { \left( \dfrac { 4 }{ 49 } +\dfrac { 9 }{ 49 } +\dfrac { 36 }{ 49 }  \right)  } =\sqrt { \dfrac { 49 }{ 49 }  } =1$

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

Perimeter of triangle whose vertices are $(0,4,0), (3,4,0)$ and $(0,4,4)$, is

  1. $10$
  2. $12$
  3. $25$
  4. $15$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

 Vertices of triangle are $(0,4,0),(3,4,0)$ and $(0,4,4)$

then perimeter=?
here AB=$\sqrt{(3-0)^2+(4-4)^2+(0-0)^2}$
$=3$
BC$=\sqrt{(3-0)^2+(4-4)^2+(0-4)^2}$
$=\sqrt{9+16}$
$=5$
CA$=\sqrt{(0-0)^2+(4-4)^2+(0-4)^2}=4$
used distance formula b/w two points
$(x _1,y _1,z _1) and (x _2,y _2,z _2)$
$=\sqrt{(x _2-x _2)^2+(y _2-y _1)^2+(z _2-z _1)^2}$
$ perimeter =ABC+BC+CA$
$=3+5+4$
$=12\ units$

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

Let the distance between vectors are given as follows :
$(i)4i +3j-6k, -2i+j-k$  be $\displaystyle \sqrt{k}$
$(ii) -2i+3j+5k, 7i-k $  be  $\displaystyle m\sqrt{n}$
Find $k-(m*n)$ ?

  1. 20

  2. 21

  3. 22

  4. 23
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

(i) Distance between $4i + 3j - 6k$ and $-2i + j - k$ is given by $\sqrt{(4 + 2)^2 + (3 - 1)^2 + (-6 + 1)^2} = \sqrt{36 + 4 + 25} = \sqrt{65}$

$\Rightarrow k = 65$

(ii) Distance between $-2i + 3j + 5k$ and $7i - k$ would be $\sqrt{(-2 - 7)^2 + (3)^2 + (5 + 1)^2} = \sqrt{81 + 9 + 36} = \sqrt{126} = 3\sqrt{14}$
$\Rightarrow m = 3, n = 14$

$\therefore k - (m*n) = 65 - (3 \times 14) = 65 - 42 = 23$

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

Find the distance between the pairs of points whose cartesian coordinates are $(2,3,-1), (2,6,2).$

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

Distance Between two points $(a,b,c)$ and $(x,y,z)$ is given by
$\sqrt { \left( a-x \right) ^{ 2 }+\left( b-y \right) ^{ 2 }+\left( c-z \right) ^{ 2 } } $
Given $(2,3,-1)$, $(2,6,2)$
Distance $=\sqrt { \left( 2-2 \right) ^{ 2 }+\left( 3-6 \right) ^{ 2 }+\left( -1-2 \right) ^{ 2 } } $
                $=\sqrt { 18 } $
                $=3\sqrt { 2 }$

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

Find the distance between the points whose position vectors are given as follows

$(-1,1,3), (0,5,6)$

  1. $\displaystyle \sqrt{118}$
  2. $8$
  3. $\displaystyle \sqrt{26}$

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

Distance Between two points $(a,b,c)$ and $(x,y,z)$ is given by
$\sqrt { \left( a-x \right) ^{ 2 }+\left( b-y \right) ^{ 2 }+\left( c-z \right) ^{ 2 } } $
Given $(-1,1,3)$, $(0,5,6)$
Distance $=\sqrt { \left( -1-0 \right) ^{ 2 }+\left( 1-6 \right) ^{ 2 }+\left( 3-6 \right) ^{ 2 } } $
                $=\sqrt { 26 } $

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

Find the distance between the points whose position vectors are given as follows

$-2\hat i+3\hat j+5\hat k, 7\hat i-\hat k$

  1. $\displaystyle 3\sqrt{14}$
  2. $\displaystyle \sqrt{54}$
  3. $\displaystyle 3\sqrt{19}$
  4. $\displaystyle \sqrt{57}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Distance Between two position vectors $a\hat i+b\hat j+c\hat k$ and $x\hat i+y\hat j+z\hat k$ is given by
$\sqrt { \left( a-x \right) ^{ 2 }+\left( b-y \right) ^{ 2 }+\left( c-z \right) ^{ 2 } } $
Given $-2\hat i+3\hat j+5\hat k$, $7\hat i-\hat k$
Distance $=\sqrt { \left( -2-7 \right) ^{ 2 }+\left( 3-0 \right) ^{ 2 }+\left( 5+1 \right) ^{ 2 } } $
                $=\sqrt { 126 } $
                $=3\sqrt { 14 } $

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

Find the distance between the points whose position vectors are given as follows

$4\hat i+3\hat j-6\hat k, -2\hat i+\hat j-\hat k$

  1. $\displaystyle \sqrt{65}$
  2. $\displaystyle \sqrt{69}$
  3. $13$

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

Distance Between two position vectors $a\hat i+b\hat j+c\hat k$ and $x\hat i+y\hat j+z\hat k$ is given by
$\sqrt { \left( a-x \right) ^{ 2 }+\left( b-y \right) ^{ 2 }+\left( c-z \right) ^{ 2 } } $
Given $4\hat i+3\hat j-6\hat k$, $-2\hat i+\hat j-\hat k$
Distance $=\sqrt { \left( 4+2 \right) ^{ 2 }+\left( 3-1 \right) ^{ 2 }+\left( -6+1 \right) ^{ 2 } } $
                $=\sqrt { 65 } $

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

Calculate the distance between the points $(-3,6,7)$ and $(2,-1,4)$ in $3D$ space.

  1. $4.36$
  2. $5.92$
  3. $7.91$
  4. $9.11$
  5. $22.25$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know the distance formula:

$d=\sqrt{(x _2-x _1)^2+(y _2-y _1)^2+(z _2-z _1)^2}$

The coordinates are $(-3, 6, 7)$ and$ (2, -1, 4)$

$d=\sqrt{(2-(-3))^2+((-1)-6)^2+(4-7)^2}$

$d=\sqrt{(5)^2+((-7)^2+(-3)^2}$

$d=\sqrt{25+49+9}$

$d=\sqrt{83}$

$d = 9.11$

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 between z-axis and the line 
$x+y+2z-3=0=2x+3y+4z-4$, is _____________

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

$x+y+2z-3=0=2x+3y+4z-4$ at $z-axis, x=y=0$

$x+y=3-2z$ and $2x+3y=4(1-z)$
solving $x$ and $y$ in function $z$,
$2x+3(3-2z-x)=4(1-z)$
$\implies 2x+9-6z-3x=4-4z$
$\implies x=9-6z-4+4z$
$\implies x=5-2z\quad equation (2)$
$\implies (5-2z)+y=3-2z$
$\implies y=3-2z-5+2z$
$y=-2\quad equation (2)$
So, $\sqrt {x^2+y^2}=\sqrt {(-2)^2+(5-2z)^2}$
at, $z=\cfrac {5}{2}$, this would be minimum.

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

The distances of the point $P(1,2,3)$ from the coordinates axes are:

  1. $\sqrt {13} ,\sqrt {10} ,\sqrt 5 $
  2. $\sqrt {11} ,\sqrt {10} ,\sqrt 5 $
  3. $\sqrt {13} ,\sqrt {20} ,\sqrt {15} $
  4. $\sqrt {23} ,\sqrt {10} ,\sqrt 5 $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The point: $(1, 2, 3)$
$\therefore$ Co-ordinates axes are $(1, 0, 0), (0, 2, 0), (0, 0, 3)$
Distance of $(1, 2, 3)$ from $(1, 0, 0)$
$= \sqrt{(1 - 1)^2 + (0 - 2)^2 + (0 - 3)^2)}$
$= \sqrt{4 + 9} = \sqrt{13}$
Distance of $(1, 2, 3)$ from $(0, 2, 0)$
$= \sqrt{(0 - 1)^2 + (2 - 2)^2 + (0 - 3)^2)}$
$= \sqrt{1 + 9} = \sqrt{10}$
Distance of $(1, 2, 3)$ from $(0, 0, 3)$
$= \sqrt{(0 - 1)^2 + (0 - 2)^2 + (3 - 3)^2)}$
$= \sqrt{5}$
Hence, the distances of the point $P(1,2,3)$ from the coordinates axes$= \sqrt{13}, \sqrt{10}, \sqrt{5}$
Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

What is the distance in space between $(1,0,5)$ and $(-3,6,3)$?

  1. $4$
  2. $6$
  3. $2\sqrt { 11 } $
  4. $2\sqrt { 14 } $
  5. $12$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The distance between the points $(1,0,5)$ and $(-3,6,3)$ is given below:

$D=\sqrt { \left( 3-5 \right) ^{ 2 }+\left( 6-0 \right) ^{ 2 }+\left( -3-1 \right) ^{ 2 } } $
$=\sqrt { \left( -2 \right) ^{ 2 }+\left( 6 \right) ^{ 2 }+\left( -4 \right) ^{ 2 } } $
$=\sqrt { 4+36+16 } =\sqrt { 56 } $
$=2\sqrt { 14 }$ 

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

$L _1:\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}$
$L _2:\dfrac{x-2}{3}=\dfrac{y-4}{2}=\dfrac{z-5}{5}$ be two given lines, point P lies on $L _1$ and Q lies on $L _2$ then distance between P and Q can be

  1. $\dfrac{1}{3}$
  2. $\dfrac{1}{9}$
  3. $15$
  4. $30$
Reveal answer Fill a bubble to check yourself
A,C,D Correct answer
Explanation

$\dfrac{|\begin{vmatrix}2-1&4-2  &5-3 \2 & 3 &4 \ 3& 2 & 5\end{vmatrix}|}{|\begin{vmatrix}i&j  &k \2 & 3 & 4\3 & 2 & 5\end{vmatrix}|}=\dfrac{1}{\sqrt{78}}$

Multiple choice mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

For hyperbola  $\dfrac{x^2}{16}-\dfrac{y^2}{25}=1$ distances between two directrices are

  1. $\dfrac{16}{\sqrt{41}}$
  2. $\dfrac{25}{\sqrt{41}}$
  3. $\pm\dfrac{32}{\sqrt{141}}$
  4. $\dfrac{32}{\sqrt{41}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given equation of hyperbola $\dfrac {x^2}{16}-\dfrac {y^2}{25}=1$
Here $a^2=16, b^2=25$
Distance between directrix is  $\dfrac{2a^2}{\sqrt{a^2+b^2}} = \dfrac{2\times16}{\sqrt{16+25}} = \dfrac{32}{\sqrt{41}}$

Multiple choice mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

For hyperbola  $-\dfrac{x^2}{16}+\dfrac{y^2}{25}=1$ distance between directrices is 

  1. $\dfrac{50}{\sqrt{41}}$
  2. $\dfrac{16}{\sqrt{41}}$
  3. $\dfrac{25}{\sqrt{41}}$
  4. $\dfrac{32}{\sqrt{41}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given equation is $-\dfrac {x^2}{16}+\dfrac {y^2}{25}=1$
$\Rightarrow \dfrac {y^2}{25}-\dfrac {x^2}{16}=1$
Distance between directrix is   $\dfrac{2a}{\sqrt{a^2+b^2}}$
Here $a=25, b=16$
So, answer is $= \dfrac{50}{\sqrt{41}}$.