Tag: distance between two points in 3d

Questions Related to distance between two points in 3d

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

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

  1. $\sqrt{17}$

  2. $\sqrt{29}$

  3. $\sqrt{34}$

  4. $\sqrt{41}$

Show answer
Correct Option: C
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 } $

The equation of the set of points which are equidistant from the points $(1, 2, 3)$ and $(3, 2, -1)$.

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

  1. $x-2z=0$

  2. $2x-z=0$

  3. $2x+y=0$

  4. $x-2y=0$

Show answer
Correct Option: A
Explanation:

Let the given points be A($1,2,3$) and B($3,2,-1$) and let the point equidistant from A and B be P($x,y,z$) 

then  $PA=PB$
$\sqrt {{{(x - 1)}^2} + {{(y - 2)}^2} + {{(z - 3)}^2}}  = \sqrt {{{(x - 3)}^2} + {{(y - 2)}^2} + {{(z + 1)}^2}} $
Squaring both sides
${(x - 1)^2} + {(y - 2)^2} + {(z - 3)^2} = {(x - 3)^2} + {(y - 2)^2} + {(z + 1)^2}$
${x^2} + 1 - 2x + {z^2} + 9 - 6z = {x^2} + 9 - 6x + {z^2} + 1 + 2z$
$-2x-6z+10=-6x+2z+10$
$4x-8z=0$
$x-2z=0$

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

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

  1. $(3, 4, 12)$

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

  3. $(4, 5, 13)$

  4. $(40, 53, 157)$

Show answer
Correct Option: C
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 $C$.

The x-coordinate of a point on the line joining the points $P(2,2,1)$ and $Q(5,1,-2)$ is $4$. Find its z-coordinate.

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

  1. $-1$

  2. $-2$

  3. $1$

  4. $2$

Show answer
Correct Option: A
Explanation:

$P(2,2,1),Q(5,1,-2)$


$\therefore$ Equation of line through $P$ & $Q$,


$\cfrac { x-2 }{ 2-5 } =\cfrac { y-2 }{ 2-1 } =\cfrac { z-1 }{ 1+2 } \\ \Rightarrow \cfrac { x-2 }{ -3 } =\cfrac { y-2 }{ 1 } =\cfrac { z-1 }{ 3 } =r$

$\therefore P$ be point of line 

$\Rightarrow P\equiv (-3r+2,r+2,3r+1)$

$ \therefore -3r+2=4$ ($\because $ since x-coordinate is $4$)

$\Rightarrow r=\cfrac { -2 }{ 3 } $

$\therefore$ z-coordinate $=3r+1=-1$

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

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

  1. 4

  2. 2

  3. 6

  4. 8

Show answer
Correct Option: C
Explanation:

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

The distance between the parallel planes given by the equations, $\vec{r}.(2\hat{i}-2\hat{j}+\hat{k})+3=0$ and $\vec{r}.(4\hat{i}-4\hat{j}+2\hat{k})+5=0$ is-

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

  1. $1/2$

  2. $1/3$

  3. $1/4$

  4. $1/6$

Show answer
Correct Option: D
Explanation:

Planes are  $2i-2j+k+3=0,4i-4j+2k+5=0,2i-2j+k+5/2=0$

distance between them is $=\cfrac{|c _1-c _2|}{\sqrt{a^2+b^2+c^2}}\=\cfrac{|3-\cfrac{5}{2}|}{\sqrt{2^2+2^2+1^2}}=\cfrac{1}{6}$

Distance between $A(4,5,6)$ from origin $O$ is

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

  1. $25\sqrt3$

  2. $\sqrt {77}$

  3. $3\sqrt5$

  4. Data Insufficient

Show answer
Correct Option: B
Explanation:

Origin is $O(0,0,0)$ and given point is $A(4,5,6)$

So, distance $=$ $\sqrt {(4-0)^2+(5-0)^2+(6-0)^2}$
$=\sqrt {4^2 + 5^2 + 6^2} = \sqrt {77}$

If the extremities of a diagonal of a square are $(1, -2, 3)$ and $(2, -3, 5)$, then area of the square is

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

  1. $6$

  2. $3$

  3. $\displaystyle \dfrac{3}{2}$

  4. $\sqrt{3}$

Show answer
Correct Option: B
Explanation:

Let the extremities of the diagonal of a square be $(1,-2,3)$ and $B(2,-3,5)$.
Then $AB$ is given by $ {({1}^{2} + {1}^{2} + {2}^{2})}^{0.5} $ = $ \sqrt{6} $
Hence, length of the side $ = \sqrt {3} $
So, area of square will be $ \sqrt{3} \times \sqrt{3}  = 3$

The point equidistant from the points $(0,0,0), (1,0,0), (0,2,0)$ and $(0,0,3)$ is

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

  1. $(1,2,3)$

  2. $\left (\displaystyle \dfrac{1}{2},1,\dfrac{3}{2}\right)$

  3. $\left (-\displaystyle \dfrac{1}{2}, -1,-\displaystyle \dfrac{3}{2}\right)$

  4. $(1,-2,3)$

Show answer
Correct Option: B
Explanation:

Equation of sphere passes through origin is given by


$x^2+y^2+z^2+ax+by+cz=0 ...(1)$

Given, it also passes through $(1,0,0),(0,2,0),0,0,3)$

$1+a=0\Rightarrow a = -1$

$4+2b=0\Rightarrow b = -2$

and $ 9+3c=0\Rightarrow c = -3$

$\therefore$the equation of the sphere becomes $x^2+y^2+z^2-x-2y-3z=0$

Comparing with $x^2+y^2+z^2+2gx+2fy+2kz+C=0$

Therefore, the point equidistant from all the given four point will be the centre of the sphere $(1)$ passing through all these points which is $\left(\dfrac{1}{2},1, \dfrac{3}{2}\right)i.e. the \  centre$.

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

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

  1. $\sqrt{21}$

  2. $\sqrt{5}$

  3. $\sqrt{17}$

  4. none of these

Show answer
Correct Option: A
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.