Mathematics

3D Geometry and Coordinate Distance

140 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 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 $B\left(\dfrac{2t^{2}}{1+t^{2}}, \dfrac{(1+t)^{2}}{1+t^{1}}\right)$

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

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

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

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

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

$=1$

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 sum of the squares of the distance of a point from the three coordinate axes be $36$, then its distance from the origin is

  1. $6$ units
  2. $3$ $\sqrt{2}$ units
  3. $2$ $\sqrt{3}$ units

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

Let $(x, y, z)$ be the point.


Given sum of the squares of distance from point to the axes is $36$. 

$ \Rightarrow (x^2+y^2) +(y^2+z^2) + (z^2+x^2) = 36 $

$ \Rightarrow 2(x^2+y^2+z^2) = 36 \Rightarrow x^2 + y^2 + z^2 = 18  $

So the distance of the point from the origin is $ = \sqrt{x^2 + y^2 + z^2} = 3\sqrt{2}$

Hence, option B; is correct.

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 of a point $(a,a,a)$ from the origin is $ \sqrt { 108 } $, then the value of $a$ is

  1. $9$
  2. $6$
  3. $-9$
  4. $-6$
Reveal answer Fill a bubble to check yourself
B,D Correct answer
Explanation
Distance between $(x _1,y _1)$ and $(x _2,y _2)$ is $\sqrt { { ({ x } _{ 1 }-{ x } _{ 2 }) }^{ 2 }+{ ({ y } _{ 1 }-{ y } _{ 2 }) }^{ 2 } } $
Distance between $(0,0,0)$ and $(a,a,a )$ is $\sqrt { {(a-0) }^{ 2 }+{ (a-0) }^{ 2 } +{ (a-0) }^{ 2 } } $

$\sqrt { 3\times { (a) }^{ 2 } } =\sqrt { 108 } =6\sqrt { 3 } \\ \Rightarrow a=\pm 6\\$
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 $P(x,\,-1)$ and $Q(3,\,2)$ is $5$ units. Find the value of $x$.

  1. $2,8$
  2. $-2,9$
  3. $1,8$
  4. $-1,7$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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


P$=(x _1,y _1)=(x,-1)$

Q$=(x _2,y _2)=(3,2)$

$25=(3-x)^2+(2+1)^2$

$25=(9+x^2-6x+9)$

$x^2-6x-7=0$

$x^2-7x+x-7=0$

$x(x-7)+1(x-7)=0$

$x=-1,7$

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

A line passes through two point $A (2, -3, -1)$ and $B (8, -1, 2)$. The coordinates of a point on this line at a distance of $14$ units from $A$ are

  1. $(14, 1, 5)$
  2. $(-10, -7, 7)$
  3. $(86, 25, 41)$

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

Given points are $A(2,-3,-1)$ and $B(8,-1,2)$
Therefore direction  ratio of $AB$ are $l = \dfrac{6}{7}, m=\dfrac{2}{7},n= \dfrac{3}{7}$ or $l=\dfrac{-6}{7},m= \dfrac{-2}{7},n= \dfrac{-3}{7}$
Hence, coordinates of a point $14$ unit from point $A$ is given as,$(2+14l,-3+14m,-1+14n)$
$\Rightarrow (14,1,5)$ or $(-10,-7,-7)$
Hence, option 'A' is correct.

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

If  $C _1:{x^2+y^2}-20x+64=0$ and $C _2:{x^2+y^2}+30x+144=0$. Then the length of the shortest line segment $PQ$  which touches $C _1$ at $P$ and  to  $C _2$ at $Q$ is

  1. $10$
  2. $15$
  3. $22$
  4. $27$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given $C _1 : x^2+y^2-20x+64=0 \Rightarrow (x-10)^2+y^2=36$
and $C _2 : x^2+y^2+30x+144=0 \Rightarrow (x+15)^2+y^2=81$
So centre and radius of $C _1$ and $C _2$ are $(10,0)$, $(-15,0)$ and $6,9$ respectively.
Then, distance between $C _1$ and $C _2$ is $\sqrt{(15+10)^{2}+(0-0)^{2}}=25$. 

$PQ$ touches $C _{1}$ at $P$ and $C _{2}$ at $Q$.
Then, shortest length of $PQ$ is $=25-(9+6)=25-15=10$

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 $(2,1,-1)$ from the line $\dfrac{x-1}{2}=\dfrac{y+1}{1}=\dfrac{z-3}{-3}$ measured parallel to the plane $x+2y+z=4$ is

  1. $\sqrt{10}$
  2. $\sqrt{20}$
  3. $\sqrt{5}$
  4. $\sqrt{30}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The distance is found by projecting the vector from the point to a point on the line onto the normal of the plane containing the line and the direction parallel to the given plane.

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 shortest distance between the line 
$\dfrac {x-1}{\alpha}=\dfrac {y+1}{-1}=\dfrac {z}{1}(\alpha \neq 1)$ and $x+y+z+1=0=2x-y+z+3$ is $\dfrac {1}{\sqrt {3}}$, then a value $\alpha$ is:

  1. $-\dfrac {16}{19}$
  2. $-\dfrac {19}{16}$
  3. $\dfrac {32}{19}$
  4. $\dfrac {19}{23}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The shortest distance between two skew lines is given by the formula involving the cross product of direction vectors and the vector connecting points on the lines.