Tag: introduction to three dimensional geometry

Questions Related to introduction to three dimensional geometry

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.

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

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

  1. $\sqrt {72}$

  2. $\sqrt {62}$

  3. $\sqrt {64}$

  4. None of these

Show answer
Correct Option: A
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.

If $O=(0,0,0),OP=5$ and the d.rs of OP are $1,2,2$ then $P _x+P _y+P _z=$

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

  1. $25$

  2. $\dfrac{25}{9}$

  3. $\dfrac{25}{3}$

  4. $\left(\dfrac{5}{3},\dfrac{10}{3},\dfrac{10}{3}\right)$

Show answer
Correct Option: D

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 $(2, -3, 1)$.

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

  1. $(32,37,71)$

  2. $(-28,-43,-69)$

  3. $(-32,-37,-71)$

  4. None of these

Show answer
Correct Option: D
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.