Tag: introduction to three dimensional geometry

Quizzes (6) Questions (119)

Questions Related to introduction to three dimensional geometry

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

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.

  1. $-1$

  2. $-2$

  3. $1$

  4. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $=3r+1=-1$

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

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-

  1. $1/2$

  2. $1/3$

  3. $1/4$

  4. $1/6$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

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

  1. $6$

  2. $3$

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

  4. $\sqrt{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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$

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

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

  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. $(1,-2,3)$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $\left(\dfrac{1}{2},1, \dfrac{3}{2}\right)i.e. the \  centre$.

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. <span>$\sqrt {62}$</span>

  3. <span>$\sqrt {64}$</span>

  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. <span>$(-28,-43,-69)$</span>

  3. <span>$(-32,-37,-71)$</span>

  4. <span>None of these</span>
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.