Mathematics

Three Dimensional Geometry Planes

198 Questions

Three dimensional geometry planes involve calculating intercepts, angles between planes, and lines of intersection. These concepts are essential for advanced mathematics assessments. The questions cover spatial relationships of planar surfaces and vector equations.

Plane interceptsLine of intersectionAngle between planesVector equations of planesPoint and plane relationships

Three Dimensional Geometry Planes Questions

Multiple choice maths introduction to three-dimensional geometry distance between two parallel planes three dimensional geometry - ii distance formula

Distance between the parallel planes $2x-3y+4z-1=0$ and $4x-6y+8z+8=0$ is

  1. $\dfrac{5}{\sqrt{29}}$
  2. $\dfrac{9}{2\sqrt{29}}$
  3. $\dfrac{1}{\sqrt{29}}$
  4. $\dfrac{9}{\sqrt{29}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Consider the given line 

Let, 

$2x-3y+4z-1=0$     -----   $(1)$

And, 

$4x - 6y + 8z + 8 = 0$

$2x - 3y + 4z + 4 = 0$   ----   $(2)$

Now, 
Distance between plane $1$ and $2$

$d=|\dfrac{d _1-d _2}{\sqrt {a^2+b^2+c^2}}|$

$=|\dfrac{-1-4}{\sqrt {2^2+(-3)^2+4^2}}|=\dfrac{5}{\sqrt {4+9+16}}$

$=\dfrac{5}{\sqrt {29}}$

Hence, distance between the planes is $\dfrac{5}{\sqrt {29}}$

So, 
Option $A$ is correct.

Multiple choice maths introduction to three-dimensional geometry distance between two parallel planes three dimensional geometry - ii distance formula

Distance between the two planes:  $2 x + 3 y + 4 z = 4$  and  $4 x + 6 y + 8 z = 12$  is

  1. $2$ units
  2. $4$ units
  3. $8$ units
  4. $\frac { 2 } { \sqrt { 29 } }$ units
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We have the equation of plane is 

$2x+3y+4z=4$ and $4x+6y+8z=12$
By the helps on these equation we get 
$2x+3y+4z=6$
Now
Let distance between planes is $d$
$d = \frac{2}{{\sqrt {29} }}$
Hence the option $D$ is the correct answer.

Multiple choice maths introduction to three-dimensional geometry distance between two parallel planes three dimensional geometry - ii distance formula

The distance between the planes $x-2y+3z=6$ and $3x-6y+9z+5=0 is $

  1. $\frac{{13}}{{3\sqrt {14} }}$
  2. $\frac{{23}}{{3\sqrt {14} }}$
  3. $\frac{{13}}{{\sqrt {14} }}$
  4. $\frac{{15}}{{42}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We have,

$\begin{array}{l} x-2y+3z=6 \ 3x-6y+9z=-5 \ x-2y+3z=6 \ x-2y+3z=\frac { { -5 } }{ 3 }  \ d=\frac { { 6+\frac { 5 }{ 3 }  } }{ { \sqrt { 1+4+9 }  } } =\frac { { 23 } }{ { 3\sqrt { 14 }  } }  \end{array}$
Distance b/wl planes $ = \frac{{23}}{{3\sqrt {14} }}$
Then, 
Option $B$ is correct answer.

Multiple choice maths introduction to three-dimensional geometry distance between two parallel planes three dimensional geometry - ii distance formula

The distance between the planes $x + 2y + 3z + 7 = 0$ and $2x + 4y + 6z + 7 = 0$ is

  1. $\displaystyle \frac{\sqrt{7}}{2 \sqrt{2}}$
  2. $\displaystyle \frac{7}{2}$
  3. $\displaystyle \frac{\sqrt{7}}{2}$
  4. $\displaystyle \frac{7}{2 \sqrt{2}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation of second plane can be rearrange as $x+2y+3z+\dfrac{7}{2}=0$. 

The distance between parallel planes $Ax+By+cZ+D _1=0$ and $Ax+By+Cz+D _2$ is given by 
$\dfrac{|D _1-D _2|}{\sqrt{A^2+B^2+C^2}}$.
Hence, for the given problem distance between planes is given by:

 $\dfrac{|7-\dfrac{7}{2}|}{\sqrt{1+4+9}}=\dfrac{\sqrt{7}}{2\sqrt{2}}$.

Multiple choice maths introduction to three-dimensional geometry distance between two parallel planes three dimensional geometry - ii distance formula

If the distance between the planes $8x + 12y - 14 z = 2$ and $4x + 6y - 7z = 2$ can be expressed in the form $\displaystyle \frac{1}{\sqrt{N}}$, where N is natural, then the value of $\displaystyle \frac{N(N + 1)}{2}$ is

  1. $4950$
  2. $5050$
  3. $5150$
  4. $5151$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given planes can be written as $4x + 6y - 7z - 1 = 0$ and $4x + 6y - 7z - 2 = 0$
Since both the planes are parallel so distance between them is given by,
$d =\left | \dfrac{(-1-2)}{\sqrt{4^2+6^2+7^2}}\right |=\dfrac{1}{\sqrt{101}}$  


$\therefore N=101$

Hence, $\dfrac{N(N+1)}{2}=101\times  51 = 5151$

Multiple choice maths introduction to three-dimensional geometry distance between two parallel planes three dimensional geometry - ii distance formula

If the distance between the planes $8x + 12y - 14z = 2$ and $4x + 6y - 7z = 2$ can be expressed in the form of $ \displaystyle \frac {1}{ \sqrt N} $ where $N$ is a natural number, then the value of $ \displaystyle \frac { N(N+1)}{2} $ is

  1. $4950$
  2. $5050$
  3. $5150$
  4. $5151$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given planes can be written as $4x + 6y - 7z - 1 = 0$ and $4x + 6y - 7z - 2 = 0$
Since both the planes are parallel so distance between them is given by,
$d =\left |  \dfrac{(-1-2)}{\sqrt{4^2+6^2+7^2}}\right |=\dfrac{1}{\sqrt{101}}$ $\therefore N=101$
Hence $\dfrac{N(N+1)}{2}=101 \times 51 = 5151$

Multiple choice maths introduction to three-dimensional geometry distance between two parallel planes three dimensional geometry - ii distance formula

If the distance between the planes $8x + 12y - 14z = 2$ and $4x + 6y - 7z = 2$ can be expressed int he form $\dfrac{1}{\sqrt{N}}$ where $N$ is natural, then the value of $\dfrac{N(N+1)}{2}$ is

  1. $4950$
  2. $5050$
  3. $5150$
  4. $5151$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given planes can be written as $4x + 6y - 7z - 1 = 0$ and $4x + 6y - 7z - 2 = 0$
Since both the planes are parallel so distance between them is given by,
$d =\left |\dfrac{(-1-2)}{\sqrt{4^2+6^2+7^2}}\right |=\dfrac{1}{\sqrt{101}}$

$\therefore N=101$
Ergo $\dfrac{N(N+1)}{2}=101\times 51 = 5151$

Multiple choice maths introduction to three-dimensional geometry distance between two parallel planes three dimensional geometry - ii distance formula

If ${ p } _{ 1 },{ p } _{ 2 },{ p } _{ 3 }$ denote the distance of the plane $2x-3y+4z+2=0$ from the planes $2x-3y+4z+6=0, 4x-6y+8z+3=0$ and $2x-3y+4z-6=0$ respectively, then 

  1. ${ p } _{ 1 }+8{ p } _{ 2 }-{ p } _{ 3 }=0$
  2. ${ { p } _{ 3 } }^{ 2 }=16{ { p } _{ 2 } }^{ 2 }$
  3. $8{ { p } _{ 2 } }^{ 2 }={ { p } _{ 1 } }^{ 2 }$
  4. ${ p } _{ 1 }+2{ p } _{ 2 }+3{ p } _{ 3 }=\sqrt { 29 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since the planes are all parallel planes,

$\displaystyle { p } _{ 1 }=\dfrac { \left| 2-6 \right|  }{ \sqrt { { 2 }^{ 2 }+{ 3 }^{ 2 }+{ 4 }^{ 2 } }  } =\dfrac { 4 }{ \sqrt { 4+9+16 }  } =\dfrac { 4 }{ \sqrt { 29 }  } $

Equation of the plane $4x-6y+8z+3=0$ can be written as $2x-3y+4z+\displaystyle\dfrac { 3 }{ 2 } =0$

So, $\displaystyle { p } _{ 2 }=\dfrac { \left| 2-\dfrac { 3 }{ 2 }  \right|  }{ \sqrt { { 2 }^{ 2 }+{ 3 }^{ 2 }+{ 4 }^{ 2 } }  } =\dfrac { 1 }{ 2\sqrt { 29 }  } $

and $\displaystyle { p } _{ 3 }=\dfrac { \left| 2+6 \right|  }{ \sqrt { { 2 }^{ 2 }+{ 3 }^{ 2 }+{ 4 }^{ 2 } }  } =\dfrac { 8 }{ \sqrt { 29 }  } $

$\Rightarrow { p } _{ 1 }+8{ p } _{ 2 }-{ p } _{ 3 }=0$

Multiple choice maths introduction to three-dimensional geometry distance between two parallel planes three dimensional geometry - ii distance formula

The distance between the planes $\displaystyle 4x - 5y + 3z = 5$ and $\displaystyle 4x - 5y + 3z + 2 = 0$ is

  1. $\displaystyle \frac{7}{2 \sqrt{5}}$
  2. $\displaystyle 7$
  3. $\displaystyle \frac{7}{5 \sqrt{2}}$
  4. $\displaystyle 3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given planes are $4x -5y+ 3z - 5 = 0$ and $4x -5y+ 3z + 2 = 0$
Since both the planes are parallel so distance between them is,
$= \left| \dfrac{(-5-2)}{\sqrt{4^2+5^2+3^2}}\right|=\dfrac{7}{5\sqrt{2}}$

Multiple choice maths introduction to three-dimensional geometry distance between two parallel planes three dimensional geometry - ii distance formula

The distance between the planes $\displaystyle 2x + y + 2z = 8$ and $\displaystyle 4x + 2y + 4z + 5 = 0$ is

  1. $\displaystyle \frac{3}{2}$
  2. $\displaystyle \frac{5}{2}$
  3. $\displaystyle \frac{7}{2}$
  4. $\displaystyle \frac{9}{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given planes are $2x + y + 2z = 8$ and $4x + 2y + 4z + 5 = 0$
Multiplying first equation by 2 we get,
$4x+2y+4z=16 , 4x+2y+4z=-5$
Distance betweeb the planes $= \dfrac{21}{\sqrt {4^2+2^2+4^2}}$
$\therefore \dfrac{21}{6}=\dfrac{7}{2}$
Multiple choice maths introduction to three-dimensional geometry distance between two parallel planes three dimensional geometry - ii distance formula

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. $\dfrac{1}{2}$
  2. $\displaystyle \frac{1}{6}$
  3. $\displaystyle \frac{\sqrt{2}}{3}$
  4. $1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\vec { r } .\left( 2\hat { i } -2\hat { j } +\hat { k }  \right) =-3$

Multiplying by $2$
$\vec { r } .\left( 4\hat { i } -4\hat { j } +2\hat { k }  \right) =-6$
Equation of other plane
$\vec { r } .\left( 4\hat { i } -4\hat { j } +2\hat { k }  \right) =-5$
Distance between the two parallel planes
$=\cfrac { { c } _{ 1 }+{ c } _{ 2 } }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } }  } $
$=\cfrac { \left( -6 \right) -\left( -5 \right)  }{ \sqrt { { 4 }^{ 2 }+{ 4 }^{ 2 }+2^{ 2 } }  } $
$=\cfrac { 1 }{ \sqrt { 16+16+4 }  } $
$=\cfrac { 1 }{ 6 } $

Multiple choice maths introduction to three-dimensional geometry distance between two parallel planes three dimensional geometry - ii distance formula

If $P _1\,,\, P _2\,

,\, P _3$ denotes the perpendicular distances of the plane $2x -3y + 4z + 2 = 0$ from the parallel planes  $2x- 3y + 4z +6 = 0, 4x -6y + 8z + 3 = 0 $ and $2x- 3y + 4z- 6 = 0$ respectively, then

  1. $P _1\, +\, 8P _2\, -\, P _3\, =\, 0$
  2. $P _3\, =\, 16P _2$
  3. $8P _2\, =\, P _1$
  4. $P _1\, +\, 2P _2\, +\, 3P _3\, =\, \sqrt{29}$
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

Since the planes are parallel planes

$\displaystyle { P } _{ 1 }=\frac { \left| 2-6 \right|  }{ \sqrt { { 2 }^{ 2 }+{ 3 }^{ 2 }+{ 4 }^{ 2 } }  } =\frac { 4 }{ \sqrt { 4+9+16 }  } =\frac { 4 }{ \sqrt { 29 }  } $
Equation of the plane $4x-6y+8z+3=0$ can be written as $2x-3y+4z+\displaystyle\frac{3}{2}=0$
So, $\displaystyle { P } _{ 2 }=\frac { \left| 2-\frac { 3 }{ 2 }  \right|  }{ \sqrt { { 2 }^{ 2 }+{ 3 }^{ 2 }+{ 4 }^{ 2 } }  } =\frac { 1 }{ 2\sqrt { 29 }  } $
and $\displaystyle { P } _{ 3 }=\frac { \left| 2+6 \right|  }{ \sqrt { { 2 }^{ 2 }+{ 3 }^{ 2 }+{ 4 }^{ 2 } }  } =\frac { 8 }{ \sqrt { 29 }  } $

So, from options:
(A) $\displaystyle { P } _{ 1 }+8{ P } _{ 2 }-{ P } _{ 3 }=\frac { 4 }{ \sqrt { 29 }  } -8\times \frac { 1 }{ 2\sqrt { 29 }  } -\frac { 8 }{ \sqrt { 29 }  } =\frac { 8 }{ \sqrt { 29 }  } -\frac { 8 }{ \sqrt { 29 }  } =0$
(B) $\displaystyle 16{ P } _{ 2 }=\frac { 16 }{ 2\sqrt { 29 }  } =\frac { 8 }{ \sqrt { 29 }  } ={ P } _{ 3 }$
(C) $\displaystyle 8{ P } _{ 2 }=\frac { 8 }{ 2\sqrt { 29 }  } =\frac { 4 }{ \sqrt { 29 }  } ={ P } _{ 1 }$
(D) $\displaystyle { P } _{ 1 }+2{ P } _{ 2 }+3{ P } _{ 3 }=\frac { 4 }{ \sqrt { 29 }  } +\frac { 2 }{ 2\sqrt { 29 }  } +3\times \frac { 8 }{ \sqrt { 29 }  } =\frac { 29 }{ \sqrt { 29 }  } =\sqrt { 29 } $

Multiple choice maths introduction to three-dimensional geometry distance between two parallel planes three dimensional geometry - ii distance formula

A line having direction ratios $3,4,5$ cuts $2$ planes $2x-3y+6z-12=0$ and $2x-3y+6z+2=0$ at point P & Q, then Find length of PQ 

  1. ${{35\sqrt 2 } \over {12}}$
  2. ${{35\sqrt 2 } \over {24}}$
  3. ${{35\sqrt 2 } \over 6}$
  4. ${{35\sqrt 2 } \over 8}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$3r,4r,5r$ be the point that cuts the two plane. 

$\therefore $ for plane $1$, $ 2x+6z-3y-12=0$
$\Rightarrow 6r+30r-12r-12=0$
$\Rightarrow r=\cfrac { 1 }{ 2 } $
for plane $2$, 
$2x-3y+6z+2=0$
$\Rightarrow 6r-12r+30r+2=0$
$\Rightarrow r=\cfrac { -1 }{ 12 } $
point of intersection at plane $1$ $\Rightarrow P(\cfrac { 3 }{ 2 } ,2,\cfrac { 5 }{ 2 } )$
point of intersection at plane $2$ $\Rightarrow Q(\cfrac { -3 }{ 12 } ,\cfrac { -1 }{ 3 } ,\cfrac { -5 }{ 12 } )$
$\therefore $ distance $(PQ)$ $==\sqrt { (\cfrac { 3 }{ 2 } +\cfrac { 3 }{ 12 } )^{ 2 }+(2+\cfrac { 1 }{ 3 } )^{ 2 }+(\cfrac { 5 }{ 2 } +\cfrac { 5 }{ 12 } )^{ 2 } } =\cfrac { 35\sqrt { 2 }  }{ 12 } $
Ans $A$