Tag: three dimensional geometry - ii

Quizzes (6) Questions (94)

Questions Related to three dimensional geometry - ii

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,
$= \left| \dfrac{(-5-2)}{\sqrt{4^2+5^2+3^2}}\right|=\dfrac{7}{5\sqrt{2}}$