Tag: three dimensional geometry - ii

Questions Related to three dimensional geometry - ii

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

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

  1. $\displaystyle \frac{3}{2}$

  2. $\displaystyle \frac{5}{2}$

  3. $\displaystyle \frac{7}{2}$

  4. $\displaystyle \frac{9}{2}$

Show answer
Correct Option: C
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 $= \dfrac{21}{\sqrt {4^2+2^2+4^2}}$
$\therefore \dfrac{21}{6}=\dfrac{7}{2}$

The distance between the planes given by $\vec{r}.\left ( i:+:2j:-:2k \right ):+:5= 0$ and $\vec{r}.\left ( i:+:2j:-:2k \right ):-:8= 0$ is

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

  1. $1$ unit

  2. $13/3$ units

  3. $13$ units

  4. none of these

Show answer
Correct Option: B
Explanation:
Given planes
$\vec{r}\cdot(\hat{i}+2\hat{j}-2\hat{k})+5=0$---------(1)
$\vec{r}\cdot(\hat{i}+2\hat{j}-2\hat{k})-8=0$---------(2)
from Here $a=1,b=2,c=-2$ and $d _{1}=-5\ and\ d _{2}=8$
$distance=\left | \dfrac{d _{1}-d _{2}}{\sqrt{a^2+b^2+c^2}} \right |$
$distance=\left | \dfrac{-5-8}{\sqrt{1^2+2^2+(-2)^2}} \right |$
$distance=\left | \dfrac{-13}{\sqrt{1+4+4}} \right |$
$distance=\left | \dfrac{-13}{\sqrt{9}} \right |$
$distance=\left | \dfrac{-13}{3} \right |$
$distance=\dfrac{13}{3}$

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 parallel planes three dimensional geometry - ii distance formula

  1. $\dfrac{1}{2}$

  2. $\displaystyle \frac{1}{6}$

  3. $\displaystyle \frac{\sqrt{2}}{3}$

  4. $1$

Show answer
Correct Option: B
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 } $

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

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

  1. $P _1\, +\, 8P _2\, -\, P _3\, =\, 0$

  2. $P _3\, =\, 16P _2$

  3. $8P _2\, =\, P _1$

  4. $P _1\, +\, 2P _2\, +\, 3P _3\, =\, \sqrt{29}$

Show answer
Correct Option: A,B,C,D
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 } $

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 

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

  1. ${{35\sqrt 2 } \over {12}}$

  2. ${{35\sqrt 2 } \over {24}}$

  3. ${{35\sqrt 2 } \over 6}$

  4. ${{35\sqrt 2 } \over 8}$

Show answer
Correct Option: A
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 $A$

 Find the distance between the parallel planes $\vec { r } .(2\hat { i } -3\hat { j } +6\hat { k } )=5$ and$\quad \vec { r } .(6\hat { i } -9\hat { j } +18\hat { k } )\quad +\quad 20\quad =\quad 0. $
maths introduction to three-dimensional geometry distance between two parallel planes three dimensional geometry - ii distance formula

  1. $\dfrac {28}{21}$

  2. $\dfrac {5}{3}$

  3. $\dfrac {5}{21}$

  4. None of these

Show answer
Correct Option: B

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

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

  1. $\displaystyle \frac{7}{2}$

  2. $\displaystyle \frac{5}{2}$

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

  4. $\displaystyle \frac{9}{2}$

Show answer
Correct Option: A
Explanation:

Distance between parallel planes $ax+by+cz+d _1$ and $ax+by+cz+d _2$ is given by $\dfrac{|d _1-d _2|}{\sqrt{a^2+b^2+c^2}}$.

The equation of first plane we can taken as $4x+2y+4z-16=0$.
Hence the distance is $\dfrac{|-16-5|}{\sqrt{4^2+2^2+4^2}}=\dfrac{7}{2}$

The angle between the line $\dfrac{x-1}{1}=\dfrac{y+2}{1}=\dfrac{z-4}{0}$ and the plane $y+z+2=0$ is

angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

  1. $\dfrac{\pi}{3}$

  2. $\dfrac{\pi}{4}$

  3. $\dfrac{\pi}{6}$

  4. $\dfrac{\pi}{2}$

Show answer
Correct Option: A
Explanation:

$\text cos\theta = \dfrac{1.0+1.1+0.1}{(√(1^{2}+1^{2}+0^{2}).(√0^{2}+1^{2}+1^{2})}$

$\text cos\theta = \dfrac{1}{√2.√2}$
$\text cos\theta = \dfrac{1}{2}$
$\theta = \dfrac{π}{3}$

The angle between the line $\dfrac{x}{2} = \dfrac{y}{3} = \dfrac{z}{4}$ and the plane $3x + 2y - 3z = 4$, is

angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

  1. $45^o$

  2. $0^o$

  3. $\cos^{-1} \left(\dfrac{24}{\sqrt{29 \times 22}}\right)$

  4. $90^o$

Show answer
Correct Option: B
Explanation:

Line$:\cfrac { x }{ 2 } =\cfrac { y }{ 3 } =\cfrac { z }{ 4 } $ has directions $(2,3,4)$

Plane$:3x+2y-3z=4$ has normal with direction ratios $(3,2,-3)$
$\therefore$ angle between plane and line be $\theta$ then angle between line and its direction will be $90-\theta$.
$\cos { (90-\theta ) } =\cfrac { 2\times 3+3\times 2+4\times -3 }{ \sqrt { ({ 2 }^{ 2 }+{ 3 }^{ 2 }+{ 4 }^{ 2 })({ 3 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }) }  } =0\ \therefore 90-\theta =90\quad \Rightarrow \theta ={ 0 }^{ \circ  }$

The projection of the line segment joining the points $(1, 2, 3)$ and $(4, 5, 6)$ on the plane $2x + y + z = 1$ is 

angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

  1. $1$

  2. $\sqrt{3}$

  3. $5$

  4. $6$

Show answer
Correct Option: B
Explanation:

Points $A(1,2,3)$ and $B(4,5,6)$ ,  Plane :$2x+y+z=1$

length of projection is distance between foot of perpendicular of $A$ & $B$ on Plane.
Directions of normal to plane $\Rightarrow 2,1,1$
let $(2r+1,r+2,r+3)$ is foot of $\bot$ from $A(1,2,3)$.
$(2r+1)2+(r+2)+(r+3)=1\ \Rightarrow r=-1$
foot of $\bot$ from $A=(-1,1,2)$
Similarly,If $(2k+4,k+5,k+6)$ is foot of $\bot$ from $B(4,5,6)$
$\therefore (2k+1)2+(k+5)+(k+6)=1\ \Rightarrow k=-3$
foot of $\bot$ from $B=(-2,2,3)$
$\therefore$ distance between foot of $\bot$ from $A$ & foot of $\bot$ from $B$.
$\Rightarrow \sqrt { { (-2-(-1)) }^{ 2 }+{ (2-1) }^{ 2 }+{ (3-2) }^{ 2 } } =\sqrt { 3 } $