Tag: the plane

Questions Related to the plane

Equation of the plane bisecting the angle between the planes $2x-y+2z+3=0$ and $3x-2y+6z+8=0$

maths the plane angle between planes angle between two planes problems involving equation of plane

  1. $5x-y-4z-45=0$

  2. $5x-y-4z-3=0$

  3. $23x+13y+32z-45=0$

  4. $23x-13y+32z+5=0$

Show answer
Correct Option: B
Explanation:

The equation of bisector is,
$\dfrac{2x-y+2z+3}{\sqrt{2^2+1^2+2^2})} = \pm \dfrac{3x-2y+6z+8}{\sqrt{3^2+2^2+6^2}}$
$\Rightarrow 7(2x-y+2z+3) = \pm 3(3x-2y+6z+8)$
$\Rightarrow 5x-y-4z-3=0$ or $23x-13y+32z+45=0$
Hence, option 'B' is correct.

Let two planes $p _{1}:2x-y+z=2$, and $p _{2}:x+2y-z=3$ are given. The equation of the acute angle bisector of planes $P _{1}$ and $P _{2}$ is

maths the plane angle between planes angle between two planes problems involving equation of plane

  1. $x-3y+2z+1=0$

  2. $3x+y-5=0$

  3. $x+3y-2z+1=0$

  4. $3x +z+7=0$

Show answer
Correct Option: B
Explanation:

Given, $2x-y+z=2$    ...$(i)$

and $x+2y-z=3$    ...$(ii)$

$\therefore$  Equation of the planes bisecting the angles between them are. $\displaystyle\dfrac { 2x-y+z-2 }{ \sqrt { 4+1+1 }  } =\pm \dfrac { x+2y-z-3 }{ \sqrt { 1+4+1 }  } $

$\Rightarrow 2x-y+z-2=\pm x+2y-z-3$  ...$(iii)$

and $3x+y-5=0$    ...$(iv)$

If $\theta $ be the angle between the plane $(iv)$ and $(ii)$, we have $\displaystyle\cos { \theta =\dfrac { 1\left( 3 \right) +2\left( 1 \right) -2\left( 0 \right)  }{ \sqrt { 1+4+1 } \quad \quad \sqrt { 9+1+25 }  }  } =\dfrac { 5 }{ \sqrt { 210 }  } $

$\displaystyle\Rightarrow \tan { \theta =\dfrac { 5 }{ \sqrt { 185 }  }  } <1$

$\therefore \quad \theta <{ 45 }^{ o }$

Hence, equation of the acute angle of bisects is $3x+y-5=0$.

Two planes are prependicular  to one another. One of them contains vector $\vec{a}, \vec{b}$ and the other contains $\vec{c}, \vec{d}$ then $(\vec{a} \times \vec{b}) . (\vec{c}\times \vec{d}) = $

maths the plane angle between planes angle between two planes problems involving equation of plane

  1. $1$

  2. $0$

  3. $[\vec{a}   \vec{b}    \vec{c} ]$

  4. $[ \vec{b}    \vec{c}    \vec{d} ]$

Show answer
Correct Option: B
Explanation:

Let plane $P$, contains $a,b$ vector
$\vec{n} _{1}=\ \vec{a}\times \vec{b}$
Plane $P _{2}$ contain $\vec{c},\vec{d}$ vector
$\vec{n} _{2}=\vec{c}\times \vec{d}$
If $ P _{1}\perp P _{2}$ than $ n _{1}\perp\ n _{2}$
$(\vec{a}\times \vec{b}).(\vec{c}\times \vec{d})=0$

Tetrahedron has Vertices at $O(0,0,0)$ , $A(1,2, 1)$ , $B(2,1,3)$ , $C(-1,1,2)$ . Then the angle between the faces $OAB$ and $ABC$ will be

maths the plane angle between planes angle between two planes problems involving equation of plane

  1. $\cos^{-1} (\dfrac{19}{35})$


  2. $\cos^{-1} (\dfrac{17}{31})$

  3. $30^{0}$

  4. $90^{0}$

Show answer
Correct Option: A
Explanation:

$n _{1}= \overrightarrow{OA}\times \overrightarrow{OB}= \begin{vmatrix}\hat{i} &\hat{j}  &\hat{k} \1  &2  &1 \2  &1  &3 \end{vmatrix}= 5\hat{i}-\hat{j}-3\hat{k}$


$n _{2}=\overrightarrow{AB}\times \overrightarrow{AC}= \begin{vmatrix}\hat{i} &\hat{j}  &\hat{k} \1  &-1  &2 \-2  &-1  &1 \end{vmatrix}= \hat{i}-5\hat{j}-3\hat{k}$

$\cos \theta = \dfrac{\vec n _{1}-\vec n _{2}}{\left | n _{1} \right |\left | n _{2} \right |}$

$\theta = \cos^{-1}\left ( \dfrac{19}{35} \right )$

Consider the planes $3x-6y+2z+5=0$ and $4x-12y+3z=3$. The plane $67x-162y+47z+44=0$ bisects the angle between the given planes which-

maths the plane angle between planes angle between two planes problems involving equation of plane

  1. Contains origin

  2. Is acute

  3. Is obtuse

  4. None of these

Show answer
Correct Option: A,B
Explanation:

For $3x-6y+2z+5=0$ and $-4x+12y-3z+3=0$ bisector are
$\displaystyle \frac { 3x-6y+2z+5 }{ \sqrt { 9+36+4 }  } =\pm \frac { -4x+12y-3z+3 }{ \sqrt { 16+144+9 }  } $
The plane which bisects the angle between the plane that contains the origin
$13\left( 3x-6y+2z+5 \right) =7\left( -4x+12y-3z+3 \right) \ \Rightarrow 67x-162y+47z+44=0$
Further $3\times \left( -4 \right) +\left( -6 \right) \times 12+2\times \left( -3 \right) <0$
Hence, the origin lies in the acute angle.

Angle between planes $2x-y+z$ $=$ $6$ and $x+y+2z$ $=$ $7,$ is -

maths the plane angle between planes angle between two planes problems involving equation of plane

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

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

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

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

Show answer
Correct Option: C
Explanation:
Plane $1$: $2x-y+z=6$
normal vector is $\bar{n _1}=2\hat{i}-\hat{j}+\hat{k}$
Plane $2$: $x+y+2z=7$
normal vector is $\bar{n _2}=\hat{i}+\hat{j}+2\hat{k}$
Angle between planes is same as the angle between their normal.
$\Rightarrow \cos\theta =\dfrac{\bar{n _1}\cdot\bar{n _2}}{|\bar{n _1}||\bar{n _2}|}$
$=\dfrac{(2\hat{i}-\hat{j}+\hat{k})\cdot(\hat{i}+\hat{j}+2\hat{k})}{(\sqrt{4+1+1})\sqrt{1+1+4}}$
$=\left|\dfrac{2-1+2}{\sqrt{6}\cdot \sqrt{6}}\right|$
$=\dfrac{3}{6}$
$=\dfrac{1}{2}$
$\Rightarrow \cos\theta =\dfrac{1}{2}$
$\Rightarrow \theta =\dfrac{2}{3}$.

The equation of the plane bisecting the angle between the planes $\displaystyle 3x +4y = 4$ and $\displaystyle 6x - 2y + 3z + 5 = 0$ that contains the origin, is

maths the plane angle between planes angle between two planes problems involving equation of plane

  1. $\displaystyle 9x - 38y + 15z + 43 = 0$

  2. $\displaystyle 51x + 18y + 15z = 3$

  3. $\displaystyle 9x + 2y + 3z + 1 = 0$

  4. $\displaystyle 17x + 9y + 15z = 26$

Show answer
Correct Option: B
Explanation:
Equation of given planes can be written as
 $3x+4y=4 , 6x−2y+3z+5=0$

formula is
  $\dfrac {a _1x+b _1y+c _1z+d _1}{\sqrt {a _1^2+b _1^2+c _1^2}}$ = +  or  - $\dfrac {a _2x+b _2y+c _2z+d _2}{\sqrt {a _2^2+b _2^2+c _2^2}}$

by substituting the values in the given formula we will get 

$\dfrac {3x+4y+0z+-41}{\sqrt {3^2+4^2+0^2}}$ = + or - $\dfrac {6x+-2y+3z+5}{\sqrt {6^2+(-2)^2+3^2}}$

$\Rightarrow$ $21x+28y-28 = +\  or\  - 30x-10y+160+25$

so when adding the above equation we will get $51x + 18y + 160z - 3 = 0$

is the plane bisecting the angle containing the origin, and when subtracting we will get $9x - 38y + 160z + 53 = 0$ is the other bisecting plane.

Hence the plane $51x + 18y + 160z - 3 = 0\  or\  51x + 18y + 160z  = 3$ bisects the acute angle and therefore origin lies in the acute angle.

The equation of the plane bisecting the obtuse angle between the planes $\displaystyle x+y+z= 1$ and $\displaystyle x+2y-4z= 5$ is

maths the plane angle between planes angle between two planes problems involving equation of plane

  1. $\displaystyle \left ( \sqrt{7}-1 \right )x+\left ( \sqrt{7}-2 \right )y+\left ( \sqrt{7}+4 \right )z+5-\sqrt{7}= 0$

  2. $\displaystyle \left ( \sqrt{7}+1 \right )x+\left ( \sqrt{7}+2 \right )y+\left ( \sqrt{7}+4 \right )z+5-\sqrt{7}= 0$

  3. $\displaystyle \left ( \sqrt{7}+1 \right )x+\left ( \sqrt{7}+2 \right )y+\left ( \sqrt{7}-4 \right )z=\sqrt{7}$

  4. None of these

Show answer
Correct Option: D
Explanation:

Given planes are  $ x+y+z-1=0.....(1)$ and $x+2y-4z-5=0.........(2)$
Therefore equation of planes bisecting these planes are
$\dfrac{x+y+z-1}{\sqrt{3}}=\pm\dfrac{x+2y-4z-5}{\sqrt{21}}$

$\Rightarrow x+y+z-1=\pm\dfrac{x+2y-4z-5}{\sqrt{7}}$

$\Rightarrow (\sqrt{7}-1) x+(\sqrt{7}-2)y+(\sqrt{7}+4)z = \sqrt{7}+5 ...(3)$ and $(\sqrt{7}+1) x+(\sqrt{7}+2)y+(\sqrt{7}-4)z = \sqrt{7}-5  ....(4)$
If $\theta$ is the angle between $(1)$ and $(3)$, then

$  \cos\theta = \dfrac{(\sqrt{7}-1).1+(\sqrt{7}-2).1+(\sqrt{7}+4).1}{(\sqrt{(\sqrt{7}-1)^2+(\sqrt{7}-2)^2+(\sqrt{7}+4)^2}).(\sqrt{3})}= \dfrac{3\sqrt{7}+2}{(\sqrt{40+2\sqrt{7}}).(\sqrt{3})}> \dfrac{1}{2}$

$\Rightarrow \theta > 45^\circ$
Hence, plane $(1)$ bisects the obtuse angle between the given planes.
Therefore equation of plane bisecting acute angle  between given plane is
$(\sqrt{7}-1) x+(\sqrt{7}-2)y+(\sqrt{7}+4)z = \sqrt{7}+5 $

Hence, option 'D' is correct.

Let two planes $p _{1}:2x-y+z=2$, and $p _{2}:x+2y-z=3$ are given. The equation of the bisector of angle of the planes  $P _{1}$ and $P _{2}$ which does not contains origin, is

maths the plane angle between planes angle between two planes problems involving equation of plane

  1. $x-3y+2z+1=0$

  2. $x+3y=5$

  3. $x+3y+2z+2=0$

  4. $3x+y=5$

Show answer
Correct Option: D
Explanation:
Given planes are $p _{1}:2x-y+z=2$ and $p _{2}:x+2y-z=3$

Normals to the planes
$N _1:\dfrac{1}{\sqrt{6}}(2,-1,1)$
$N _2:\dfrac{1}{\sqrt{6}}(1,2,-1)$

Let $N$ be the normal vector of angle bisector
$N=  N _1+N _2$ or $ N _1-N _2$
$N = (3,1,0)$ or $(1,-3,2)$

The equation of plane is
$P = P _1+ \lambda P _2$
$P= 2x-y+z-2 + \lambda (x+2y-z -3) $

If $N = (3,1,0)$, then $\lambda = 1$,
Equation of Plane $=  P = 3x+y- 5$
It does not pass through origin.

Hence, option D is correct.