Angles Between Planes and Bisectors

Given planes are $x-y+2{z}+1=0,2{x}+y+z+2=0$

Given ,

When the corner of a square is folded such that two planes are perpendicular, the geometry dictates that the angle between the original lines OP and QR becomes 90 degrees.

$\begin{array}{l} { \pi _{ 1 } }=ax+by+cz=0 \ a+b+c=0 \ 3a+2b+c=0 \ \frac { a }{ { -1 } } =\frac { { -b } }{ { 1-3 } } =\frac { c }{ { 2-3 } }  \ \frac { a }{ { -1 } } =\frac { b }{ 2 } =\frac { c }{ { -1 } }  \ -x+2y-z=0 \ { \pi _{ 1 } }:, x-2y+z=0 \ and, \ { \pi _{ 2 } }=ax+by+cz=0 \ a+b+c=0 \ 3a+b+2c=0 \ \frac { a }{ 1 } =\frac { { -b } }{ { 2-3 } } =\frac { c }{ { 1-3 } }  \ \frac { a }{ 1 } =\frac { b }{ 1 } =\frac { c }{ { -2 } }  \ { \pi _{ 2 } }:, x+y-2z=0 \ Now, \ \cos  \theta =\frac { { \left( { 1-2-2 } \right)  } }{ { \sqrt { 6 } \sqrt { 6 }  } } =\frac { { -3 } }{ 6 } =\frac { { -1 } }{ 2 }  \ \therefore \theta ={ 120^{ \circ  } } \ Hence,, the, option, C, is, the, correct, answer. \end{array}$

Angle between two planes is the angle between their normal vectors.

The given planes are $x+y+x+I=0$ and $2x-2y+2z+I=0$ 

${ P } _{ 1 }:2x-3y-6z=5\ { P } _{ 2 }:6x+2y-9z=4$

The orthogonality condition (dot product = 0) and the obtuse angle condition (dot product with z-axis < 0) constrain the value of alpha. Solving the equations leads to the specified interval.

Plane ${P} _{1}$ is parallel to $\overrightarrow{a}$ and $\overrightarrow{b}$.
The normal to ${P} _{1}$ is along $\overrightarrow{a}\times \overrightarrow{b}$.
Plane ${P} _{2}$ is parallel to $\overrightarrow{c}$ and $\overrightarrow{d}$.
The normal to ${P} _{2}$ is along $\overrightarrow{c}\times \overrightarrow{d}$.
$\overrightarrow{A}$ is along the line of intersection of planes ${P} _{1}$ and ${P} _{2}$.
$\therefore \overrightarrow{A}$ is along $\left(\overrightarrow{a}\times\overrightarrow{b}\right)\times\left(\overrightarrow{c}\times\overrightarrow{d}\right)$
$\overrightarrow{a}\times\overrightarrow{b}=\left|\begin{matrix} \hat{i} &\hat{j}  &\hat{k}  \ 0 & 2 & 3 \ 0 &4  &-3  \end{matrix}\right|$
$=\left(-6-12\right)\hat{i}-0.\hat{j}+0.\hat{k}$ on simplification
$=-18\hat{i}$
$\left(\overrightarrow{a}\times\overrightarrow{b}\right)\times\left(\overrightarrow{c}\times\overrightarrow{d}\right)$
$\overrightarrow{c}\times\overrightarrow{d}=\left|\begin{matrix} \hat{i} &\hat{j}  &\hat{k}  \ 0 & 1 & -1 \ 3 &3  &0 \end{matrix}\right|$
$=\left(0+3\right)\hat{i}-\left(0+3\right)\hat{j}+\left(0+3\right)\hat{k}$ on simplification
$=3\hat{i}-3\hat{j}-3\hat{k}$
$=3\left(\hat{i}-\hat{j}-\hat{k}\right)$
The angle between $\overrightarrow{A}$ and $2\hat{i}+\hat{j}-2\hat{k}$ is  $\theta$
$\cos{\theta}=\dfrac{\overrightarrow{A}}{\left|\overrightarrow{A}\right|}.\dfrac{\left(2\hat{i}+\hat{j}-2\hat{k}\right)}{3\sqrt{2}}$
$\pm\dfrac{\left(\hat{j}-\hat{k}\right).\left(2\hat{i}+\hat{j}-2\hat{k}\right)}{3\sqrt{2}}$
$\pm\dfrac{1}{\sqrt{2}}$
and $\cos{\theta}=\pm\dfrac{1}{\sqrt{2}}$
$\Rightarrow \theta=\dfrac{\pi}{4},\dfrac{3\pi}{4}$

Plane ${p} _{1}$ is parallel to $\overrightarrow{a}$ and $\overrightarrow{b}$ the normal to ${p} _{1}$ is along $\overrightarrow{a}\times \overrightarrow{b}$ 
Plane ${p} _{2}$ is parallel to $\overrightarrow{c}$ and $\overrightarrow{d}$ the normal to ${p} _{2}$ is along $\overrightarrow{c}\times \overrightarrow{d}$
$\overrightarrow{A}$ is along the line of intersection of planes ${p} _{1}$ and ${p} _{2}$ 
$\therefore \overrightarrow{A}$ is along $\left(\overrightarrow{a}\times \overrightarrow{b}\right)\times \left(\overrightarrow{c}\times \overrightarrow{d}\right)$
$\overrightarrow{a}\times \overrightarrow{b}=\left[\begin{matrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 2 &  3\ 0 & 4 & -3 \end{matrix}\right]$
$=\hat{i}\left(-6-12\right)-\hat{j}\left(0-0\right)+\hat{k}\left(0\right)$
$=-18\hat{i}$
$\overrightarrow{c}\times \overrightarrow{d}=\left|\begin{matrix} \hat{i} & \hat{j} & \hat{k} \ 0  & 1 & -1 \ 3 & 3 & 0 \end{matrix}\right|$
$=\hat{i}\left(0+3\right)-\hat{j}\left(0+3\right)+\hat{k}\left(0-3\right)$
$=3\hat{i}-3\hat{j}-3\hat{k}$
$=3\left(\hat{i}-\hat{j}-\hat{k}\right)$
$ \therefore \overrightarrow{A}$ is along $\hat{i}\times \left(\hat{i}-\hat{j}-\hat{k}\right)=\hat{j}-\hat{k}$
The angle between $\overrightarrow{A}$ and $2\hat{i}+\hat{j}-2\hat{k}$ is $\theta$
$\cos{\theta}=\dfrac{\overrightarrow{A}}{\left|\overrightarrow{A}\right|}.\dfrac{\left(2\hat{i}+\hat{j}-2\hat{k}\right)}{3}$
$   =\pm \dfrac{\left(\hat{j}-\hat{k}\right)\left(2\hat{i}+\hat{j}-2\hat{k}\right)}{3\sqrt{2}}$
$=\pm\dfrac{\left(1+2\right)}{3\sqrt{2}} = \pm \dfrac{1}{\sqrt{2}}$
and $\cos{\theta}=\pm \dfrac{1}{\sqrt{2}}$
$\Rightarrow \theta=\dfrac{\pi}{4},\dfrac{3\pi}{4}$

The plane ${p} _{1}$ contains the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ into normal is along $\overrightarrow{a}\times \overrightarrow{b}$
The normal to plane ${p} _{2}$ is along  $\overrightarrow{c}\times \overrightarrow{d}$.
$\left(\overrightarrow{a}\times \overrightarrow{b}\right)\times \left(\overrightarrow{c}\times \overrightarrow{d}\right)=0$
$\Rightarrow$ two normals are parallel
$\therefore$ the angle between the planes is zero

Plane1 :$3x+4y-5z+1=0$

The equations of the plane which is equally inclined to the planes $x-y+z-3=0$ and $x+y+z+4=0$ is/are- 
$\dfrac { x-y+z-3 }{ \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 }+{ 1 }^{ 2 } }  } \pm \dfrac { x+y+z+4 }{ \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 }+{ 1 }^{ 2 } }  } =0$
$\Rightarrow x+z=-1$ and $y=\dfrac { -7 }{ 2 } $
If the plane contains origin
Then desired planes are $x+z=0$ & $y=0$

Ans: A,D

$cos \theta = \dfrac{ a _1 , a _2 + b _1 , b _2 + c _1 , c _2}{\sqrt{a _1^2 + b _1^2 + c _1^2} \sqrt{a _2^2 + b^2 _2 + c _2^2}}$

Equation of the planes is $2x-4y+2z+3=0$ and $3x-2y+6z+8=0$
Then equation of the plane bisection the angles between them are
$\displaystyle \frac { 2x-4y+2z+3 }{ \sqrt { 4+16+4+9 }  } =\pm \frac { 3x-2y+6z+8 }{ \sqrt { 9+4+36+64 }  } $
$\displaystyle \Rightarrow \frac { 2x-4y+2z+3 }{ \sqrt { 33 }  } =\pm \frac { 3x-2y+6z+8 }{ \sqrt { 113 }  } $
$\Rightarrow 5x-y-4z-22=0$ and $23x-13y+32z+26=0$

The angle between two intersecting planes is equal to the acute angle determined by the normal vectors of the two planes.

I am using the constant $\lambda$ instead of $k$ to avoid the confusion. 

Consider $P : bx+cz+d=0$
a) Equation of $YOZ$ plane is $x=0$
Since, $(i).(bj+ck)=0$
Therefore, $P$ is perpendicular to $YOZ$

b)  Equation of $ZOX$ plane is $y=0$
Since, $(j).(bj+ck)=b \neq 0$
Therefore, $P$ is not perpendicular to $ZOX$

c)  Equation of $XOY$ plane is $z=0$
Since, $(k).(bj+ck)=c \neq 0$
Therefore, $P$ is not perpendicular to $XOY$

d)  Consider. $z=k$
Since, $(k).(bj+ck)=c \neq 0$
Therefore, $P$ is not perpendicular to $z=k$

Ans: A

Since, the planes $2x-y+\lambda z-5=0$ & $x+4y+2z-7=0$ are perpendicular to each other
Therefore, $\left( 2i-j+\lambda k \right) .\left( i+4j+2k \right) =0$
$\Rightarrow 2-4+2\lambda =0$
$\Rightarrow \lambda =1$

Ans: A

Since, the planes $\vec { r } .(2\widehat { i } -\widehat { j } +2\widehat { k } )=4$ & $\vec{r}. (3\widehat{i}+ 2\widehat{j}+\lambda\widehat{k})= 3\ $ are perpendicular to each other
Therefore, $\left( 2i-j+2 k  \right) .\left( 3i+2j+\lambda k \right) =0$
$\Rightarrow 6-2+2\lambda =0$
$\Rightarrow \lambda =-2$

Ans: B

Angle between $2x-y+z=6$ & $x+y+2z=5$ is
$\theta =\cos ^{ -1

}{ \left[ \dfrac { \left( 2i-j+k \right) .\left( i+j+2k \right)  }{

\sqrt { \left( { 2 }^{ 2 }+{ 1 }^{ 2 }+{ 1 }^{ 2 } \right) \left( 1^{ 2

}+{ 1 }^{ 2 }+2^{ 2 } \right)  }  }  \right]  } =\dfrac { \pi  }{ 3 } $

Ans: B

To determine if the bisecting plane contains the origin, check the signs of the expressions for the two given planes at the origin. The bisector that contains the origin is the one where the signs of the constants match the signs of the plane equations.

The angle between $x+y+z=0$ & $3x-4y+5z=0$ is
$\theta =\cos ^{ -1

}{ \left[ \dfrac { \left(\vec  i+\vec j+\vec k \right) .\left( 3\vec i-4\vec j+5\vec k \right)  }{

\sqrt { \left( { 1 }^{ 2 }+{ 1 }^{ 2 }+{ 1 }^{ 2 } \right) \left( 3^{ 2

}+{ 4 }^{ 2 }+5^{ 2 } \right)  }  }  \right]  } $

The bisector of planes $4x+3y-5z=0$ & $5x-12y+13z=0$ can be given by

$\dfrac { 4x+3y-5z }{ \sqrt { { 4 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 } }  } \pm \dfrac { 5x-12y+13z }{ \sqrt { { 5 }^{ 2 }+12^{ 2 }+13^{ 2 } }  } =0$

$\Rightarrow 52x+39y-65z\pm \left( 25x-60y+65z \right) =0$

$\Rightarrow 11x-3y=0$ and $27x+99y-130z=0$

Ans: A

Given planes are  $ x-y+z-1=0.....(1)$ and $x+y+z-2=0.........(2)$
Therefore equation of plane bisecting these planes are
$\dfrac{x-y+z-1}{\sqrt{3}}=\pm\dfrac{x+y+z-2}{\sqrt{3}}$
$\Rightarrow x+z = \dfrac{3}{2}.......(3)$ and $y = \dfrac{1}{2}.......(4)$
If $\theta$ is the angle between (2) and (4) then,
$  \cos\theta = \dfrac{1/2}{(1/2).(\sqrt{3})}=\dfrac{1}{\sqrt{3}}$
$\Rightarrow \theta > 45^\circ$
Hence plane (4) bisects the obtuse angle between the given planes.
Therefore equation of plane bisecting acute angle  between given plane is
$x+z = \dfrac{3}{2}$

The normal vector of the plane is n = (1, 1, 1). The direction vector of the line of intersection of the two planes can be found by taking the cross product of their normal vectors: (3, 4, 1) and (5, 8, 2). Cross product: i(8 - 8) - j(6 - 5) + k(24 - 20) = (0, -1, 4), direction vector b = (0, -1, 4). Let theta be the angle between the line and the plane. Then sin(theta) = |n dot b| / (|n| |b|). Dot product n dot b = 1(0) + 1(-1) + 1(4) = 3. Magnitude of n = sqrt(1^2 + 1^2 + 1^2) = sqrt(3). Magnitude of b = sqrt(0^2 + (-1)^2 + 4^2) = sqrt(17). Therefore, sin(theta) = 3 / (sqrt(3) * sqrt(17)) = 3 / sqrt(51) = sqrt(9 / 51) = sqrt(3 / 17), so theta = sin^(-1)(sqrt(3 / 17)).

Given planes are $\bar { r } \cdot \bar { n _{ 1 } } =\left| \bar { { d } _{ 1 } }  \right| $ and $\bar { r } \cdot \bar { n _{ 2 } } =\left| \bar { { d } _{ 2 } }  \right| $ 

Concept using the angle between the  phases is equal to their normals.
$\therefore$ vector $\perp$ to the face $OAB$ is $\overline{OA}\times \overline{OB}=5\hat{i}-\hat{j}-3\hat{k}$
and vector $\perp$ to the face $ABC$ is $\overline{AB}\times \overline{AC}=\hat{i}-5\hat{j}-3\hat{k}$
$\therefore$ Let $\theta$ be the angle between the faces $OAB$ and $ABC$ 
$\displaystyle \therefore \cos \theta =\frac{\left ( 5\hat{i}-\hat{j}-3\hat{k} \right )\left ( \hat{i}-5\hat{j}-3\hat{k} \right )}{\left | 5\hat{i}-\hat{j}-3\hat{k} \right |\left | \hat{i}-5\hat{j}-3\hat{k} \right |}$
$\displaystyle \therefore \cos \theta =\frac{19}{35}$

$\overrightarrow { AD } =-2\hat { i } +2\hat { j } -\hat { k } ,\overrightarrow { Ac } =\hat { i } +2\hat { j } +2\hat { k } ,\overrightarrow { AB } =3\hat { j } +4\hat { k } : \ \overrightarrow { n _{ 1 } } =\overrightarrow { AD } \times \overrightarrow { AC } =\begin{vmatrix} \hat { i }  & \hat { j }  & \hat { k }  \ -2 & 2 & -1 \ 1 & 2 & 2 \end{vmatrix}=6\hat { i } +3\hat { j } -6\hat { k } =3\left( 2\hat { i } +\hat { j } -2\hat { k }  \right) \ \overrightarrow { n _{ 2 } } =\overrightarrow { AC } \times \overrightarrow { AB } =\begin{vmatrix} \hat { i }  & \hat { j }  & \hat { k }  \ 1 & 2 & 2 \ 0 & 3 & 4 \end{vmatrix}=2\hat { i } -4\hat { j } +3\hat { k } : \ \left| \overrightarrow { n _{ 1 } } \times \overrightarrow { n _{ 2 } }  \right| =3\begin{vmatrix} \hat { i }  & \hat { j }  & \hat { k }  \ 2 & 1 & -2 \ 2 & -4 & 3 \end{vmatrix}=3\left( 5\hat { i } -10\hat { j } -10\hat { k }  \right) \ \sin  \theta =\dfrac { 5 }{ \sqrt { 29 }  } \left( \because \sin  \theta =\dfrac { \left| \overrightarrow { n _{ 1 } } \times \overrightarrow { n _{ 2 } }  \right|  }{ \left| \overrightarrow { n _{ 1 } }  \right| \left| \overrightarrow { n _{ 2 } }  \right|  }  \right) $

The equation of the given planes are

$3x-4y+12z-26=0$   ...$(1)$

$x+2y-2z-9=0$   ...$(2)$

$\therefore $the equations of the planes bisecting the angles between them are $\displaystyle\dfrac { 3x-4y+12z-26 }{ \sqrt { 9+16+144 }  } =\pm \dfrac { x+2y-2z-9 }{ \sqrt { 1+4+4 }  } $

$\Rightarrow 3\left( 3x-4y+12z-26 \right) =\pm 13\left( x+2y-2z-9 \right) $

$\Rightarrow 4x+38y-62z-39=0$   ...$(3)$

and $22x+14y+102-195=0$   ...$(4)$

If $\theta $ isthe angle between the planes $(4)$ and $(2)$, we have 

$\displaystyle\cos { \theta  } =\dfrac { 1\left( 22 \right) +2\left( 14 \right) -2\left( 10 \right)  }{ \sqrt { 1+4+4 } .\sqrt { 484+196+100 }  } =\sqrt { \dfrac { 5 }{ 39 }  } $

$\displaystyle\Rightarrow \sin { \theta  } =\sqrt { 1-\cos ^{ 2 }{ \theta  }  } =\sqrt { 1-\dfrac { 5 }{ 39 }  } =\sqrt { \dfrac { 34 }{ 39 }  } $

$\displaystyle\Rightarrow \tan { \theta  } =\sqrt { \dfrac { 34 }{ 5 }  } >1\Rightarrow \theta >{ 45 }^{ O }$

Hence, the plane $(4)$ bisects the obtuse angle between the given plane. Thus the other plane $(3)$ bisects the acute angle.

$\therefore 4x+38y-62z-36=0$

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.

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 $3x+y-5=0$.

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}$
$(\vec{a}\times \vec{b}).(\vec{c}\times \vec{d})=0$

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

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.

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