Tag: the plane

Questions Related to the plane

Which of the following planes is equally inclined to the planes $\displaystyle 4x + 3y - 5z = 0$ and $\displaystyle 5x - 12y + 13z = 0$?

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

  1. $\displaystyle 11x - 3y = 0$

  2. $\displaystyle 3x + 11y = 0$

  3. $\displaystyle 3x + 11y = 65z$

  4. none of these

Show answer
Correct Option: A
Explanation:

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

The equation of the plane bisecting the acute angle between the planes $\displaystyle x - y + z - 1 = 0$ and $\displaystyle x + y + z = 2$ is

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

  1. $\displaystyle x + z = \frac{3}{2}$

  2. $\displaystyle 2y = 1$

  3. $\displaystyle x - y - z = 3$

  4. None of these

Show answer
Correct Option: A
Explanation:

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
$x+z = \dfrac{3}{2}$

An angle between the plane, x+y+z=5 and the line of intersection of the planes, 3x+4y+z-1=0 and 5x+8y+2z+14=0, is

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

  1. $

    \sin ^ { - 1 } ( 3 / \sqrt { 17 } )

    $

  2. $

    \cos ^ { - 1 } ( \sqrt { 3 / 17 } )

    $

  3. $

    \sin ^ { - 1 } ( \sqrt { 3 / 17 } )

    $

  4. $

    \cos ^ { - 1 } ( 3 / \sqrt { 17 } )

    $

Show answer
Correct Option: A

The angle between the planes $\bar { r } \cdot \bar { n _{ 1 } } =\left| \bar { { d } _{ 1 } }  \right| $ and $\bar { r } \cdot \bar { n _{ 2 } } =\left| \bar { { d } _{ 2 } }  \right| $

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

  1. $\cos^{-1}\left(\displaystyle \frac{\bar{n _{1} }\cdot\bar{d} _{1}}{\left | \bar{d} _{1}\times \bar{d} _{2} \right |}\right)$

  2. $\cos^{-1}\left(\displaystyle \frac{\bar{n} _{1}.\bar{n} _{2}}{\left |\bar{n} _{1} \right |\left | \bar{n} _{2} \right |}\right)$

  3. $\cos^{-1}\left(\displaystyle \frac{\bar{n} _{1}\bar{n} _{2}}{\bar{n} _{1}\times \bar{n} _{2} }\right)$

  4. $\cos^{-1}\left(\displaystyle \frac{\bar{n} _{1}\cdot \left | \bar{d} _{2} \right |}{\left | \bar{n} _{1} \right |\left | \bar{n} _{2} \right |}\right)$

Show answer
Correct Option: B
Explanation:

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| $ 

Angle between the planes is same as the angle between the normal vectors.
Hence the angle  $\theta=\cos^{-1}\left(\dfrac{\bar{n} _1.\bar{n} _2}{|\bar{n} _1||\bar{n} _2|}\right)$

The tetrahedron has vertices $0\left ( 0,0,0 \right ),A\left ( 1,2,1 \right ),B\left ( 2,1,3 \right )$ and $C\left ( -1,1,2 \right )$, 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. $\displaystyle \cos ^{-1}\frac{17}{31}$

  2. $30^{0}$

  3. $90^{0}$

  4. $\displaystyle \cos ^{-1}\frac{19}{35}$

Show answer
Correct Option: D
Explanation:

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 $ABC$ 
$\displaystyle \therefore \cos \theta =\frac{\left ( 5\hat{i}-\hat{j}-3\hat{k} \right )\left ( \hat{i}-5\hat{j}-3\hat{k} \right )}{\left | 5\hat{i}-\hat{j}-3\hat{k} \right |\left | \hat{i}-5\hat{j}-3\hat{k} \right |}$
$\displaystyle \therefore \cos \theta =\frac{19}{35}$

Let $A(0,0,0),B(1,1,1),C(3,2,1)$ and $D(3,1,2)$ be four points. The angle between the planes through the points $A,B,C$ and through the points $A,B,D$ is

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

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

  2. $\displaystyle \dfrac { \pi  }{ 6 } $

  3. $\displaystyle \dfrac { \pi  }{ 4 } $

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

Show answer
Correct Option: D
Explanation:
Let ${n} _{1}$ and ${n} _{2}$ be the vectors normal to the palnes $ABC$ and $ABD$ respectively.
${ n } _{ 1 }=AB\times AC=-i+2j-k\\ { n } _{ 2 }=AB\times AD=i+j-2k$
Let $\theta$ be the acute angle between the planes, then $\theta$ is the acute angle between their normals ${n} _{1}$ and ${n} _{2}$
$\displaystyle \therefore \cos { \theta  } =\dfrac { \left| -1+2+2 \right|  }{ \sqrt { 6 } .\sqrt { 6 }  } =\dfrac { 3 }{ 2 } =\dfrac { 1 }{ 2 } =\cos { \dfrac { \pi  }{ 3 }  } \Rightarrow \theta =\dfrac { \pi  }{ 3 } $

The angle between two planes $\displaystyle r.n=q$ and $\displaystyle r.n'=q'$ is

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

  1. $\displaystyle \sin ^{-1}\left ( \frac{n.n'}{nn'} \right )$

  2. $\displaystyle \cos ^{-1}\left ( \frac{n.n'}{nn'} \right )$

  3. $\displaystyle \tan ^{-1}\left ( \frac{n.n'}{nn'} \right )$

  4. None of these

Show answer
Correct Option: B
Explanation:
Given planes
$r\cdot n=q$------(1)
$r\cdot {n}'={q}'$-------(2)
Angle between two planes is between their normal vector 
$\left | n \right |\left | {n}' \right |cos\alpha=n \cdot {n}'$
$cos\alpha=\dfrac{n \cdot {n}'}{\left | n \right |\left | {n}' \right |}$
$\alpha=\cos^{-1}(\dfrac{n \cdot {n}'}{\left | n \right |\left | {n}' \right |})$

The sine of angle formed by the lateral face ADC and plane of the base ABC of the tetrahedron ABCD where $\displaystyle a\equiv (3, -2, 1); B\equiv (3, 1, 5); C\equiv (4, 0, 3)and D\equiv (1, 0, 0)is$

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

  1. $\displaystyle \frac{2}{\sqrt{29}}$

  2. $\displaystyle \frac{5}{\sqrt{29}}$

  3. $\displaystyle \frac{3\sqrt3}{\sqrt{29}}$

  4. $\displaystyle \frac{-2}{\sqrt{29}}$

Show answer
Correct Option: B
Explanation:

$\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 a plane bisecting the angle between the plane $2x -y + 2z + 3 = 0$ and $3x- 2y + 6z + 8 = 0$ is

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,C
Explanation:
Given planes
$2x-y+2z+3=0,\quad 3x-2y+6z+8=0$ for any set of planes, plane bisecting the two planes is obtained by
$\cfrac { { \Pi  } _{ 1 } }{ \left| { \Pi  } _{ 1 } \right|  } =\pm \cfrac { { \Pi  } _{ 2 } }{ \left| { \Pi  } _{ 2 } \right|  } $
By substituting
$\cfrac { 2x-y+2z+3 }{ \sqrt { 4+1+{ 2 }^{ 2 } }  } =\pm \cfrac { 3x-2y+6z+8 }{ \sqrt { 9+{ 2 }^{ 2 }+36 }  } \\ \cfrac { 2x-y+2z+3 }{ 3 } =\pm \cfrac { 3x-2y+6z+8 }{ 7 } $
First Case:
$7(2x-y+2z+3)=3(3x-2y+6z+8)$
On solving: $5x-y-4z-3=0$
Second Case:
$7(2x-y+2z+3)=-3(3x-2y+6z+8)$
We get: $23x-13y+32z+45=0$

Equation of the plane bisecting the acute angle between the planes  $x+2y-2z-9=0,\ 3x-4y+12z-26=0$ is

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

  1. $2(4x+17y-31z)+36=0$

  2. $8x-16y+4z+27=0$

  3. $16x-32y+8z-27=0$

  4. None of these

Show answer
Correct Option: D
Explanation:

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.

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