Tag: the plane

Quizzes (3) Questions (79)

Questions Related to the plane

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

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

  1. $\displaystyle 11x - 3y = 0$
  2. $\displaystyle 3x + 11y = 0$
  3. $\displaystyle 3x + 11y = 65z$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

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

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

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
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
$x+z = \dfrac{3}{2}$

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

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| $

  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. $\cos^{-1}\left(\displaystyle \frac{\bar{n} _{1}\cdot \left | \bar{d} _{2} \right |}{\left | \bar{n} _{1} \right |\left | \bar{n} _{2} \right |}\right)$
Reveal answer Fill a bubble to check yourself
B Correct answer
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  $\theta=\cos^{-1}\left(\dfrac{\bar{n} _1.\bar{n} _2}{|\bar{n} _1||\bar{n} _2|}\right)$

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

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

  1. $\displaystyle \cos ^{-1}\frac{17}{31}$
  2. $30^{0}$
  3. $90^{0}$
  4. $\displaystyle \cos ^{-1}\frac{19}{35}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $ABC$ 
$\displaystyle \therefore \cos \theta =\frac{\left ( 5\hat{i}-\hat{j}-3\hat{k} \right )\left ( \hat{i}-5\hat{j}-3\hat{k} \right )}{\left | 5\hat{i}-\hat{j}-3\hat{k} \right |\left | \hat{i}-5\hat{j}-3\hat{k} \right |}$
$\displaystyle \therefore \cos \theta =\frac{19}{35}$

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

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

  1. $\displaystyle \dfrac { \pi }{ 2 } $
  2. $\displaystyle \dfrac { \pi }{ 6 } $
  3. $\displaystyle \dfrac { \pi }{ 4 } $
  4. $\displaystyle \dfrac { \pi }{ 3 } $
Reveal answer Fill a bubble to check yourself
D Correct answer
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 } $
Multiple choice maths the plane angle between planes angle between two planes problems involving equation of plane

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

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given planes
$r\cdot n=q$------(1)
$r\cdot {n}'={q}'$-------(2)
Angle between two planes is between their normal vector 
$\left | n \right |\left | {n}' \right |cos\alpha=n \cdot {n}'$
$cos\alpha=\dfrac{n \cdot {n}'}{\left | n \right |\left | {n}' \right |}$
$\alpha=\cos^{-1}(\dfrac{n \cdot {n}'}{\left | n \right |\left | {n}' \right |})$
Multiple choice maths the plane angle between planes angle between two planes problems involving equation of plane

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

  1. $\displaystyle \frac{2}{\sqrt{29}}$
  2. $\displaystyle \frac{5}{\sqrt{29}}$
  3. $\displaystyle \frac{3\sqrt3}{\sqrt{29}}$
  4. $\displaystyle \frac{-2}{\sqrt{29}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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) $

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

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

  1. $5x - y - 4z - 45 = 0$
  2. $5x - y - 4z -3 = 0$
  3. $23x - 13y + 32z + 45 = 0$
  4. $23x - 13y + 32z + 5 = 0$
Reveal answer Fill a bubble to check yourself
B,C Correct answer
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: $23x-13y+32z+45=0$

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

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

  1. $2(4x+17y-31z)+36=0$
  2. $8x-16y+4z+27=0$
  3. $16x-32y+8z-27=0$

  4. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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