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Questions Related to maths

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.

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

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:

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


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

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

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