Tag: the plane

Questions Related to the plane

What is the cosine of angle between the planes $x + y + z + I = 0$ and $2x-2y+2x+I=0$ ?

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

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

  2. $\dfrac{1}{3}$

  3. $\dfrac{2}{3}$

  4. None of the above

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Correct Option: B
Explanation:

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

For two planes,  $a _{ 1 }x+b _{ 1 }y+c _{ 1 }z+d _{ 1 }=0$ and $ a _{ 2 }x+b _{ 2 }y+c _{ 2 }z+d _{ 2 }=0$ the cosine of the angle between them is,

$\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} }  } $

So, for the given planes we have
$\cos\theta =\dfrac { 1\times 2+1\times (-2)+1\times 2 }{ \sqrt { 3 } \sqrt { 12 }  } =\dfrac { 2 }{ 6 } =\dfrac { 1 }{ 3 } $
Hence, option B is correct.

The angle between the planes $2x-3y-6z=5$ and $6x+2y-9z=4$ is

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

  1. ${\cos ^{ - 1}}\left( {\dfrac{{30}}{{77}}} \right)$

  2. ${\cos ^{ - 1}}\left( {\dfrac{{40}}{{77}}} \right)$

  3. ${\cos ^{ - 1}}\left( {\dfrac{{50}}{{77}}} \right)$

  4. ${\cos ^{ - 1}}\left( {\dfrac{{60}}{{77}}} \right)$

Show answer
Correct Option: D
Explanation:

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


Angle between plane is angle between normals.


$\therefore \cos { \theta  } =\cfrac { 2\times 6+(-3)\times 2+(-6)(-9) }{ \sqrt { { 2 }^{ 2 }+{ 3 }^{ 2 }+{ 6 }^{ 2 } } \sqrt { { 6 }^{ 2 }+{ 2 }^{ 2 }+{ 9 }^{ 2 } }  } =\cfrac { 60 }{ 77 } $

$ \theta =\cos ^{ -1 }{ \left (\cfrac { 60 }{ 77 } \right ) } $

A line lies in $YZ-$plane and makes angle of $30^o$ with the $Y-$axis, then its inclination to the $Z-$axis is 

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

  1. $30^o$ or $60^o$

  2. $60^o$ or $90^o$

  3. $60^o$ or $120^o$

  4. $30^o$ or $150^o$

Show answer
Correct Option: C
Explanation:

since line lies on $y-z$ plane $\alpha ={ 90 }^{ 0 }$

$\beta ={ 30 }^{ 0 }$
$\therefore \cos ^{ 2 }{ \alpha  } +\cos ^{ 2 }{ \beta  } +\cos ^{ 2 }{ \gamma  } =1$
$\therefore \cos ^{ 2 }{ { 30 }^{ 0 } } +\cos ^{ 2 }{ { 30 }^{ 0 } } +\cos ^{ 2 }{ \gamma  } =1$
$\cos ^{ 2 }{ \gamma  } =\cfrac { 1 }{ 4 } \Rightarrow \cos { \gamma  } =\pm \cfrac { 1 }{ 2 } $
$\gamma ={ 60 }^{ 0 },{ 120 }^{ 0 }$
Ans: $C$

If vectors $\bar{b}=\left(\tan\alpha, -1 2\sqrt{\sin \dfrac{\alpha}{2}}\right)$ and $\bar{c}=\left(\tan \alpha , \tan\alpha -\dfrac{3}{\sqrt{\sin \alpha/2}}\right)$ are orthogonal and vector $\bar{a}=(1, 3, \sin 2\alpha)$ make an obtuse angle with the z-axis, then?

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

  1. $\alpha =\tan^{-1}(-2)$

  2. $\alpha =\tan^{-1}(-3)$

  3. $\alpha =\tan^{-1}(2)$

  4. $-2 < \alpha < 0$

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Correct Option: D

Let $\overrightarrow{A}$ be vector parallel to the line of intersection of planes ${p} _{1}$ and ${p} _{2}$ through the origin. ${p} _{1}$ is parallel to the vectors $\overrightarrow{a}=2\hat{j}+3\hat{k}$ and $\overrightarrow{b}=4\hat{j}-3\hat{k}$ and ${p} _{2}$ is parallel to the vectors $\overrightarrow{c}=\hat{j}-\hat{k}$ and $\overrightarrow{d}=3\hat{i}+3\hat{j}$. The angle between $\overrightarrow{A}$ and $2\hat{i}+\hat{j}-2\hat{k}$ is 

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

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

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

  3. $\dfrac{\pi}{6}$

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

Show answer
Correct Option: B,D
Explanation:

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 $\cos{\theta}=\pm\dfrac{1}{\sqrt{2}}$
$\Rightarrow \theta=\dfrac{\pi}{4},\dfrac{3\pi}{4}$

Let $\overrightarrow{A}$ be vector parallel to the line of intersection of planes ${p} _{1}$ and ${p} _{2}$ through the origin. ${p} _{1}$ is parallel to the vectors $\overrightarrow{a}=2\hat{j}+3\hat{k}$ and $\overrightarrow{b}=4\hat{j}-3\hat{k}$ and ${p} _{2}$ is parallel to the vectors $\overrightarrow{c}=\hat{j}-\hat{k}$ and $\overrightarrow{d}=3\hat{i}+3\hat{j}$. The angle between $\overrightarrow{A}$ and $2\hat{i}+\hat{j}-2\hat{k}$ is:

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

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

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

  3. $\dfrac{\pi}{6}$

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

Show answer
Correct Option: B,D
Explanation:

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 $\cos{\theta}=\pm \dfrac{1}{\sqrt{2}}$
$\Rightarrow \theta=\dfrac{\pi}{4},\dfrac{3\pi}{4}$

Let $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c},\overrightarrow{d}$ are such that $\left(\overrightarrow{a}\times \overrightarrow{b}\right)\times \left(\overrightarrow{c}\times \overrightarrow{d}\right)=0$.Let ${p} _{1}$ and ${p} _{2}$ be the planes determined by the pairs of vectors $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c},\overrightarrow{d}$ respectively . The angle between the planes ${p} _{1}$ and ${p} _{2}$ is

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

  1. $0$

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

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

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

Show answer
Correct Option: A
Explanation:

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

The equation of the bisector of the obtuse angle between the planes $3x+4y-5z+1=0, 5x+12y-13z=0$ is

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

  1. $11x+4y-3z=0$

  2. $14x-8y+13=0$

  3. $2x+8y-8z-1=0$

  4. $13x-7z+18=0$

Show answer
Correct Option: C
Explanation:

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

Plane2 :$5x+12y-12z=0$
let us construct a $||$gm $ABCD$ with $AB$ & $AD$ in direction of normal to plane $\bot$ & plane2 respectively.
$\overrightarrow { AB } =3\hat { i } +4\hat { j } -5\hat { k } \ \overrightarrow { AD } =5\hat { i } +12\hat { j } -13\hat { k } $
$\therefore \overrightarrow { AC } $ will be the acute angle bisector whereas $\overrightarrow { BD } $ will be in direction of obtuse angle bisector to the normals.
$\overrightarrow { AC } =\overrightarrow { AB } +\overrightarrow { AD } $ (by $||$gm law of addition )
$\overrightarrow { BD } =\overrightarrow { AB } -\overrightarrow { AD } $ (by $\triangle$ law of addition)
$\therefore \overrightarrow { BD } =-2\hat { i } -8\hat { j } +8\hat { k } $ is the direction of the normal to the plane through obtuse angle bisector plane1 & plane2.
$\therefore$ Equation of plane through the line of  intersection of plane1 & plane2
$(3x+4y-5z+1)+\lambda (5x+12y-13z)=0\ (3+5\lambda )x+(4+12\lambda )y+(-5-13\lambda )+1=0$
The above plane should be parallel to the plane formed  as it is normal.
$\therefore \dfrac { 3+5\lambda  }{ -2 } =\dfrac { 4+12\lambda  }{ -8 } =\dfrac { -5-13\lambda  }{ 8 } \ \Rightarrow \lambda =-1$
$\therefore $ The required plane is ,
$-2x-8y+8z+1=0\ \Rightarrow 2x+8y-8z-1=0$

The equations of the plane which passes through $(0, 0, 0)$ and which is equally inclined to the planes $x-y+z-3=0$ and $x+y+z+4=0$ is/are-

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

  1. $y=0$

  2. $x=0$

  3. $x+y=0$

  4. $x+z=0$

Show answer
Correct Option: A,D
Explanation:

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

The angle between planes $\overline { r } .\left( 2\overline { i } -3\overline { j } +4\overline { k }  \right) +11=0$ and $\overline { r } .\left( 3\overline { i } -2\overline { j } -3\overline { k }  \right) +27=0$ is

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

  1. $\cfrac{\pi}{6}$

  2. $\cfrac{\pi}{4}$

  3. $\cfrac{\pi}{3}$

  4. $\cfrac{\pi}{2}$

Show answer
Correct Option: D
Explanation:

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

$\Rightarrow cos \theta = \dfrac{6 + 6 - 12}{\sqrt{4 + 9 + 16} \sqrt{9 + 4 + 9}}$
$\therefore cos \theta = 0$
$\therefore \theta = \dfrac{\pi}{2}$