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

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

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

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  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. ${\cos ^{ - 1}}\left( {\dfrac{{60}}{{77}}} \right)$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 ) } $

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

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

  1. $30^o$ or $60^o$
  2. $60^o$ or $90^o$
  3. $60^o$ or $120^o$
  4. $30^o$ or $150^o$
Reveal answer Fill a bubble to check yourself
C Correct answer
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: $C$

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

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?

  1. $\alpha =\tan^{-1}(-2)$
  2. $\alpha =\tan^{-1}(-3)$
  3. $\alpha =\tan^{-1}(2)$
  4. $-2 < \alpha < 0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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.

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

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 

  1. $\dfrac{\pi}{2}$
  2. $\dfrac{\pi}{4}$
  3. $\dfrac{\pi}{6}$
  4. $\dfrac{3\pi}{4}$
Reveal answer Fill a bubble to check yourself
B,D Correct answer
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 $\cos{\theta}=\pm\dfrac{1}{\sqrt{2}}$
$\Rightarrow \theta=\dfrac{\pi}{4},\dfrac{3\pi}{4}$

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

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:

  1. $\dfrac{\pi}{2}$
  2. $\dfrac{\pi}{4}$
  3. $\dfrac{\pi}{6}$
  4. $\dfrac{3\pi}{4}$
Reveal answer Fill a bubble to check yourself
B,D Correct answer
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 $\cos{\theta}=\pm \dfrac{1}{\sqrt{2}}$
$\Rightarrow \theta=\dfrac{\pi}{4},\dfrac{3\pi}{4}$

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

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

  1. $0$
  2. $\dfrac{\pi}{4}$
  3. $\dfrac{\pi}{3}$
  4. $\dfrac{\pi}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

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

  1. $11x+4y-3z=0$
  2. $14x-8y+13=0$
  3. $2x+8y-8z-1=0$
  4. $13x-7z+18=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 ,
$-2x-8y+8z+1=0\ \Rightarrow 2x+8y-8z-1=0$

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

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-

  1. $y=0$
  2. $x=0$
  3. $x+y=0$
  4. $x+z=0$
Reveal answer Fill a bubble to check yourself
A,D Correct answer
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

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

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

  1. $\cfrac{\pi}{6}$
  2. $\cfrac{\pi}{4}$
  3. $\cfrac{\pi}{3}$
  4. $\cfrac{\pi}{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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