Tag: angle between two planes

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Questions Related to angle between two planes

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

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

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

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

Find the equation of the bisector planes of the angles between the planes $2x - y + 2z + 3 = 0$ and $3x - 2y + 6z + 8 = 0$.

  1. $ 5x-y-4z-22=0$
  2. $ 23x-13y+32z+26 = 0 $
  3. $ 19x-y-4z+26 = 0 $

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

Equation of the planes is $2x-4y+2z+3=0$ and $3x-2y+6z+8=0$
Then equation of the plane bisection the angles between them are
$\displaystyle \frac { 2x-4y+2z+3 }{ \sqrt { 4+16+4+9 }  } =\pm \frac { 3x-2y+6z+8 }{ \sqrt { 9+4+36+64 }  } $
$\displaystyle \Rightarrow \frac { 2x-4y+2z+3 }{ \sqrt { 33 }  } =\pm \frac { 3x-2y+6z+8 }{ \sqrt { 113 }  } $
$\Rightarrow 5x-y-4z-22=0$ and $23x-13y+32z+26=0$

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

The angle between two planes is equal to

  1. the angle between the tangents to them from any point

  2. the angle between the normals to them from any point

  3. the angle between the lines parallel to the planes from any point

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

The angle between two intersecting planes is equal to the acute angle determined by the normal vectors of the two planes.

This is different from the angle between the normals to the planes from any point.

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

lf the planes $ x+2y-z+5=0,\ 2x-ky+4z+3=0$ are perpendicular, then $ {k} $ is

  1. $1$
  2. $-1$
  3. $0$
  4. $2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

I am using the constant $\lambda$ instead of $k$ to avoid the confusion. 

The normals to the planes are given by $i+2j-k$ and $2i-\lambda j+4k$, respectively. 
Since, they are perpendicular dot product between normals are zero. 
Thus, $(i+2j-k).(2i-\lambda+4k)=0$. 
$\Rightarrow2-2\lambda-4=0 \Rightarrow \lambda=-1$

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

In the space the equation $by+ cz+ d= 0$ represents a plane perpendicular to the plane:

  1. $YOZ$
  2. $ZOX$
  3. $XOY$
  4. $Z= k$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Consider $P : bx+cz+d=0$
a) Equation of $YOZ$ plane is $x=0$
Since, $(i).(bj+ck)=0$
Therefore, $P$ is perpendicular to $YOZ$

b)  Equation of $ZOX$ plane is $y=0$
Since, $(j).(bj+ck)=b \neq 0$
Therefore, $P$ is not perpendicular to $ZOX$

c)  Equation of $XOY$ plane is $z=0$
Since, $(k).(bj+ck)=c \neq 0$
Therefore, $P$ is not perpendicular to $XOY$

d)  Consider. $z=k$
Since, $(k).(bj+ck)=c \neq 0$
Therefore, $P$ is not perpendicular to $z=k$

Ans: A