Tag: angle between planes

Questions Related to angle between planes

Find the planes bisecting the acute angle between the planes $x-y+2x+1=0$ and $2x+y+z+2=0$

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

  1. $x+z-1=0$

  2. $x+z+1=0$

  3. $x-z-1=0$

  4. None of these

Show answer
Correct Option: B
Explanation:

Given planes are $x-y+2{z}+1=0,2{x}+y+z+2=0$

The plane bisecting the acute angle between the planes will be $\dfrac{x-y+2{z}+1}{\sqrt{1+1+4}}=-\dfrac{2{x}+y+z+2}{\sqrt{4+1+1}}$
$\implies x+z+1=0$

The planes $x-3y+4z-1=0$ and $kx-4y+3z-5=0$ are perpendicular then value of $k$ is

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

  1. $24$

  2. $-24$

  3. $12$

  4. $0$

Show answer
Correct Option: B
Explanation:
Let direction ratios of the perpendicular to the plane 
$x-3y+4z-1=4$ are $a _{2}=1, b _{1}=-3, c _{1}=4$
and that of planes will be perpendicular if 
$a _{1}a _{2}+b _{1}b _{2}+c _{1}c _{2}=0$
$k+(-3) \times (-4)+4 \times{3}=0$
$k+12+12=0$
$k=-24$








The equation of the plane which bisects the angle between the planes $3x-6y+2z+5=0$ and $4x-12y+3z-3=0$ which contains the origin is ?

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

  1. $33x-13y+32z+45=0$

  2. $x-3y+z-5=0$

  3. $33x+13y+32z+45=0$

  4. $None\ of\ these$

Show answer
Correct Option: D
Explanation:

Given ,

The required equation of plane bisects the given two planes.
$\begin{array}{l} \therefore \frac { { 3x-6y+2z+5 } }{ { \sqrt { { 3^{ 2 } }+{ { \left( { -6 } \right)  }^{ 2 } }+{ { \left( 2 \right)  }^{ 2 } } }  } } =\pm \frac { { 4x-12y+3z-3 } }{ { \sqrt { { 4^{ 2 } }+{ { \left( { -12 } \right)  }^{ 2 } }+{ 3^{ 2 } } }  } }  \ \frac { { 3x-6y+2z+5 } }{ { \sqrt { 49 }  } } =\pm \frac { { 4x-12y+3z-3 } }{ { \sqrt { 169 }  } }  \ \frac { { 3x-6y+2z+5 } }{ 7 } =\pm \frac { { 4x-12y+3z-3 } }{ { 13 } }  \ 39x-78y+26z+65=\pm 28x-84y+21z-21 \end{array}$
Now, solving for the positive value, we get
$39x - 78y + 26z + 65$.........(i)
or,  $11x + 6y + 5z + 36 = 0$
And for negative value, we get
$\begin{array}{l} 39x-78y+26z+65=-\left( { 28x-84y+21z-21 } \right)  \ or,\, \, 67x-162y+47z+44=0.......\left( { ii } \right)  \end{array}$
$\because $None of the answer matches with the given equation.
Hence,
Option $D$ is correct  in this case.

The corner of a square OPQR is folded up so that the plane OPQ is perpendicular to the plane OQR, the angle between OP and QR is 

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

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

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

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

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

Show answer
Correct Option: A

The angle between the plane passing through the points $A(0,\ 0,\ 0),\ B(1,\ 1,\ 1),\ C(3,\ 2,\ 1)$ & the plane passing through $A(0,\ 0,\ 0),\ B(1,\ 1,\ 1),\ D(3,\ 1,\ 2)$ is

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

  1. $90^{o}$

  2. $45^{o}$

  3. $120^{o}$

  4. $30^{o}$

Show answer
Correct Option: C
Explanation:

$\begin{array}{l} { \pi _{ 1 } }=ax+by+cz=0 \ a+b+c=0 \ 3a+2b+c=0 \ \frac { a }{ { -1 } } =\frac { { -b } }{ { 1-3 } } =\frac { c }{ { 2-3 } }  \ \frac { a }{ { -1 } } =\frac { b }{ 2 } =\frac { c }{ { -1 } }  \ -x+2y-z=0 \ { \pi _{ 1 } }:\, x-2y+z=0 \ and, \ { \pi _{ 2 } }=ax+by+cz=0 \ a+b+c=0 \ 3a+b+2c=0 \ \frac { a }{ 1 } =\frac { { -b } }{ { 2-3 } } =\frac { c }{ { 1-3 } }  \ \frac { a }{ 1 } =\frac { b }{ 1 } =\frac { c }{ { -2 } }  \ { \pi _{ 2 } }:\, x+y-2z=0 \ Now, \ \cos  \theta =\frac { { \left( { 1-2-2 } \right)  } }{ { \sqrt { 6 } \sqrt { 6 }  } } =\frac { { -3 } }{ 6 } =\frac { { -1 } }{ 2 }  \ \therefore \theta ={ 120^{ \circ  } } \ Hence,\, the\, option\, C\, is\, the\, correct\, answer. \end{array}$

The angle between the planes
$\vec{r}(\hat{i}+2\hat{j}+\hat{k})=4$ and $\vec{r}(\hat{-i}+\hat{j}+2\hat{k})=9$

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

  1. $30^{\mathrm{o}}$

  2. $60^{\mathrm{o}}$

  3. $45^{\mathrm{o}}$

  4. $90^{0}$

Show answer
Correct Option: B
Explanation:

Angle between two planes is the angle between their normal vectors.

For the first plane, normal vector is $\vec{n _0}=(1,2,1)$
For second plane, normal vector is $\vec{n _1}=(-1,1,2)$
Let $\theta$ be the angle between the planes, it is also the angle between their normals.
$\implies \cos \theta = \dfrac{{n} _{1}.{n} _{2}}{|n _1||n _2|} $

$\implies \cos \theta = \dfrac{(1,2,1)\cdot (-1,1,2)}{\sqrt{1^2+2^2+1^2}\sqrt{(-1)^2+1^2+2^2}} $

$\implies \cos \theta = \dfrac{-1+2+2}{6}=\dfrac{1}{2}$
$\implies \theta $ = $ {60}^{o}$

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

Show answer
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. $-2 < \alpha < 0$

Show answer
Correct Option: D