Tag: angle between planes

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Questions Related to angle between planes

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

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

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

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $\dfrac{x-y+2{z}+1}{\sqrt{1+1+4}}=-\dfrac{2{x}+y+z+2}{\sqrt{4+1+1}}$
$\implies x+z+1=0$

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

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

  1. $24$
  2. $-24$
  3. $12$
  4. $0$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 
$a _{1}a _{2}+b _{1}b _{2}+c _{1}c _{2}=0$
$k+(-3) \times (-4)+4 \times{3}=0$
$k+12+12=0$
$k=-24$








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

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 ?

  1. $33x-13y+32z+45=0$
  2. $x-3y+z-5=0$
  3. $33x+13y+32z+45=0$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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 

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

When the corner of a square is folded such that two planes are perpendicular, the geometry dictates that the angle between the original lines OP and QR becomes 90 degrees.

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

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

  1. $90^{o}$
  2. $45^{o}$
  3. $120^{o}$
  4. $30^{o}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

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

  1. $30^{\mathrm{o}}$
  2. $60^{\mathrm{o}}$
  3. $45^{\mathrm{o}}$
  4. $90^{0}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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}$
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.