Mathematics

Three Dimensional Geometry Planes

198 Questions

Three dimensional geometry planes involve calculating intercepts, angles between planes, and lines of intersection. These concepts are essential for advanced mathematics assessments. The questions cover spatial relationships of planar surfaces and vector equations.

Plane interceptsLine of intersectionAngle between planesVector equations of planesPoint and plane relationships

Three Dimensional Geometry Planes Questions

Multiple choice maths the plane equation of a plane in intercept form equation of a plane in different forms different forms of equation of a plane

If a plane passes through a fixed point $\left ( 2, 3, 4 \right )$ and meets the axes of reference in $A$, $B$ and $C$, the point of intersection of the planes through $A$, $B$, $C$ parallel to the coordinate planes can be

  1. $\left ( 6, 9, 12 \right )$
  2. $\left ( 4, 12, 16 \right )$
  3. $\left ( 1, 1, -1 \right )$
  4. $\left ( 2, 3, -4 \right )$
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

Let us say a plane P $ax+by+cz=k$ passes through $\left( 2,3,4 \right) $ so $2a+3b+4c=k \quad -(1)$

$A\left( \dfrac { k }{ a } ,0,0 \right) ,\quad B\left( 0,\dfrac { k }{ b } ,0 \right) ,\quad C\left( 0,0,\dfrac { k }{ c }  \right) $

Points of intersection will be $\left< \dfrac { k }{ a } ,\dfrac { k }{ b } ,\dfrac { k }{ c }  \right> $

Let $\dfrac { k }{ a } =x\quad \dfrac { k }{ b } =y\quad \dfrac { k }{ c } =z$ so in $(1)$

$\dfrac { 2k }{ x } +\dfrac { 3k }{ y } +\dfrac { 4k }{ z } =k$

$\dfrac { 2 }{ x } +\dfrac { 3 }{ y } +\dfrac { 4 }{ z } =1\quad -(1)$

$(a)$ if $(x,y,z) = (6,9,12)$

$\dfrac { 2 }{ 6 } +\dfrac { 3 }{ 9 } +\dfrac { 4 }{ 12 } =\dfrac { 1 }{ 3 } +\dfrac { 1 }{ 3 } +\dfrac { 1 }{ 3 } =1$ Hence true.

$(b)$ $\left< 4,12,16 \right> $

$\dfrac { 2 }{ 4 } +\dfrac { 3 }{ 12 } +\dfrac { 4 }{ 16 } =\dfrac { 1 }{ 2 } +\dfrac { 1 }{ 4 } +\dfrac { 1 }{ 4 } =1$ Hence correct

$(c)$ $\left< 1,1,-1 \right> $

$\dfrac { 2 }{ 1 } +\dfrac { 3 }{ 1 } +\dfrac { 4 }{ -1 } =1$ Hence this is also correct.

$(d)$ $\left< 2,3,-4 \right> $

$\dfrac { 2 }{ 2 } +\dfrac { 3 }{ 3 } +\dfrac { 4 }{ -4 } =2-1=1$ This is also correct.

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 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

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

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

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

If the planes $ 2x-y+ \lambda z- 5=0$ and $x+4y+2z- 7= 0$ are perpendicular, then $\lambda=$

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

Since, the planes $2x-y+\lambda z-5=0$ & $x+4y+2z-7=0$ are perpendicular to each other
Therefore, $\left( 2i-j+\lambda k \right) .\left( i+4j+2k \right) =0$
$\Rightarrow 2-4+2\lambda =0$
$\Rightarrow \lambda =1$

Ans: A