Mathematics

Three Dimensional Geometry Planes

254 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 angle between planes angle between two planes problems involving equation of plane

An angle between the plane, x+y+z=5 and the line of intersection of the planes, 3x+4y+z-1=0 and 5x+8y+2z+14=0, is

  1. $

    \sin ^ { - 1 } ( 3 / \sqrt { 17 } )

    $
  2. $

    \cos ^ { - 1 } ( \sqrt { 3 / 17 } )

    $
  3. $

    \sin ^ { - 1 } ( \sqrt { 3 / 17 } )

    $
  4. $

    \cos ^ { - 1 } ( 3 / \sqrt { 17 } )

    $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The normal vector of the plane is n = (1, 1, 1). The direction vector of the line of intersection of the two planes can be found by taking the cross product of their normal vectors: (3, 4, 1) and (5, 8, 2). Cross product: i(8 - 8) - j(6 - 5) + k(24 - 20) = (0, -1, 4), direction vector b = (0, -1, 4). Let theta be the angle between the line and the plane. Then sin(theta) = |n dot b| / (|n| |b|). Dot product n dot b = 1(0) + 1(-1) + 1(4) = 3. Magnitude of n = sqrt(1^2 + 1^2 + 1^2) = sqrt(3). Magnitude of b = sqrt(0^2 + (-1)^2 + 4^2) = sqrt(17). Therefore, sin(theta) = 3 / (sqrt(3) * sqrt(17)) = 3 / sqrt(51) = sqrt(9 / 51) = sqrt(3 / 17), so theta = sin^(-1)(sqrt(3 / 17)).

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

The tetrahedron has vertices $0\left ( 0,0,0 \right ),A\left ( 1,2,1 \right ),B\left ( 2,1,3 \right )$ and $C\left ( -1,1,2 \right )$, then  the angle between the faces $OAB$ and $ABC$ will be

  1. $\displaystyle \cos ^{-1}\frac{17}{31}$
  2. $30^{0}$
  3. $90^{0}$
  4. $\displaystyle \cos ^{-1}\frac{19}{35}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Concept using the angle between the  phases is equal to their normals.
$\therefore$ vector $\perp$ to the face $OAB$ is $\overline{OA}\times \overline{OB}=5\hat{i}-\hat{j}-3\hat{k}$
and vector $\perp$ to the face $ABC$ is $\overline{AB}\times \overline{AC}=\hat{i}-5\hat{j}-3\hat{k}$
$\therefore$ Let $\theta$ be the angle between the faces $OAB$ and $ABC$ 
$\displaystyle \therefore \cos \theta =\frac{\left ( 5\hat{i}-\hat{j}-3\hat{k} \right )\left ( \hat{i}-5\hat{j}-3\hat{k} \right )}{\left | 5\hat{i}-\hat{j}-3\hat{k} \right |\left | \hat{i}-5\hat{j}-3\hat{k} \right |}$
$\displaystyle \therefore \cos \theta =\frac{19}{35}$

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

Let $A(0,0,0),B(1,1,1),C(3,2,1)$ and $D(3,1,2)$ be four points. The angle between the planes through the points $A,B,C$ and through the points $A,B,D$ is

  1. $\displaystyle \dfrac { \pi }{ 2 } $
  2. $\displaystyle \dfrac { \pi }{ 6 } $
  3. $\displaystyle \dfrac { \pi }{ 4 } $
  4. $\displaystyle \dfrac { \pi }{ 3 } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Let ${n} _{1}$ and ${n} _{2}$ be the vectors normal to the palnes $ABC$ and $ABD$ respectively.
${ n } _{ 1 }=AB\times AC=-i+2j-k\\ { n } _{ 2 }=AB\times AD=i+j-2k$
Let $\theta$ be the acute angle between the planes, then $\theta$ is the acute angle between their normals ${n} _{1}$ and ${n} _{2}$
$\displaystyle \therefore \cos { \theta  } =\dfrac { \left| -1+2+2 \right|  }{ \sqrt { 6 } .\sqrt { 6 }  } =\dfrac { 3 }{ 2 } =\dfrac { 1 }{ 2 } =\cos { \dfrac { \pi  }{ 3 }  } \Rightarrow \theta =\dfrac { \pi  }{ 3 } $
Multiple choice maths the plane angle between planes angle between two planes problems involving equation of plane

The sine of angle formed by the lateral face ADC and plane of the base ABC of the tetrahedron ABCD where $\displaystyle a\equiv (3, -2, 1); B\equiv (3, 1, 5); C\equiv (4, 0, 3)and D\equiv (1, 0, 0)is$

  1. $\displaystyle \frac{2}{\sqrt{29}}$
  2. $\displaystyle \frac{5}{\sqrt{29}}$
  3. $\displaystyle \frac{3\sqrt3}{\sqrt{29}}$
  4. $\displaystyle \frac{-2}{\sqrt{29}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\overrightarrow { AD } =-2\hat { i } +2\hat { j } -\hat { k } ,\overrightarrow { Ac } =\hat { i } +2\hat { j } +2\hat { k } ,\overrightarrow { AB } =3\hat { j } +4\hat { k } : \ \overrightarrow { n _{ 1 } } =\overrightarrow { AD } \times \overrightarrow { AC } =\begin{vmatrix} \hat { i }  & \hat { j }  & \hat { k }  \ -2 & 2 & -1 \ 1 & 2 & 2 \end{vmatrix}=6\hat { i } +3\hat { j } -6\hat { k } =3\left( 2\hat { i } +\hat { j } -2\hat { k }  \right) \ \overrightarrow { n _{ 2 } } =\overrightarrow { AC } \times \overrightarrow { AB } =\begin{vmatrix} \hat { i }  & \hat { j }  & \hat { k }  \ 1 & 2 & 2 \ 0 & 3 & 4 \end{vmatrix}=2\hat { i } -4\hat { j } +3\hat { k } : \ \left| \overrightarrow { n _{ 1 } } \times \overrightarrow { n _{ 2 } }  \right| =3\begin{vmatrix} \hat { i }  & \hat { j }  & \hat { k }  \ 2 & 1 & -2 \ 2 & -4 & 3 \end{vmatrix}=3\left( 5\hat { i } -10\hat { j } -10\hat { k }  \right) \ \sin  \theta =\dfrac { 5 }{ \sqrt { 29 }  } \left( \because \sin  \theta =\dfrac { \left| \overrightarrow { n _{ 1 } } \times \overrightarrow { n _{ 2 } }  \right|  }{ \left| \overrightarrow { n _{ 1 } }  \right| \left| \overrightarrow { n _{ 2 } }  \right|  }  \right) $

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

The equation of a plane bisecting the angle between the plane $2x -y + 2z + 3 = 0$ and $3x- 2y + 6z + 8 = 0$ is

  1. $5x - y - 4z - 45 = 0$
  2. $5x - y - 4z -3 = 0$
  3. $23x - 13y + 32z + 45 = 0$
  4. $23x - 13y + 32z + 5 = 0$
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Explanation
Given planes
$2x-y+2z+3=0,\quad 3x-2y+6z+8=0$ for any set of planes, plane bisecting the two planes is obtained by
$\cfrac { { \Pi  } _{ 1 } }{ \left| { \Pi  } _{ 1 } \right|  } =\pm \cfrac { { \Pi  } _{ 2 } }{ \left| { \Pi  } _{ 2 } \right|  } $
By substituting
$\cfrac { 2x-y+2z+3 }{ \sqrt { 4+1+{ 2 }^{ 2 } }  } =\pm \cfrac { 3x-2y+6z+8 }{ \sqrt { 9+{ 2 }^{ 2 }+36 }  } \\ \cfrac { 2x-y+2z+3 }{ 3 } =\pm \cfrac { 3x-2y+6z+8 }{ 7 } $
First Case:
$7(2x-y+2z+3)=3(3x-2y+6z+8)$
On solving: $5x-y-4z-3=0$
Second Case:
$7(2x-y+2z+3)=-3(3x-2y+6z+8)$
We get: $23x-13y+32z+45=0$

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

Equation of the plane bisecting the acute angle between the planes  $x+2y-2z-9=0,\ 3x-4y+12z-26=0$ is

  1. $2(4x+17y-31z)+36=0$
  2. $8x-16y+4z+27=0$
  3. $16x-32y+8z-27=0$

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

The equation of the given planes are

$3x-4y+12z-26=0$   ...$(1)$

$x+2y-2z-9=0$   ...$(2)$

$\therefore $the equations of the planes bisecting the angles between them are $\displaystyle\dfrac { 3x-4y+12z-26 }{ \sqrt { 9+16+144 }  } =\pm \dfrac { x+2y-2z-9 }{ \sqrt { 1+4+4 }  } $

$\Rightarrow 3\left( 3x-4y+12z-26 \right) =\pm 13\left( x+2y-2z-9 \right) $

$\Rightarrow 4x+38y-62z-39=0$   ...$(3)$

and $22x+14y+102-195=0$   ...$(4)$

If $\theta $ isthe angle between the planes $(4)$ and $(2)$, we have 

$\displaystyle\cos { \theta  } =\dfrac { 1\left( 22 \right) +2\left( 14 \right) -2\left( 10 \right)  }{ \sqrt { 1+4+4 } .\sqrt { 484+196+100 }  } =\sqrt { \dfrac { 5 }{ 39 }  } $

$\displaystyle\Rightarrow \sin { \theta  } =\sqrt { 1-\cos ^{ 2 }{ \theta  }  } =\sqrt { 1-\dfrac { 5 }{ 39 }  } =\sqrt { \dfrac { 34 }{ 39 }  } $

$\displaystyle\Rightarrow \tan { \theta  } =\sqrt { \dfrac { 34 }{ 5 }  } >1\Rightarrow \theta >{ 45 }^{ O }$

Hence, the plane $(4)$ bisects the obtuse angle between the given plane. Thus the other plane $(3)$ bisects the acute angle.

$\therefore 4x+38y-62z-36=0$

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

Equation of the plane bisecting the angle between the planes $2x-y+2z+3=0$ and $3x-2y+6z+8=0$

  1. $5x-y-4z-45=0$
  2. $5x-y-4z-3=0$
  3. $23x+13y+32z-45=0$
  4. $23x-13y+32z+5=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation of bisector is,
$\dfrac{2x-y+2z+3}{\sqrt{2^2+1^2+2^2})} = \pm \dfrac{3x-2y+6z+8}{\sqrt{3^2+2^2+6^2}}$
$\Rightarrow 7(2x-y+2z+3) = \pm 3(3x-2y+6z+8)$
$\Rightarrow 5x-y-4z-3=0$ or $23x-13y+32z+45=0$
Hence, option 'B' is correct.

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

Let two planes $p _{1}:2x-y+z=2$, and $p _{2}:x+2y-z=3$ are given. The equation of the acute angle bisector of planes $P _{1}$ and $P _{2}$ is

  1. $x-3y+2z+1=0$
  2. $3x+y-5=0$
  3. $x+3y-2z+1=0$
  4. $3x +z+7=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, $2x-y+z=2$    ...$(i)$

and $x+2y-z=3$    ...$(ii)$

$\therefore$  Equation of the planes bisecting the angles between them are. $\displaystyle\dfrac { 2x-y+z-2 }{ \sqrt { 4+1+1 }  } =\pm \dfrac { x+2y-z-3 }{ \sqrt { 1+4+1 }  } $

$\Rightarrow 2x-y+z-2=\pm x+2y-z-3$  ...$(iii)$

and $3x+y-5=0$    ...$(iv)$

If $\theta $ be the angle between the plane $(iv)$ and $(ii)$, we have $\displaystyle\cos { \theta =\dfrac { 1\left( 3 \right) +2\left( 1 \right) -2\left( 0 \right)  }{ \sqrt { 1+4+1 } \quad \quad \sqrt { 9+1+25 }  }  } =\dfrac { 5 }{ \sqrt { 210 }  } $

$\displaystyle\Rightarrow \tan { \theta =\dfrac { 5 }{ \sqrt { 185 }  }  } <1$

$\therefore \quad \theta <{ 45 }^{ o }$

Hence, equation of the acute angle of bisects is $3x+y-5=0$.

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

Tetrahedron has Vertices at $O(0,0,0)$ , $A(1,2, 1)$ , $B(2,1,3)$ , $C(-1,1,2)$ . Then the angle between the faces $OAB$ and $ABC$ will be

  1. $\cos^{-1} (\dfrac{19}{35})$
  2. $\cos^{-1} (\dfrac{17}{31})$
  3. $30^{0}$
  4. $90^{0}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$n _{1}= \overrightarrow{OA}\times \overrightarrow{OB}= \begin{vmatrix}\hat{i} &\hat{j}  &\hat{k} \1  &2  &1 \2  &1  &3 \end{vmatrix}= 5\hat{i}-\hat{j}-3\hat{k}$


$n _{2}=\overrightarrow{AB}\times \overrightarrow{AC}= \begin{vmatrix}\hat{i} &\hat{j}  &\hat{k} \1  &-1  &2 \-2  &-1  &1 \end{vmatrix}= \hat{i}-5\hat{j}-3\hat{k}$

$\cos \theta = \dfrac{\vec n _{1}-\vec n _{2}}{\left | n _{1} \right |\left | n _{2} \right |}$

$\theta = \cos^{-1}\left ( \dfrac{19}{35} \right )$

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

Consider the planes $3x-6y+2z+5=0$ and $4x-12y+3z=3$. The plane $67x-162y+47z+44=0$ bisects the angle between the given planes which-

  1. Contains origin

  2. Is acute

  3. Is obtuse

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

For $3x-6y+2z+5=0$ and $-4x+12y-3z+3=0$ bisector are
$\displaystyle \frac { 3x-6y+2z+5 }{ \sqrt { 9+36+4 }  } =\pm \frac { -4x+12y-3z+3 }{ \sqrt { 16+144+9 }  } $
The plane which bisects the angle between the plane that contains the origin
$13\left( 3x-6y+2z+5 \right) =7\left( -4x+12y-3z+3 \right) \ \Rightarrow 67x-162y+47z+44=0$
Further $3\times \left( -4 \right) +\left( -6 \right) \times 12+2\times \left( -3 \right) <0$
Hence, the origin lies in the acute angle.

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

Angle between planes $2x-y+z$ $=$ $6$ and $x+y+2z$ $=$ $7,$ is -

  1. $\dfrac { \pi }{ 4 } $
  2. $\dfrac { \pi }{ 2 } $
  3. $\dfrac { \pi }{ 3 } $
  4. $\dfrac {- \pi }{ 4 } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Plane $1$: $2x-y+z=6$
normal vector is $\bar{n _1}=2\hat{i}-\hat{j}+\hat{k}$
Plane $2$: $x+y+2z=7$
normal vector is $\bar{n _2}=\hat{i}+\hat{j}+2\hat{k}$
Angle between planes is same as the angle between their normal.
$\Rightarrow \cos\theta =\dfrac{\bar{n _1}\cdot\bar{n _2}}{|\bar{n _1}||\bar{n _2}|}$
$=\dfrac{(2\hat{i}-\hat{j}+\hat{k})\cdot(\hat{i}+\hat{j}+2\hat{k})}{(\sqrt{4+1+1})\sqrt{1+1+4}}$
$=\left|\dfrac{2-1+2}{\sqrt{6}\cdot \sqrt{6}}\right|$
$=\dfrac{3}{6}$
$=\dfrac{1}{2}$
$\Rightarrow \cos\theta =\dfrac{1}{2}$
$\Rightarrow \theta =\dfrac{2}{3}$.
Multiple choice maths the plane angle between planes angle between two planes problems involving equation of plane

The equation of the plane bisecting the angle between the planes $\displaystyle 3x +4y = 4$ and $\displaystyle 6x - 2y + 3z + 5 = 0$ that contains the origin, is

  1. $\displaystyle 9x - 38y + 15z + 43 = 0$
  2. $\displaystyle 51x + 18y + 15z = 3$
  3. $\displaystyle 9x + 2y + 3z + 1 = 0$
  4. $\displaystyle 17x + 9y + 15z = 26$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Equation of given planes can be written as
 $3x+4y=4 , 6x−2y+3z+5=0$

formula is
  $\dfrac {a _1x+b _1y+c _1z+d _1}{\sqrt {a _1^2+b _1^2+c _1^2}}$ = +  or  - $\dfrac {a _2x+b _2y+c _2z+d _2}{\sqrt {a _2^2+b _2^2+c _2^2}}$

by substituting the values in the given formula we will get 

$\dfrac {3x+4y+0z+-41}{\sqrt {3^2+4^2+0^2}}$ = + or - $\dfrac {6x+-2y+3z+5}{\sqrt {6^2+(-2)^2+3^2}}$

$\Rightarrow$ $21x+28y-28 = +\  or\  - 30x-10y+160+25$

so when adding the above equation we will get $51x + 18y + 160z - 3 = 0$

is the plane bisecting the angle containing the origin, and when subtracting we will get $9x - 38y + 160z + 53 = 0$ is the other bisecting plane.

Hence the plane $51x + 18y + 160z - 3 = 0\  or\  51x + 18y + 160z  = 3$ bisects the acute angle and therefore origin lies in the acute angle.

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

The equation of the plane bisecting the obtuse angle between the planes $\displaystyle x+y+z= 1$ and $\displaystyle x+2y-4z= 5$ is

  1. $\displaystyle \left ( \sqrt{7}-1 \right )x+\left ( \sqrt{7}-2 \right )y+\left ( \sqrt{7}+4 \right )z+5-\sqrt{7}= 0$
  2. $\displaystyle \left ( \sqrt{7}+1 \right )x+\left ( \sqrt{7}+2 \right )y+\left ( \sqrt{7}+4 \right )z+5-\sqrt{7}= 0$
  3. $\displaystyle \left ( \sqrt{7}+1 \right )x+\left ( \sqrt{7}+2 \right )y+\left ( \sqrt{7}-4 \right )z=\sqrt{7}$

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

Given planes are  $ x+y+z-1=0.....(1)$ and $x+2y-4z-5=0.........(2)$
Therefore equation of planes bisecting these planes are
$\dfrac{x+y+z-1}{\sqrt{3}}=\pm\dfrac{x+2y-4z-5}{\sqrt{21}}$

$\Rightarrow x+y+z-1=\pm\dfrac{x+2y-4z-5}{\sqrt{7}}$

$\Rightarrow (\sqrt{7}-1) x+(\sqrt{7}-2)y+(\sqrt{7}+4)z = \sqrt{7}+5 ...(3)$ and $(\sqrt{7}+1) x+(\sqrt{7}+2)y+(\sqrt{7}-4)z = \sqrt{7}-5  ....(4)$
If $\theta$ is the angle between $(1)$ and $(3)$, then

$  \cos\theta = \dfrac{(\sqrt{7}-1).1+(\sqrt{7}-2).1+(\sqrt{7}+4).1}{(\sqrt{(\sqrt{7}-1)^2+(\sqrt{7}-2)^2+(\sqrt{7}+4)^2}).(\sqrt{3})}= \dfrac{3\sqrt{7}+2}{(\sqrt{40+2\sqrt{7}}).(\sqrt{3})}> \dfrac{1}{2}$

$\Rightarrow \theta > 45^\circ$
Hence, plane $(1)$ bisects the obtuse angle between the given planes.
Therefore equation of plane bisecting acute angle  between given plane is
$(\sqrt{7}-1) x+(\sqrt{7}-2)y+(\sqrt{7}+4)z = \sqrt{7}+5 $

Hence, option 'D' is correct.

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

Let two planes $p _{1}:2x-y+z=2$, and $p _{2}:x+2y-z=3$ are given. The equation of the bisector of angle of the planes  $P _{1}$ and $P _{2}$ which does not contains origin, is

  1. $x-3y+2z+1=0$
  2. $x+3y=5$
  3. $x+3y+2z+2=0$
  4. $3x+y=5$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given planes are $p _{1}:2x-y+z=2$ and $p _{2}:x+2y-z=3$

Normals to the planes
$N _1:\dfrac{1}{\sqrt{6}}(2,-1,1)$
$N _2:\dfrac{1}{\sqrt{6}}(1,2,-1)$

Let $N$ be the normal vector of angle bisector
$N=  N _1+N _2$ or $ N _1-N _2$
$N = (3,1,0)$ or $(1,-3,2)$

The equation of plane is
$P = P _1+ \lambda P _2$
$P= 2x-y+z-2 + \lambda (x+2y-z -3) $

If $N = (3,1,0)$, then $\lambda = 1$,
Equation of Plane $=  P = 3x+y- 5$
It does not pass through origin.

Hence, option D is correct.
Multiple choice

What is the relationship between a point, a line, and a plane in Euclidean geometry?

  1. A point is a zero-dimensional object, a line is one-dimensional, and a plane is two-dimensional.

  2. A point is a one-dimensional object, a line is two-dimensional, and a plane is three-dimensional.

  3. A point is a two-dimensional object, a line is three-dimensional, and a plane is four-dimensional.

  4. A point is a three-dimensional object, a line is four-dimensional, and a plane is five-dimensional.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In Euclidean geometry, a point has no dimensions, a line has one dimension (length), and a plane has two dimensions (length and width).