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

If the planes $\vec{r}. (2\widehat{i}- \widehat{j}+ 2\widehat{k})= 4$ and $\vec{r}. (3\widehat{i}+ 2\widehat{j}+\lambda\widehat{k})= 3$ are perpendicular, then $\lambda =$

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

Since, the planes $\vec { r } .(2\widehat { i } -\widehat { j } +2\widehat { k } )=4$ & $\vec{r}. (3\widehat{i}+ 2\widehat{j}+\lambda\widehat{k})= 3\ $ are perpendicular to each other
Therefore, $\left( 2i-j+2 k  \right) .\left( 3i+2j+\lambda k \right) =0$
$\Rightarrow 6-2+2\lambda =0$
$\Rightarrow \lambda =-2$

Ans: B

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

The angle between the planes, $\vec{r}.(2\widehat{i}- \widehat{j}+\widehat {k})=6$ and $\vec{r}.(\widehat{i}+ \widehat{j}+2\widehat {k})=5$ , is:

  1. $\dfrac{\pi}{3}$
  2. $\dfrac{2\pi}{3}$
  3. $\dfrac{\pi}{6}$
  4. $\dfrac{5\pi}{6}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Angle between $2x-y+z=6$ & $x+y+2z=5$ is
$\theta =\cos ^{ -1

}{ \left[ \dfrac { \left( 2i-j+k \right) .\left( i+j+2k \right)  }{

\sqrt { \left( { 2 }^{ 2 }+{ 1 }^{ 2 }+{ 1 }^{ 2 } \right) \left( 1^{ 2

}+{ 1 }^{ 2 }+2^{ 2 } \right)  }  }  \right]  } =\dfrac { \pi  }{ 3 } $

Ans: B

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

The angle between the planes $ 3x-6y+2z+5=0 $ 7 $ 4x-12y+3z=3 $.Which is bisected by the plane
$ 67x-162y+47z+44 = 0 $is the angle which-

  1. contains origin

  2. is acute

  3. is obtuse

  4. is right angle
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To determine if the bisecting plane contains the origin, check the signs of the expressions for the two given planes at the origin. The bisector that contains the origin is the one where the signs of the constants match the signs of the plane equations.

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

A plane$ P _{1}$ has the equation $2x-y+z=4$ and the plane $P _{2}$ has the equation $x+ny+2z=11.$ If the angle between $P _{1}$ and $P _{2}$ is $\pi /3$ then the value (s) of '$n$' is (are)

  1. $7/2$
  2. $17,-1$
  3. $-17,1$
  4. $-7/2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
A plane$ P _{1}$ has the equation $2x-y+z=4$ and the plane $P _{2}$ has the equation $x+ny+2z=11.$  
The direction vector of the normal of the first plane is $2i-j+k$ and second plane is $i+nj+2k$.
The angle between $P _{1}$ and $P _{2}$ is $\pi /3$
$\cos { \dfrac { \pi  }{ 3 }  } =\dfrac { \left( 2i-j+k \right) .\left( i+nj+2k \right)  }{ \sqrt { \left( { 2 }^{ 2 }+{ 1 }^{ 2 }+{ 1 }^{ 2 } \right) \left( { 1 }^{ 2 }+{ n }^{ 2 }+{ 2 }^{ 2 } \right)  }  } $
$\Rightarrow \dfrac { 1 }{ 2 } =\dfrac { 4-n }{ \sqrt { 6\left( 5+{ n }^{ 2 } \right)  }  } $
$\Rightarrow \cos^2(\dfrac{\pi}{3})=\dfrac{(4-n)^2}{6(5+n^2)} $
$\Rightarrow n^2+16n-17=0\Rightarrow (n+17)(n-1)=0$
$\Rightarrow n=-17,1$
Multiple choice maths the plane angle between planes angle between two planes problems involving equation of plane

The angle between the planes $\displaystyle x + y + z = 0$ and $\displaystyle 3x - 4y + 5z = 0$ is

  1. $\displaystyle \cos ^{-1}\left ( \frac{1}{5} \sqrt{\frac{2}{5}} \right )$
  2. $\displaystyle \frac{\pi }{2}$
  3. $\displaystyle \frac{\pi }{3}$
  4. $\displaystyle \cos ^{-1}\left ( \frac{2}{5} \sqrt{\frac{2}{3}} \right )$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The angle between $x+y+z=0$ & $3x-4y+5z=0$ is
$\theta =\cos ^{ -1

}{ \left[ \dfrac { \left(\vec  i+\vec j+\vec k \right) .\left( 3\vec i-4\vec j+5\vec k \right)  }{

\sqrt { \left( { 1 }^{ 2 }+{ 1 }^{ 2 }+{ 1 }^{ 2 } \right) \left( 3^{ 2

}+{ 4 }^{ 2 }+5^{ 2 } \right)  }  }  \right]  } $

$= \cos ^{ -1 } \left( \dfrac { 2 }{ 5 } \sqrt { \dfrac { 2 }{ 3 }  }  \right) $

Ans: D

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

Which of the following planes is equally inclined to the planes $\displaystyle 4x + 3y - 5z = 0$ and $\displaystyle 5x - 12y + 13z = 0$?

  1. $\displaystyle 11x - 3y = 0$
  2. $\displaystyle 3x + 11y = 0$
  3. $\displaystyle 3x + 11y = 65z$

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

The bisector of planes $4x+3y-5z=0$ & $5x-12y+13z=0$ can be given by

$\dfrac { 4x+3y-5z }{ \sqrt { { 4 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 } }  } \pm \dfrac { 5x-12y+13z }{ \sqrt { { 5 }^{ 2 }+12^{ 2 }+13^{ 2 } }  } =0$

$\Rightarrow 52x+39y-65z\pm \left( 25x-60y+65z \right) =0$

$\Rightarrow 11x-3y=0$ and $27x+99y-130z=0$

Ans: A

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

The equation of the plane bisecting the acute angle between the planes $\displaystyle x - y + z - 1 = 0$ and $\displaystyle x + y + z = 2$ is

  1. $\displaystyle x + z = \frac{3}{2}$
  2. $\displaystyle 2y = 1$
  3. $\displaystyle x - y - z = 3$

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

Given planes are  $ x-y+z-1=0.....(1)$ and $x+y+z-2=0.........(2)$
Therefore equation of plane bisecting these planes are
$\dfrac{x-y+z-1}{\sqrt{3}}=\pm\dfrac{x+y+z-2}{\sqrt{3}}$
$\Rightarrow x+z = \dfrac{3}{2}.......(3)$ and $y = \dfrac{1}{2}.......(4)$
If $\theta$ is the angle between (2) and (4) then,
$  \cos\theta = \dfrac{1/2}{(1/2).(\sqrt{3})}=\dfrac{1}{\sqrt{3}}$
$\Rightarrow \theta > 45^\circ$
Hence plane (4) bisects the obtuse angle between the given planes.
Therefore equation of plane bisecting acute angle  between given plane is
$x+z = \dfrac{3}{2}$

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 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

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.
$\Rightarrow \cos\theta =\dfrac{\bar{n _1}\cdot\bar{n _2}}{|\bar{n _1}||\bar{n _2}|}$
$=\dfrac{(2\hat{i}-\hat{j}+\hat{k})\cdot(\hat{i}+\hat{j}+2\hat{k})}{(\sqrt{4+1+1})\sqrt{1+1+4}}$
$=\left|\dfrac{2-1+2}{\sqrt{6}\cdot \sqrt{6}}\right|$
$=\dfrac{3}{6}$
$=\dfrac{1}{2}$
$\Rightarrow \cos\theta =\dfrac{1}{2}$
$\Rightarrow \theta =\dfrac{2}{3}$.