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

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

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
$x+z = \dfrac{3}{2}$