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 equation of a plane in intercept form equation of a plane in different forms different forms of equation of a plane

lf a plane meets the coordinate axes at $A,B,C$ , then equation of plane is such that centroid of triangle $ABC$ is $\left (\displaystyle \dfrac{1}{3}\dfrac{2} {3},\dfrac{4}{3}\right)$

  1. $4x+2y+z=4$
  2. $4x+2y+z=3$
  3. $x+y+z=3$
  4. $x+y+z=9$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Assume equation of plane is, $ax+by+cz=d$
Now this plane intersect axes at $A,B$ and $C$,
$\Rightarrow A = \left (\dfrac{d}{a}, 0, 0\right), B =  \left (0, \dfrac{d}{b}, 0\right)$ and $ C =\left (0, 0, \dfrac{d}{c}\right)$
So the centroid of triangle $ABC$ is, $\left (\dfrac{d}{3a}, \dfrac{d}{3b}, \dfrac{d}{3c}\right)$
Comparing this with given value $ a=d, b = 2d$ and $ c = 4d$
Hence equation of the plane is 

$4x+2y+z = 4$

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 from a point $P(a,b,c)$ perpendicular $PA$ and $PB$ are drawn to $yz$ and $zx$ planes, find the equation of the plane $OAB$:

  1. $\displaystyle \dfrac { x }{ a } +\dfrac { y }{ b } -\dfrac { z }{ c } =0$
  2. $\displaystyle \dfrac { x }{ a } +\dfrac { y }{ b } +\dfrac { z }{ c } =0$

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

The coordinates of $A$ and $B$ are $(0,b,c)$ and $(a,0,c)$ respectively.

The equation of the plane passing through $O(0,0,0),A(0,b,0)$ and $B(a,0,c)$ is given by 
$\displaystyle \begin{vmatrix} x-0 & y-0 & z-0 \ 0-0 & b-0 & c-0 \ a-0 & 0-0 & c-0 \end{vmatrix}=0\Rightarrow bcx+acy-abz=0$
$\displaystyle \Rightarrow \frac { x }{ a } +\frac { y }{ b } -\frac { z }{ c } =0$

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

The equation of the plane which is parallel to y-axis and cuts off intercepts of length 2 and 3 from x-axis and z-axis is :

  1. $ 3x + 2z = 1$
  2. $ 3x+ 2z = 6 $
  3. $ 2x+ 3z = 6 $
  4. $ 3x+ 2z = 0 $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation of plane parallel to $y-axis$ is, $ax+bz+1=0$      ----- ( 1 )


Here,  $x=2$ and $z=3$


Substituting $x=2$ and $z=0$ in equation ( 1 ),

$\Rightarrow$  $2a+0+1=0$

$\Rightarrow$  $x=\dfrac{-1}{2}$

Substituting $x=0$ and $z=3$ in equation ( 1 ),

$\Rightarrow$  $0+3b+1=0$

$\Rightarrow$  $b=\dfrac{-1}{3}$

Substituting value of $a$ and $b$ equation (  1 ) we get,

$\Rightarrow$  $\dfrac{-1}{2}x-\dfrac{1}{3}z+1=0$

$\Rightarrow$  $3x+2z=6$

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

The sum of Y and Z intercepts of the plane $3x+4y-6z=12$ is ___________.

  1. $10$
  2. $4$
  3. $1$
  4. $5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$3x+4y-6z=12$
$\therefore\dfrac{3x}{12}+\dfrac{4y}{12}+\dfrac{(-6)z}{12}=1$
$\therefore \dfrac{x}{4}+\dfrac{y}{3}+\dfrac{z}{(-2)}=1$
$\therefore$ y intercept $b=3$ and z intercept $c=-2$
$\therefore$ y intercept $+$ z intercept $=3+(-2)=1$

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

The plane $ax+by+cz=1$ meets the coordinate axes in $A, B$ and $C$. The centroid of the triangle is:

  1. $(3a, 3b, 3c)$
  2. $\left( \dfrac { a }{ 3 } ,\dfrac { b }{ 3 } ,\dfrac { c }{ 3 } \right)$
  3. $\left( \dfrac { 3 }{ a } ,\dfrac { 3 }{ b }, \dfrac { 3 }{ c } \right)$
  4. $\left( \dfrac { 1 }{ 3a } ,\dfrac { 1 }{ 3b } ,\dfrac { 1 }{ 3c } \right)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The plane $ax + by + cz = 1$ meets the coordinate axis in $A,B,C$, then the coordinates will be,

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

The equation of the plane in intercept form is,

$\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{c}{z} = 1$

The intercepts that the plane make on the axis is $\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}$

Let C denotes the centroid, then,

$C = \left( {\dfrac{{\dfrac{1}{a} + 0 + 0}}{3},\dfrac{{0 + \dfrac{1}{b} + 0}}{3},\dfrac{{0 + 0 + \dfrac{1}{c}}}{3}} \right)$

Therefore, the coordinates of the centroid will be $\left( {\dfrac{1}{{3a}},\dfrac{1}{{3b}},\dfrac{1}{{3c}}} \right)$.

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

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: $C$