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 line of intersection of two planes line and a plane vectors, lines and planes three dimensional geometry maths

The line of intersection of the planes $\overrightarrow { r } .\left( 3i-j+k \right) =1$ and $\overrightarrow { r } .\left( i+4j-2k \right) =2$ is parallel to the vector:

  1. $2i+7j+13k$
  2. $-2i-7j+13k$
  3. $2i+7j-13k$
  4. $-2i+7j+13k$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The line of intersection of the planes $\overrightarrow { r } .\left( 3i-k+k \right) =1$ and $\overrightarrow { r } .\left( i+4j-2k \right) =2$ is perpendicular to each, if the normal vector $\overrightarrow { { n } _{ 1 } } =3i-j+k$ and $\overrightarrow { { n } _{ 2 } } =i+4j-2k$.

$\therefore$ It is parallel to the vector, 
$\overrightarrow { { n } _{ 1 } } \times \overrightarrow { { n } _{ 2 } } =\left( 3i-j+k \right) \times \left( i+4j-2k \right) =2i+7j+13k$

Multiple choice line of intersection of two planes line and a plane vectors, lines and planes three dimensional geometry maths

Consider the planes  $3x - 6y - 2z = 15$  and  $2x + y - 2z = 5$.  Which of the following vectors is parallel to the line of intersection of given plane

  1. $13i + 2j + 15k$
  2. $14i + 2j + 13k$
  3. $13i + 3j + 15k$
  4. $14i + 2j + 15k$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Consider the problem 

Let 
$\begin{array}{l} 3x-6y{ { - } }2z=15 \ 2x+y-2z=5 \end{array}$

For $z=0$
we get 
$x=3,\;y=-1$
Direction ratios of the plane 

$3,-6,-2$ and $2,1,-2$
and 
direction ratios of intersected line
$14,2,15$
Therefore,

$\dfrac{{x - 3}}{{14}} = \dfrac{{y + 1}}{2} = \dfrac{{z - 0}}{2} = \lambda $

of planes So, vectors which parallel to the intersection plane 
$14\hat i + 2\hat j + 15\hat k$

Hence option $D$ is the correct answer.

Multiple choice line of intersection of two planes line and a plane vectors, lines and planes three dimensional geometry maths

The equations of the line of intersection of the planes $\displaystyle x + y + z = 2$ and $\displaystyle 3x - y + 2z = 5$ in symmetric form are

  1. <p class="MsoNormal">$\displaystyle \dfrac{x - \dfrac{7}{4}}{4} = \dfrac{y - \dfrac{1}{4}}{-1} = \dfrac{z}{-3}$</p>
  2. <p class="MsoNormal">$\displaystyle \dfrac{x}{3} = \dfrac{y + \dfrac{1}{3}}{1} = \dfrac{z - \dfrac{7}{4}}{-4}$</p>
  3. $\displaystyle \frac{x}{1} = \frac{3y + 1}{1} = \frac{3z - 7}{-4}$

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

Given, $\vec{n _{1}}=\hat i+\hat j+\hat k$
$\vec{n _{2}}=3\hat i-\hat j+2\hat k$

Therefore the direction vector of the line is parallel to 
$\vec{n _{1}}\times \vec{n _{2}}$
$=(\hat i+\hat j+\hat k)\times(3\hat i-\hat j+2\hat k)$
$=3\hat i+\hat j-4\hat k$
Hence, the equation of the line will be of the form
$\dfrac{x-\alpha}{3}=\dfrac{y-\beta}{1}=\dfrac{z-\gamma}{-4}$

Multiple choice line of intersection of two planes line and a plane vectors, lines and planes three dimensional geometry maths

Consider the planes $\displaystyle 3x-6y-2z=15$ and $\displaystyle 2x+y-2z=5.$ 


Assertion: The parametric equations of the line of intersection of the given planes are $\displaystyle x=3+14t, y=1+2t, z=15t.$ because  

Reason: The vector $\displaystyle 14\hat{i}+2\hat{j}+15\hat{k}$ is parallel to the line of intersection of given planes.

  1. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

  2. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

  3. Assertion is correct but Reason is incorrect

  4. Both Assertion and Reason are incorrect
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$3x-6y-2z=15$

$2x+y-2z=5$
For $z=0$ we get $x=3,y=-1$
Direction vectors of plane are $<3 -6 -2 >$ and $<2, 1 ,-2>$
Then the dr's of line of intersection of planes  is $<14, 2 ,15>$
$\therefore \dfrac{x-3}{14}=\dfrac{y+1}{2}=\dfrac{z-0}{15}=\lambda$
$\Longrightarrow x=14\lambda+3y=2\lambda-1z=15\lambda$
hence Both Assertion and Reason are incorrect 

Multiple choice line of intersection of two planes line and a plane vectors, lines and planes three dimensional geometry maths

The line of intersection of the planes $\displaystyle \bar r (3\hat i - \hat j + \hat k) = 1$ and $\displaystyle \bar r (\hat i + 4\hat j - 2\hat k) = 2$ is parallel to the vector

  1. $\displaystyle -2\hat i + 7\hat j + 13\hat k$
  2. $\displaystyle 2\hat i - 7\hat j - 13\hat k$
  3. $\displaystyle 2\hat i + 7\hat j + 13\hat k$
  4. $\displaystyle 2\hat i + 2\hat j + 13\hat k$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
We know that $\vec{r} = {x}\hat{i} +{y}\hat{j}+{z}\hat{k}$
Hence
The equations of the planes can be written as $3x - y + z - 1 = 0$ and $x + 4y - 2z - 2 = 0$
We find the coordinates of a point lying on their intersection line.
Let $z = t$
$\therefore 3x - y = 1 - t \ ...(1)$
$x + 4y = 2t + 2 \ ...(2)$
Multiplying equation (1) by 4 and adding that to equation (2), we get 
$13x = 4 - 4t + 2t + 2 = 6 - 2t$
$\therefore x = \cfrac{6 - 2t}{13}$
Substituting this in equation (1), we get 
$y = 3x - 1 + t = \cfrac{18 - 6t - 13 + 13t}{13} = \cfrac{5 + 7t}{13}$
Thus, the point can be written as $\left ( \cfrac{6 - 2t}{13}, \cfrac{5 + 7t}{13}, t \right )$
i.e. $\left ( \cfrac{6}{13}, \cfrac{5}{13}, 0 \right ) + \left ( \cfrac{-2}{13}, \cfrac{7}{13}, 1 \right ) t$
$\Rightarrow$ the line of intersection is parallel to $-2\hat{i} + 7\hat{j} + 13\hat{k}$
Multiple choice line of intersection of two planes line and a plane vectors, lines and planes three dimensional geometry maths

Consider three planes$P _1: x-y+z=1$$P _2: x+y-z=-1$$P _3: x-3y+3z=2$Let $L _1, L _2, L _3$ be the lines of intersection of the planes ${P} _{2}$ and ${P} _{3},\ {P} _{3}$ and ${P} _{1}$, and ${P} _{1}$ and ${P} _{2}$, respectively.
STATEMENT-$1$ : At least two of the lines ${L} _{1},\ {L} _{2}$ and ${L} _{3}$ are non-parallel.
and 
STATEMENT -$2$ : The three planes do not have a common point.

  1. Statement-1 is True, Statement -2 is True; Statement-2 is a correct explanation for Statement-1

  2. Statement -1 is True, Statement -2 is True; Statement-2 is NOT a correct explanation for Statement-1

  3. Statement -1 is True, Statement -2 is False

  4. Statement -1 is False, Statement -2 is True
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given three planes are 
${ P } _{ 1 }:x-y+z=1$   ...$(1)$
${ P } _{ 1 }:x+y-z=-1$    ....$(2)$
and ${ P } _{ 1 }:x-3y+3z=2$    ...$(3)$
Solving Eqs. $(1)$ and $(2)$, we have 
$x=0,z=1+y$
which does not satisfy Eq. $(3)$
As, $x-3y+3z=0-3y+3\left( 1+y \right) =3\left( \neq 2 \right) $
$\therefore$ Statement II is true. 
Nest,since we know that direction ratio's of line of intersection of planes ${ a } _{ 1 }x+{ b } _{ 1 }y+{ c } _{ 1 }z+{ d } _{ 1 }=0$ and
${ a } _{ 2 }x+{ b } _{ 2 }y+{ c } _{ 2 }z+{ d } _{ 2 }=0$
$b _{ 1 }{ c } _{ 2 }-{ b } _{ 2 }{ c } _{ 1 },{ c } _{ 1 }{ a } _{ 2 }-{ a } _{ 1 }{ c } _{ 2 },{ a } _{ 1 }{ b } _{ 1 }$
Using above result, we get
Direction ratio's of lines ${ L } _{ 1 },{ L } _{ 2 }$ and ${ L } _{ 3 }$ are $o,2,2;0,-4,-4;0,-2,-2$ respectively. 
$\Rightarrow $ All the three lines ${ L } _{ 1 },{ L } _{ 2 }$ and ${ L } _{ 3 }$ are parallel pairwise. 
$\therefore $ Statement I is false. 

Multiple choice line of intersection of two planes line and a plane vectors, lines and planes three dimensional geometry maths

Let L be the line of intersection of the planes $2x + 3y + z = 1$ and $x + 3y + 2z = 2$. If L makes an angle $\alpha$ with the positive x-axis, then $\cos \alpha$ equals

  1. $\dfrac{1}{\sqrt{3}}$
  2. $\dfrac{1}{2}$
  3. $1$
  4. $\dfrac{1}{\sqrt{2}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
We have 
$2x+3y+z=1$
$\Longrightarrow 2x+3y=1-z$      ...(i)
$x+3y+2z=2$
$\Longrightarrow x+3y=2-2z$    ...(ii)
$(i)-(ii)$
$\Longrightarrow 2x+3y-x-3y=1-z-2+2z$
$\Longrightarrow x=z-1$
$\Longrightarrow z=\cfrac{x+1}{1}$
Putting values of $z$ in (ii), we get,
$z-1+3y=2-2z$
$\Longrightarrow 3y=2-2z-z+1$
$3y=-3(z-1)$
$\Longrightarrow y=-(z-1)=-z+1$
$\therefore z=\cfrac{y-1}{1}$
Hence, we have $\cfrac{x+1}{1}=\cfrac{y-1}{1}=\cfrac{z}{1}$
thus $\cos\alpha=\cfrac{a}{(\sqrt {(a^2+b^2+c^2)})}$
$\therefore \cos\alpha=\cfrac{1}{\sqrt3}$
Multiple choice line of intersection of two planes line and a plane vectors, lines and planes three dimensional geometry maths

Find the angle between the line of intersection of the planes $\overrightarrow { r } .\left( i+2j+3k \right) =0$ and $\overrightarrow { r } .\left( 3i+2j+3k \right) =0$ with coordinate axes

  1. with $x$-axis $\displaystyle \dfrac { \pi }{ 2 } $
  2. with $y$-axis $\displaystyle \cos ^{ -1 }{ \left( \dfrac { 3 }{ \sqrt { 13 } } \right) } $
  3. with $y$-axis $\displaystyle \cos ^{ -1 }{ \left( \dfrac { 2 }{ \sqrt { 13 } } \right) } $

  4. all of these
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

Direction ratios: $\displaystyle \begin{vmatrix} i & j & k \ 1 & 2 & 3 \ 3 & 2 & 3 \end{vmatrix}=\left( 0,6,4 \right) $ or $(-,3,-2)$

Therefore angle with $x$-axis: $\displaystyle \dfrac { \pi  }{ 2 } $
with $y$- axis $\displaystyle \cos ^{ -1 }{ \left( \dfrac { 3 }{ \sqrt { 13 }  }  \right)  } $
with $z$-axis $\displaystyle \cos ^{ -1 }{ \left( \dfrac { -2 }{ \sqrt { 13 }  }  \right)  } $

Multiple choice maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

The cartesian equation of the plane which is at a distance of 10 unite from the original and perpendicular to the vector i + 2j -2k is 

  1. x+2y+2z = 30

  2. x - 2y - 2z =30

  3. x - 2y + 2z = 30

  4. x+2y-2z = 30
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The equation of a plane at distance d from origin with normal vector n is r dot (n/|n|) = d. Here n = i + 2j - 2k, so |n| = sqrt(1+4+4) = 3. The equation is x + 2y - 2z = 10 * 3 = 30.

Multiple choice maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

Find the equation of the plane through the points $(1, 0, -1), (3, 2, 2)$ and parallel to the line $\dfrac{x-1}{1}=\dfrac{y-1}{-2}=\dfrac{z-2}{3}$.

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

$L:\cfrac { x-1 }{ 1 } =\cfrac { y-1 }{ -2 } =\cfrac { z-2 }{ 3 } \ P(1,0,-1),Q(3,2,2)$

$ \therefore$ Direction of  $\overrightarrow { PQ } =2\hat { i } +2\hat { j } +3\hat { k } $
Plane is parallel to line $L$ & contains $\overrightarrow { PQ } $. (normal to plane $\bot$ to $PQ$ & $L$)
$\therefore \overrightarrow { n } =\left| \begin{matrix} \hat { i }  & \hat { j }  & \hat { k }  \ 2 & 2 & 3 \ 1 & -2 & 3 \end{matrix} \right| =(12)\hat { i } -3\hat { j } +(-6)\hat { k } $
direction ratios of normal are $(4,-1,-2)$
$\therefore$ Equation plane passing through $(1,0,-1)$ and having directions of normal $(4,-1,-2)$.
$4x-y-2z=6$

Multiple choice maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

The equation of the plane passing through the straight line $\dfrac{x-1}{2}=\dfrac{y+1}{-1}=\dfrac{z-3}{4}$ and perpendicular to plane $x+2y +z=12$ is:

  1. $9x+2y-5z+8 =0$
  2. $9x +2y -5z +10=0$
  3. $9x-2y +5z +6=0$
  4. $9x -2y -5z+4=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Let the DR of the normal of the plane be $<a, b, c>$
Since it passes through a line.
$\therefore$ Normal of the plane must be perpendicular to the line
$\therefore a + 2b + c = 0$ ...... $(1)$ ($\perp$ to another plane)
$2a - b + 4c = 0$ ...... $(2)$
On solving :
$\dfrac{a}{8 + 1} = \dfrac{-b}{4 - 2} = \dfrac{c}{-1 - 4}$
$\Rightarrow \dfrac{a}{9} = \dfrac{b}{-2} = \dfrac{c}{-5}$
$a = 9, b = -2, c = -5$
Any point on the straight line
$(2\alpha + 1, -\alpha - 1, 4\alpha + 3)$
putting $\alpha = 1$
$(3, -2, 7)$
$\therefore$ Equation of the line plane with normal DR $(9, -2, -5)$ and passing through $(3, -2, 7)$
$\therefore 9(x - 3) - 2(y + 2) - 5(z - 7)=0$
$\Rightarrow 9x - 27 - 2y - 4 - 5z + 35 = 0$
$\Rightarrow 9x - 2y - 5z + 4 = 0$

Multiple choice maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

Equation of the plane containing the straight lines $\dfrac{x}{2} = \dfrac{y}{3} = \dfrac{z}{4}$ and perpendicular to the plane containing the straight lines  $\dfrac{x}{3} = \dfrac{y}{4} = \dfrac{z}{2}$ and $\dfrac{x}{4} = \dfrac{y}{2} = \dfrac{z}{3}$

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

Vector normal to plane $P _1$ is $\overrightarrow{n _1}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\3&4&2\4&2&3\end{vmatrix}$


                                                       $=\hat{i}(12-4)-\hat{j}(9-8)+\hat{k}(6-8)$
                                                       $=8\hat{i}-\hat{j}-10\hat{k}$
Plane $P _2$ is perpendicular to this plane and it leaking the line $\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{2}{4}$

$\overrightarrow{n _2}$ is normal to

$\overrightarrow{n _2}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\8&-1&-10\2&3&4\end{vmatrix}$

       $=\hat{i}(-4+30)-\hat{j}(32+20)+\hat{k}(24+2)$

       $=26\hat{i}-52\hat{j}+26\hat{k}$

This plane $T _2$ having normal $n _2$ passing through $(0,0,0)$ is $26x-52y+26z=0$ $i.e.$ $x-2y+z=0$

Multiple choice maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

The direction cosines of the normal to the plane $x+2y-3z+4=0$ are

  1. $\cfrac { -1 }{ \sqrt { 14 } } ,\cfrac { -2 }{ \sqrt { 14 } } ,\cfrac { 3 }{ \sqrt { 14 } } $
  2. $\cfrac { 1 }{ \sqrt { 14 } } ,\cfrac { 2 }{ \sqrt { 14 } } ,\cfrac { 3 }{ \sqrt { 14 } } $
  3. $\cfrac { -1 }{ \sqrt { 14 } } ,\cfrac { 2 }{ \sqrt { 14 } } ,\cfrac { 3 }{ \sqrt { 14 } } $
  4. $\cfrac { 1 }{ \sqrt { 14 } } ,\cfrac { -2 }{ \sqrt { 14 } } ,\cfrac { -3 }{ \sqrt { 14 } } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Clearly, the normal to the plane has DR's $\equiv(1, 2, -3)$

$\therefore $ DCs  are $\equiv \pm \left(\dfrac{1}{\sqrt{1^2 + 2^2 + (-3)^2}} , \dfrac{2}{\sqrt{14}} , \dfrac{-3}{\sqrt{14}}\right)$
$= \pm \left(\dfrac{1}{\sqrt{14}} , \dfrac{2}{\sqrt{14}} , \dfrac{-3}{\sqrt{14}} \right)$
$\therefore A$