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 mathematics and statistics inverse of a matrix and linear equations application of determinants area of triangle and collinearity of three points applications of determinants

The vertices of the triangle $ABC$ are $(2, 1, 1), (3, 1, 2), (-4, 0, 1)$. The area of triangle is

  1. $\displaystyle \frac{3\sqrt{38}}{2}$
  2. $\sqrt{38}$
  3. $\displaystyle \frac{\sqrt{38}}{2}$
  4. $4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The vertices of the triangle $ABC$ are $(2,1,1),(3,1,2),(-4,0,1)$
$\overrightarrow { AB } =i+k$ and $\overrightarrow { AC } =-6i-j$
now, $\displaystyle \triangle =\frac { \left| \overrightarrow { AB } \times \overrightarrow { AC }  \right|  }{ 2 } =\frac { \left| \left( i+k \right) \times \left( -6i-j \right)  \right|  }{ 2 } =\frac { \left| i-6j-k \right|  }{ 2 } $
Therefore, $\triangle =\dfrac { \sqrt { 38 }  }{ 2 } $

Ans: C

Multiple choice business maths pair of straight lines condition for perpendicular and coincident lines and bisectors of angles pair of straight lines through origin analytical geometry

Equation of line in the place $P=\equiv 2x-y+z-4=0$ which is perpendicular to the line I whose equation is $\dfrac{x-2}{1}=\dfrac{y-2}{-1}=\dfrac{z-3}{-2}$ and which passes through point of intersection of I and P is

  1. $\dfrac{x-2}{3}=\dfrac{y-1}{5}=\dfrac{z-1}{-1}$
  2. $\dfrac{x-1}{3}=\dfrac{y-3}{5}=\dfrac{z-5}{-1}$
  3. $\dfrac{x+2}{2}=\dfrac{y+1}{-1}=\dfrac{z+1}{1}$
  4. $\dfrac{x-2}{2}=\dfrac{y-1}{-1}=\dfrac{z-1}{1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The line must pass through the intersection of the given line and plane, and be perpendicular to the given line. Solving the system of equations for the intersection point and applying the cross product of the normal vector of the plane and the direction vector of the line yields the direction ratios.

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 $\overrightarrow { r } .\left( 3\hat { i } -\hat { j } +\hat { k }  \right) =1$ and $\overrightarrow { r } .\left( \hat { i } +4\hat { j } -2\hat { k }  \right) =2$ is parallel to vector

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

The line of intersection of two planes is parallel to the cross product of their normal vectors. The normals are n1 = (3, -1, 1) and n2 = (1, 4, -2). The cross product is (-1*-2 - 1*4, 1*1 - 3*-2, 3*4 - -1*1) = (2-4, 1+6, 12+1) = (-2, 7, 13).

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

There are two different planes, one passing though the x-axis and the other passing through y-axis. The angle between the planes is $\cfrac{\pi}{4}$. Then locus of a point on the line of intersection of the planes in.

  1. $(x^2+y^2+z^2)x^2=y^2z^2$
  2. $(x^2+y^2+z^2)z^2=x^2y^2$
  3. $(x^2+y^2+z^2)y^2=x^2z^2$

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

This is a classic problem involving the intersection of planes passing through axes. The derivation leads to the relationship between the coordinates and the angle, resulting in the given locus.

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 
$r.\left( {3\hat i - \hat j + \hat k} \right) = 1$ and $r.\left( {\hat i + 4\hat j - 2\hat k} \right) = 2$ is parallel to the vector

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

$\overrightarrow { r } .(3\hat { i } -\hat { j } +\hat { k } )=1$ and $\overrightarrow { r } .(\hat { i } +4\hat { j } -2\hat { k } )=2$

direction vector of normal to plane are,
${ \overrightarrow { b }  } _{ 1 }=3\hat { i } -\hat { j } +\hat { k } ,\quad { \overrightarrow { b }  } _{ 2 }=\hat { i } +4\hat { j } -2\hat { k } $
Let $lmn$ be directions of line of intersection.
$3l-m+n=0$
$l+4m-2n=0$
$\therefore \cfrac { l }{ -2 } =\cfrac { -m }{ -7 } =\cfrac { n }{ 13 } \quad \Rightarrow (l,m,n)=(-2,7,13)$
$\therefore$ vector parallel to line of intersection of planes is,
$\overrightarrow { r } =-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

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. $\dfrac {1}{2}$
  2. $1$
  3. $\dfrac {1}{\sqrt 2}$
  4. $\dfrac {1}{\sqrt 3}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The direction vector of the line of the intersection of the planes $2x+3y+z=1$ and $x+3y+2x=2$ is given by

$n _{1}\times n _{2}$
$=(2i+3j+k)\times (i+3j+2k)$
$=3i-3j+3k$
Hence the unit vector along the direction of the line will be 
$-\dfrac{i-j+k}{\sqrt{3}}$
Thus 
$cos\alpha=\dfrac{1}{\sqrt{3}}$ where $\alpha$ is the angle that the line makes with x axis.

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

The planes $bx-ay=n,cy-bz=1,az-cx=m$ intersect in a line if

  1. $al+bm+cn=0$
  2. $al-bm+cn=0$
  3. $al-bm-cn+1=0$
  4. $al+bm+cn=1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Since the planes

$bx-ay=n$,    $\Rightarrow x=\dfrac { n+ay }{ b } \quad \longrightarrow \left( 1 \right) $

$cy-bz=l$,    $\Rightarrow z=\dfrac { cy-l }{ b } \quad \longrightarrow \left( 2 \right) $
and, $az-cx=m$
intersect in a line, eliminating $x,y,z$ we will get the desired condition.
Substitute $(1)$ and $(2)$ in $(3)$
$a\left( \dfrac { cy-l }{ b }  \right) -c\left( \dfrac { n+ay }{ b }  \right) =m$
$\Rightarrow acy-al-cn-acy=mb$
$\Rightarrow al+cn+bm=0$

$\Rightarrow al+bm+cn=0$

Answer : Option B.
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 $x+3y+2z=2$.

  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

The question is incomplete as it asks for a value without specifying what property of the line L is being calculated (e.g., direction cosines, distance). However, based on the options, it appears to be a calculation related to the direction of the line.

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

The equation of plane through the line of intersection of the planes $2x+3y+4z-7=0, x+y+z-1=0$ and perpendicular to the plane $x-5y+3z-6=0$ is

  1. $x+2y+3z=6$
  2. $x-2y+z=6$
  3. $2x+y+z=5$
  4. $x+2y+6z=3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation of a plane through the intersection of two planes is (2x+3y+4z-7) + k(x+y+z-1) = 0. By using the condition that it is perpendicular to x-5y+3z-6=0, we find k and the resulting equation.

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

The direction cosines of a line parallel to the planes $\displaystyle 3x + 4y + z = 0$ and $\displaystyle x - 2y - 3z = 5$ are

  1. $\displaystyle \left ( -1, \: 1, \: -1 \right )$
  2. $\displaystyle \left ( -\frac{1}{\sqrt{3}}, \: -\frac{1}{\sqrt{3}}, \: \frac{1}{\sqrt{3}} \right )$
  3. $\displaystyle \left ( -\frac{1}{\sqrt{3}}, \: \frac{1}{\sqrt{3}}, \: \frac{-1}{\sqrt{3}} \right )$

  4. no line possible
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given equations of the planes are $\displaystyle 3x + 4y + z = 0$ and $\displaystyle x - 2y - 3z = 5$
required line is parallel to the given palnes,i.e perpendicular to the normals to the planes whose direction ratios are
$(3,4,1)$ and $(1,-2,-3)$ respectively
let $(a,b,c)$ be direction ratios of the line.
$\Rightarrow 3a+4b+c=0 and a-2b-3c=0$
$\Rightarrow a=-b=c$
$\therefore$ direction cosines of the line are $\displaystyle \left ( -\frac{1}{\sqrt{3}}, : \frac{1}{\sqrt{3}}, : -\frac{1}{\sqrt{3}} \right )$ or $\displaystyle \left ( \frac{1}{\sqrt{3}}, : -\frac{1}{\sqrt{3}}, : \frac{1}{\sqrt{3}} \right )$   

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

The variable plane $\displaystyle \left ( 2 \lambda + 1 \right )x + \left ( 3 - \lambda \right )y + z = 4$ always passes through the line 

  1. $\displaystyle \frac{x}{0} = \frac{y}{0} = \frac{x + 4}{1}$
  2. $\displaystyle \frac{x}{1} = \frac{y}{2} = \frac{z}{-3}$
  3. $\displaystyle \frac{x}{1} = \frac{y}{2} = \frac{z - 4}{-7}$

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

Given equation of plane is $ (2 \lambda+1)x+(3- \lambda)y+z=4$


Here $\lambda$ is a variable, we have to eliminate it.


The given equation can be written as $ \lambda(2x-y)+x+3y+z=4$

Now observe that $\lambda$ will be eliminated from the equation, if $2x-y=0$

$\Rightarrow 2x=y$

$\Rightarrow x=\dfrac{y}{2}$

The given equation becomes $x+3y+z=4$

$\Rightarrow 7x+z=4$

$\Rightarrow x=\dfrac{z-4}{-7}$

So, the given plane always passes through $ \dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z-4}{-7}$ irrespective of $\lambda$

Therefore, option $C$ is correct.

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

The equation of the plane which contains the origin and the line of intersection of the planes $\vec r.\vec a=\vec p$ and $\vec r.\vec b=\vec q$ is

  1. $\vec r.\left( \vec p\vec a-\vec q\vec b \right) =0$
  2. $\vec r.\left(\vec p\vec a+\vec q\vec b \right) =0$
  3. $\vec r.\left(\vec q\vec a+\vec p\vec b \right) =0$
  4. $\vec r.\left( \vec q\vec a-\vec p\vec b \right) =0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Any plane through the inetersection of $\vec r.\vec a=\vec p$ and $\vec r.\vec b=\vec q$ is

$r.\left( \vec a-\lambda \vec b \right) =\vec p-\lambda \vec q$   ...(1)
Since it passes through the origin,
$\displaystyle \therefore 0.\left( \vec a-\lambda \vec b \right) =\vec p-\lambda \vec q$
$\Rightarrow \vec p-\lambda \vec q=0$
$\Rightarrow \lambda =\dfrac { \vec p }{ \vec q } $
Putting this value of $\lambda$ in (1), we get
$\displaystyle \vec r.\left( \vec a-\frac { \vec p }{\vec  q } \vec b \right) =\vec p-\frac {\vec  p }{ \vec q } \vec q=0\Rightarrow \vec r.\left( \vec a\vec q-\vec p\vec b \right) =0$
This is the required equation.

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

Consider a plane $x+2y+3z=15$ and a line $\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-2}{4}$ then find the distance of origin from point of intersection of line and plane.

  1. $\dfrac{1}{2}$
  2. $\dfrac{9}{2}$
  3. $\dfrac{5}{2}$
  4. $4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-2}{4}=\lambda$ $\Rightarrow x=2\lambda +1, y=3\lambda -1, z=4\lambda +2$
Now substitution in $x+2y+3z=15$
$\Rightarrow (2\lambda +1)+2(3\lambda -1)+3(4\lambda +2)=15$
$\Rightarrow 2\lambda +1+6\lambda -2+12\lambda +6=15$ $\Rightarrow 20\lambda +5=15$ $\Rightarrow \lambda =\dfrac{1}{2}$
Hence point of intersection is $(2, \dfrac{1}{2}, 4)$
Hence distance from origin is $\sqrt{4+\dfrac{1}{4}+16}=\sqrt{\dfrac{81}{4}}=\dfrac{9}{2}$.

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. $1$
  2. $\displaystyle \frac{1}{\sqrt{2}}$
  3. $\displaystyle \frac{1}{\sqrt{3}}$
  4. $\displaystyle \frac{1}{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given planes are $2x+3y+z=1$ and $x+3y+2z=2$


Direction ratios of line $L$ through their intersection is given by the cross product of direction cosines of plane.


Let the direction ratios of line are represented by $\vec r$, then

$\vec { r } =(2i+3j+k)\times (i+3j+2k)$

$ \vec { r } =\left| \begin{matrix} i & j & k \\ 2 & 3 & 1 \\ 1 & 3 & 2 \end{matrix} \right| $

$ \vec { r } =i(6-3)-j(4-1)+k(6-3)$

$ \vec { r } =3i-3j+3k$

Direction ratio of $x$ axis is $\vec a =i$

$\vec { r } .\vec { a } =\left| \vec { r }  \right| \left| \vec { a }  \right| \cos { \alpha  } $

$ (3i-3j+3k).(i)=(3\sqrt { 3 } )(1)\cos { \alpha  } $

$ 3-0+0=3\sqrt { 3 } \cos { \alpha  } $

$ \cos { \alpha  } =\dfrac { 3 }{ 3\sqrt { 3 }  } =\dfrac { 1 }{ \sqrt { 3 }  } $

So, option C is correct.

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

The vector equation of the line of intersection of the planes $r.(i+2j+3k)=0$ and $r.(3i+2j+k)=0$ is

  1. $r=\lambda (i+2j+k)$
  2. $r=\lambda (i-2j+k)$
  3. $r=\lambda (i+2j-3k)$

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

The line of intersection of the planes $r.(i+2j+3k)=0$ and $r.(3i+2j+k)=0$ is parallel to the vector 

$\left( i+2j+3k \right) \times \left( 3i+2j+k \right) =-4i+8j-4k$
Since both the planes pass through the origin, therefore their line of intersection will also pass through the origin.
Thus, the required line passes through the origin and is parallel to the vector$-4i+8j-4k$
Hence, its equation is
$r=0+\lambda '\left( -4i+8j-4k \right) \Rightarrow r=\lambda \left( i-2j+k \right) $ where $\lambda =-4\lambda'$