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

The vector equation of the plane passing through the planes $r.(i+j+k)=6$ and $r.(2i+3j+4k)=-5$ and the point $(1,1,1)$ is

  1. $r.(20i+23j+26k) = 69$
  2. $r.(2i+23j+26k) = 69$
  3. $r.(2i+2j+3k) = 69$
  4. $r.(20i+3j+26k) = 69$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The vector equation of plane passing through the intersection of planes $\vec r.\vec {n _1}=d _1$ and $\vec r.\vec {n _2}$ and also through the point $x _1,y _1,z _1$ is 


$\vec r.(\vec {n _1}+\lambda \vec {n _2})=d _1+\lambda d _2$

According to question plane passes through 

$\vec r.(\hat i+\hat j+\hat k)=6$

comparing it with $\vec r.\vec {n _1}=d _1$

$\vec {n _1}=\hat i+\hat j+\hat k$
And 
$d _1=6$
Now, other plane by it also passes 

$\vec r.(2\hat i+3\hat j+4\hat k)=-5$

$=\vec r.(-2\hat i-3\hat j - 4\hat k)=5$
Comparing it with $\vec r.\vec {n _2}=d _2$

$\vec {n _2}=-2\hat i-3\hat j-4\hat k$
And 
$d _2=5$
Now, equation of the required plane 

$\vec r.[(\hat i+\hat j+\hat k)-\lambda (2\hat i+3\hat j+4\hat k)]=5\lambda +6$   ----- (i)
Now, 
Putting $\vec r=x\hat i+y\hat j+z\hat k$

$(x\hat i+y\hat j+z\hat k).[(\hat i+\hat j+\hat k)-\lambda (2\hat i+3\hat j+4\hat k)]=5\lambda +6$

$(1-2\lambda)x+(1-3\lambda)y+(1-4\lambda)z=5\lambda +6$   ----   (ii)

Since it passes through $(1,1,1)$ 
Therefore,

$(1-2\lambda)1+(1-3\lambda)1+(1-4\lambda)1=5\lambda +6$
Hence 
$\lambda =\dfrac{-3}{14}$

Put the value of $\lambda $ in (i)

$\vec r.[(\hat i+\hat j+\hat k)-(\dfrac{-3}{14}) (2\hat i+3\hat j+4\hat k)]=5(\dfrac{-3}{14}) +6$

$\vec r.[(1+\dfrac{6}{14})\hat i+(1+\dfrac{9}{14})\hat j+(1+\dfrac{12}{14}\hat k)]=\dfrac{69}{14}$

$\vec r.(20\hat i + 23\hat j + 26\hat k) = 69$

So, the required equation of plane is $\vec r.(20\hat i + 23\hat j + 26\hat k) = 69$

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 plane $\bar{r}.(2, -3, 4) = 5$ is _____

  1. $3y - 2x -4z + 5 =0$
  2. $2x - 3y + 4z =0$
  3. $2x - 3y + 4z +5 =0$
  4. $\displaystyle \frac{x - 1}{2} = \frac{y-1}{-3} = \frac{z-1}{4}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$

$\vec{r}.(2\hat{i}-3\hat{j}+4\hat{k})=5$
$\implies 2x-3y+4z=5$
$\implies 3y-2x-4z+5=0$

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

The equation(s) of the plane,  which is/are equally inclined to the lines $\dfrac {x-1}{2}=\dfrac {y}{-2}=\dfrac {z+2}{-1}$ and $\dfrac {x+3}{8}=\dfrac {y-4}{1}=\dfrac {z}{-4}$ and passing through the origin is/are

  1. $14x-5y-7z=0$
  2. $2x+7y-z=0$
  3. $3x-4y-z=0$
  4. $x+2y-5z=0$
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

The planes equally inclined to both the lines will have their normals along the angle bisectors of the two lines.  
The unit vector along the first line is $ \dfrac{2 \hat{i} -2 \hat{j} -\hat{k} } { 3 }$.
The unit vector along the second line is  $ \dfrac{ 8\hat{i}+ \hat{j} -4\hat{k} }{ 9} $.

The vectors along the angle bisectors can be written as : 
$ (\dfrac{2}{3} +\dfrac{8}{9}) \hat{i}  + ( \dfrac{-2}{3} + \dfrac{1}{9} )\hat{j} + (\dfrac{-1}{3} + \dfrac{-4}{9} )\hat{k}  = \dfrac{1}{9} (14\hat{i} -5\hat{j} -7\hat{k}) =0 $ 
and 
$ (\dfrac{2}{3} -\dfrac{8}{9}) \hat{i}  + ( \dfrac{-2}{3} - \dfrac{1}{9} )\hat{j} + (\dfrac{-1}{3} - \dfrac{-4}{9} )\hat{k} = \dfrac{1}{9} (-2\hat{i} -7\hat{j} +\hat{k} ) $.

Hence, options A and B represent equations of planes which have normals along the angle bisector and pass through the origin. 

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

A plane through the line $\displaystyle \frac{x - 1}{1} = \frac{y + 1}{-2} = \frac{z}{1}$ has the equation

  1. $\displaystyle x + y + z = 0$
  2. $\displaystyle 3x + 2y - z = 1$
  3. $\displaystyle 4x + y - 2z = 3$
  4. $\displaystyle 3x + 2y + z = 0$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation
A plane eqn through the line $\dfrac{x-1}{1}=\dfrac{y+1}{-2}=\dfrac{z}{1}$      ...(i)
is given by $A(x-1) +B(y+1)+Cz=0$       ...(ii)
where $A,B,C$  are direction ratio which will perpence to line 
hence 
$A-2B+C=0$
so possible volves of $(A,B,C) $ are $(1,1,1)$ and $(4,1,-2)$
so eq of plane from (ii)
$\Rightarrow  1(x-1)+1(y+1)+1(z)=0$
$\Rightarrow x-1+y+1+z=0$
$\Rightarrow x+y+z=0$
          $ or $
$4(x-1)+(y+1)-2z=0$
$\Rightarrow 4x-4+y+1-2z=0$
$\Rightarrow 4x+y-2z-3=0$
$4x+y-2z=3$
So, here eq of plane is $x+y+z=0$ 
or $4x+y-2z=3$

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 a plane through the line $\displaystyle \frac{x\, -\, 1}{2}= \frac{y\, -\, 2}{3}= \frac{z\, -\, 3}{4}$ and parallel to a coordinate axis is

  1. $4y \:-\:3z\:+\:1 =\:0$
  2. $2x\:-\:z\:+\:1 =\:0$
  3. $3x\:-\:2y\:+\:1 =\:0$
  4. $2x\:+\:3y\:+\:1=\:0$
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation
General equation of a plane is
$ax+by+cz+d=0$.........where a,b,c,d are constants taking any value

Let us consider the plane parallel to x-axis. 

Hence, the equation of the plane can be written as $ay+bz =1 $. 

The plane passes through $(1,2,3)$ and the normal to the plane is perpendicular to the vector along the line. 

Hence, 
$ 2a+3b =1$ and $ 3a+4b =0 $

On solving we get, $a= -4$ and $b=3$. 

Hence, $-4y+3z =1$ or $4y-3z +1 =0 $.

Let us consider the plane parallel to the y-axis.

Hence, the equation of the plane can be written as: 

$ cx+dz = 1 $

$c+3d =1 $ and $ 2c+4d =0 $. 

$\Rightarrow c = -2 $ and $d=1$ 

Hence, the equation of the plane is $ 2x-z+1 =0 $.

Similarly, we can find the equation for the plane parallel to the z-axis. The equation of the plane can be written as $ex+fy = 1 $.

After substituting the value of the point $(1,2,3)$ and using the information of the normal to the plane being perpendicular to the line, we get the equation of the plane as $ 3x-2y+1 =0$.

Hence, all three options are correct.
Multiple choice maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

Consider the plane passing through the points $A(2, 2, 1), B(3, 4, 2)$ and $C(7, 0, 6)$.
Which one of the following points lies on the plane?

  1. $(1, 0, 0)$
  2. $(1, 0, 1)$
  3. $(0, 0, 1)$

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

The plane passing through the points $A(2,2,1),B(3,4,2) and C(7,0,6)$.


We can get two vectors in the plane by subtracting pairs of points in the
plane :
$[ 2, 2, 1 ] - [ 3 ,4, 2 ] = [ -1, -2, -1 ]$

$[ 7 ,0, 6 ] - [ 3, 4 ,2 ] = [ 4, -4 ,4 ]$

The cross product of these two vectors will be in the unique direction orthogonal to both, and hence in the direction of the normal vector
to the plane


$ [ -1 ,-2, -1 ] \times  [ 4 ,-4 ,4 ] = [ 1, 0 ,0 ] $

hence the points lies on the plane of $(1 , 0 , 0)$



Multiple choice maths linear graphs quadrants the cartesian plane the cartesian system

A Cartesian plane consists of two mutually _____ lines intersecting at their zeros.  

  1. perpendicular

  2. parallel
  3. at angle of $60^o$
  4. at angle of $30^o$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The Cartesian plane consists of two perpendicular and directed lines whose intersection point is the zero point for both the lines.

Option A is correct.

Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

An angle between the plane , $x+y+z=5$ and the line of intersection of the planes, $3x+4y+x-1=0$ and $5x+8y+2z+14=0$

  1. $\sin^{-1}(\sqrt{3/17})$
  2. $\cos^{-1}(\sqrt{3/17})$
  3. $\cos^{-1}(3/\sqrt{17})$
  4. $\sin^{-1}(3/\sqrt{17})$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We have

$\overrightarrow {{\pi _1}} :x + y + z = 5$
$\overrightarrow {{r _1}} :3x + 4y + z - 1 = 0$
$\overrightarrow {{r _2}} :5x + 8y + 2z + 14 = 0$
Line of intersection of planes $\parallel \,\,to\,\,\overrightarrow {{r _1}}  \times \overrightarrow {{r _1}} $
By the helps of determinate
$\left| { \begin{array} { *{ 20 }{ c } }{ \widehat { i }  } & { \widehat { j }  } & { \widehat { k }  } \ 3 & 4 & 1 \ 4 & 8 & 2 \end{array} } \right| $
$ = \widehat i\left( 0 \right) - \widehat j\left( {6 - 5} \right) + \widehat k\left( {24 - 20} \right)$
$ =  - \widehat j + 4\widehat k$
Now,
$\sin \theta  = \frac{{ - 1 + 4}}{{\sqrt 3 \sqrt {17} }} = \sqrt {\frac{3}{{17}}} $
$\therefore \theta  = {\sin ^{ - 1}}\sqrt {\frac{3}{{17}}} $
Hence the option $A$ is the correct answer.

Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

The line $\dfrac {x - 2}{3} = \dfrac {y - 3}{4} = \dfrac {z - 4}{5}$ is parallel to the plane.

  1. $3x + 4y + 5z = 7$
  2. $2x + y - 2z=0$
  3. $x + y - z = 2$
  4. $2x + 3y$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Direction ratios of given line are $3,4,5$ direction ratios of the perpendicular to the  plane $2x+y-2z=0$ are $2,1,-2$


Now
$\begin{array}{l} 2\times 3+1\times 4+\left( { -2 } \right) \times 5 \ =6+4-10 \ =0 \end{array}$

perpendicular to the plane is perpendicular to the given line
so, the plane $2x+y-2z$ is parallel to the given line.

Hence, the correct option is $B$

Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

The line $\cfrac{x+3}{3}=\cfrac{y-2}{-2}=\cfrac{z+1}{1}$ and the plane $4x+5y+3z-5=0$ intersect at a point

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

Let $\dfrac{x+3}{3}=\dfrac{y-2}{-2}=\dfrac{z+1}{1}=k$


On solving, we get

$\Rightarrow x=3k-3$

$\Rightarrow y=-2k+2$ 

$\Rightarrow z=k-1$

On substituing these values in the given plane equation we get,

$\Rightarrow 4x+5y+3z-5=0$

$\Rightarrow 4(3k-3)+5(-2k+2)+3(k-1)-5=0$

On simpliying we get,

$\Rightarrow 5k=10$

$\Rightarrow k=2$

Substituting this value of $k$ in equations of $x,y,z$ we get

$\Rightarrow x=3,y=-2,z=1$

Hence point of intersection is $(3,-2,1)$

Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

If the line $\cfrac{x-1}{2}=\cfrac{y+3}{1}=\cfrac{z-5}{-1}$ is parallel to the plane $px+3y-z+5=0$, then the value of $p$

  1. $2$
  2. $-2$
  3. $\cfrac{1}{2}$
  4. $\cfrac{1}{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
line $11$ plane 
$\therefore$ line $\bot$ normal to plane 
$\therefore (2)(P)+(1)(3)+(-1)(-1)=0$  
$\therefore 2p + 3 + 1 =0$
$\therefore P=-2$
Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

The angle between the plane $2 x - y + z = 6$ and a perpendiculars to the planes $x + y + 2 z = 7$ and $x - y = 3$ is

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

The normal to the plane 2x - y + z = 6 is n1 = (2, -1, 1). The perpendiculars to the other two planes are their normals: n2 = (1, 1, 2) and n3 = (1, -1, 0). The cross product n2 x n3 gives the direction vector of the line perpendicular to both planes, which is (2, 2, -2). The dot product of n1 and (2, 2, -2) is 4 - 2 - 2 = 0, meaning the plane is parallel to the line, but the question asks for the angle between the plane and the perpendiculars. Given the orthogonality, the angle is pi/2.

Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

Gives the line $\displaystyle L:\frac { x-1 }{ 3 } =\frac { y+1 }{ 2 } =\frac { z-3 }{ -1 } $ and the plane $\pi :x-2y=0$. Of the following assertions, the only one that is always true is:

  1. $L$ is $\bot$ to $\pi$
  2. $L$ lies in $\pi$
  3. $L$ is parallel to $\pi$

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

Since $3\left( 1 \right) +2\left( -2 \right) +\left( -1 \right) \left( -1 \right) =3-4+1=0$

$\therefore$ given line is $\bot$ to the normal to the plane i.e. given line is parallel to the given plane.
Also, $(1,-1,3)$ lies on the plane $x-2y-z=0$
$1-2\left( -1 \right) -3=0\Rightarrow 1+2-3=0$
which is true
$\therefore L$ lies in plane $\pi$

Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

Consider a plane $x + y - z = 1$ and the point $A(1, 2, -3)$. A line $L$ has the equation $x = 1 + 3r$, $y = 2 - r$, $z = 3 + 4r$

The coordinate of a point $B$ of line $L$, such that $AB$ is parallel to the plane, is

  1. $(10, -1, 15)$
  2. $(-5, 4, -5)$
  3. $(4, 1, 7)$
  4. $(-8, 5, -9)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let $\vec { OB } =\left( 1+3r \right)\hat i+\left( 2-r \right)\hat j+\left( 3+4r \right)\hat k$
$\vec { AB } =\vec { OB } -\vec { OA } =\left( 1+3r \right)\hat i+\left( 2-r \right)\hat j+\left( 3+4r \right)\hat k-\hat i-2\hat j+3\hat k=3r\hat i-r\hat j+\left( 6+4r \right)\hat k$
Since, $\vec { AB }$ is parallel to $x+y-z=1$
Therefore, $\vec { AB } .\left(\hat i+\hat j-\hat k \right) =0$
$\Rightarrow \left( 3r\hat i-r\hat j+\left( 6+4r \right)\hat k \right) .\left(\hat i+\hat j-\hat k \right)=0 $
$\Rightarrow 3r-r-6-4r=0$
$\Rightarrow r=-3$
Therefore, $\vec { OB } =-8i+5j-9k$

Ans: D