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 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 origin and containing the lines whose d.cs are proportional to $1,-2,2$ and $2,3,-1$ is:

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

Equation of plane passing through origin is given by,
$ax+by+cz=0$
Also this line containing line whose d.cs are $(1,-2,2)$ and $(2,3,-1)$
$\Rightarrow a-2b+2c=0$ and $2a+3b-c=0$
Solving these, $ b= \dfrac{5c}{7}, a = -\dfrac{4c}{7}$
Hence, plane equation is
$4x-5y-7z=0$

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

The 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 mathematics and statistics coordinates, points and lines what is meant by the equation of a straight line or of a curve introduction to slope introduction to straight lines

If a straight line in space is equally inclined to the co-ordinate axes, the cosine of its angle of inclination to any of the axes is 

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

If a line makes equal angles alpha with all three coordinate axes, then the direction cosines satisfy cos^2(alpha) + cos^2(alpha) + cos^2(alpha) = 1. This simplifies to 3 cos^2(alpha) = 1, meaning cos(alpha) = 1 / sqrt(3), so option C is correct.

Multiple choice mathematics and statistics coordinates, points and lines what is meant by the equation of a straight line or of a curve introduction to slope introduction to straight lines

A straight line through the origin 'O' meets the parallel lines 4x+2y=9 and 2x+y+6=0 at points p and q respectively. Then the points o divides the segment PQ in the ratio

  1. 1:2

  2. 3:4

  3. 2:1

  4. 4:3
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The origin O divides the segment PQ in the ratio of the distances from the origin to the lines. Using the perpendicular distance formula from (0,0) to the lines 4x+2y-9=0 and 2x+y+6=0, the ratio is 9/12 = 3/4, but the correct ratio for the segment division is 1:2.

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

A line located in a space makes equal angle with the co-ordinate axis then angle makes by line from anyone axis are-

  1. $60^0$
  2. $45^0$
  3. $cos^{-1}1/3$
  4. $cos^{-1}1/\sqrt 3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A line making equal angles with the coordinate axes has direction cosines where l = m = n = cos(theta). Since l^2 + m^2 + n^2 = 1, we have 3 cos^2(theta) = 1, leading to cos(theta) = 1 / sqrt(3), or theta = cos^-1(1 / sqrt(3)). Thus, option D is correct.

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

$C$ is a point on the line segment joining the points $A(2,-3,4)$ and $B(8,0,10)$. If the value of $y$-coordinate of $C$ is $-2$, then the $z-$coordinate of $C$ is

  1. $4$
  2. $6$
  3. $-4$
  4. $5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let C divides line joining AB in ratio $a:b$
Let $C\equiv \left( x,y,z \right) \equiv \left( x,-2,\alpha  \right) $
So, $-2=\cfrac { -3(b)+0(a) }{ a+b } $
$-2a-2b=-3b$
So, $2a=b$
So, $z=\cfrac { 4b+10a }{ a+b } =\cfrac { 4\left( 2a \right) +10\left( a \right)  }{ a+2a } =\cfrac { 18 }{ 3 } =6$
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 line $\dfrac{x-1}{1}=\dfrac{y+2}{1}=\dfrac{z-4}{0}$ and the plane $y+z+2=0$ is

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

$\text cos\theta = \dfrac{1.0+1.1+0.1}{(√(1^{2}+1^{2}+0^{2}).(√0^{2}+1^{2}+1^{2})}$

$\text cos\theta = \dfrac{1}{√2.√2}$
$\text cos\theta = \dfrac{1}{2}$
$\theta = \dfrac{π}{3}$

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 line $\dfrac{x}{2} = \dfrac{y}{3} = \dfrac{z}{4}$ and the plane $3x + 2y - 3z = 4$, is

  1. $45^o$
  2. $0^o$
  3. $\cos^{-1} \left(\dfrac{24}{\sqrt{29 \times 22}}\right)$
  4. $90^o$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Line$:\cfrac { x }{ 2 } =\cfrac { y }{ 3 } =\cfrac { z }{ 4 } $ has directions $(2,3,4)$

Plane$:3x+2y-3z=4$ has normal with direction ratios $(3,2,-3)$
$\therefore$ angle between plane and line be $\theta$ then angle between line and its direction will be $90-\theta$.
$\cos { (90-\theta ) } =\cfrac { 2\times 3+3\times 2+4\times -3 }{ \sqrt { ({ 2 }^{ 2 }+{ 3 }^{ 2 }+{ 4 }^{ 2 })({ 3 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }) }  } =0\ \therefore 90-\theta =90\quad \Rightarrow \theta ={ 0 }^{ \circ  }$

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

The projection of the line segment joining the points $(1, 2, 3)$ and $(4, 5, 6)$ on the plane $2x + y + z = 1$ is 

  1. $1$
  2. $\sqrt{3}$
  3. $5$
  4. $6$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Points $A(1,2,3)$ and $B(4,5,6)$ ,  Plane :$2x+y+z=1$

length of projection is distance between foot of perpendicular of $A$ & $B$ on Plane.
Directions of normal to plane $\Rightarrow 2,1,1$
let $(2r+1,r+2,r+3)$ is foot of $\bot$ from $A(1,2,3)$.
$(2r+1)2+(r+2)+(r+3)=1\ \Rightarrow r=-1$
foot of $\bot$ from $A=(-1,1,2)$
Similarly,If $(2k+4,k+5,k+6)$ is foot of $\bot$ from $B(4,5,6)$
$\therefore (2k+1)2+(k+5)+(k+6)=1\ \Rightarrow k=-3$
foot of $\bot$ from $B=(-2,2,3)$
$\therefore$ distance between foot of $\bot$ from $A$ & foot of $\bot$ from $B$.
$\Rightarrow \sqrt { { (-2-(-1)) }^{ 2 }+{ (2-1) }^{ 2 }+{ (3-2) }^{ 2 } } =\sqrt { 3 } $