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 introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

The equation of plane passing through $(-1,0,-1)$ parallel to $xz$ plane is

  1. $y=-2$
  2. $y=0$
  3. $-x-z=0$

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

Given that the plane is parallel to $xz$ plane and the plane passes through $(-1,0,-1)$

Since the plane is parallel to $xz$ plane , the $y$ coordinate should be constant
Given that it passes through point $(-1,0,-1)$ , therefore the plane lies on $xz$ plane
Therefore the equation of plane is $y=0$
The correct options are $B$

Multiple choice maths introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

The planes $2x-y+4z=5$ and $5x-2.5y+10z=6$ are

  1. Parallel

  2. Perpendicular

  3. Intersect
  4. intersect $x$ axis
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Planes are $2x-y+4z=5$ 

and $5x-2.5y+10z=6$
Multiply both sides by 2 to the second equation
$\Rightarrow 10x-5y+20=12$
Now divide both sides by $2$
$\Rightarrow 2x-y+4z=\dfrac{12}{5}$

Clearly both planes are parallel 

Multiple choice maths introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

In a three-dimensional space, the equation $3x - 4y = 0$ represents.

  1. A plane containing $Z-axis$
  2. A plane containing $X-axis$
  3. A plane containing $Y-axis$
  4. Passing through $(0, 0)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If we consider the z-part also, then the equation is $3x-4y+0z=0$

Which means on putting any value of z equation will have no effect as $0\times z=0$
$\therefore$ it will be a plane containing $Z-axis$.($\because$ it will pass through all points z if it satisfy condition for$ (x,y) $)
(D) is not right because its plane and not a line so it will pass through $(0,0,0)$ not $(0,0)$.
Hence, $(A)$


Multiple choice maths introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

Which of the following is true for a plane?

  1. A locus is called a plane if the line joining any two arbitrary points on the locus is also a part of the locus.
  2. Value of $y$ in a $zx$ plane is non-zero.
  3. Value of $z$ in a $xy$ plane is zero.

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

Option A and C are correct 

A locus is called a plane if the line joining any two arbitrary points on the locus is also a part of the locus. and also Value of z in a xy plane is zero.

Multiple choice maths vectors: lines in two and three dimensions distance from a point to line perpendicular distance of a point from a plane the distance from a point to a line

The length of the perpendicular drawn from the points $(5,4,-1)$ to the line $\overline r  = \widehat i + \lambda \left( {2\widehat i + 9\widehat i + 5\widehat k} \right)$ is

  1. $\dfrac{{\sqrt {2190} }}{{110}}$
  2. $\sqrt { \frac { { 2199 } }{ { 110 } } } $
  3. $\sqrt { \frac { { 2109 } }{ { 110 } } } $
  4. $\dfrac{{\sqrt {23190} }}{{110}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

According to the question:

$\begin{array}{l} let\, the\, point\, (5,4,-1)\, \, be\, \, P\, and\, the\, point\, through\, which\, the\, \, line\, passes\, \, be\, \, Q\, (1,0,0).\, \, \, \,  \ the\, line\, is\, parallel\, to\, the\, vector:\, \, \overrightarrow { r } =\left( { 2\hat { i } +9\hat { i } +5\hat { k }  } \right)  \ Now, \ \overrightarrow { PQ } =-4\hat { i } -4\widehat { j } +\hat { k }  \ \therefore \, \, \, \overrightarrow { r\,  } \, \times \overrightarrow { PQ } =\left| \begin{array}{l} \, \, \hat { i } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \widehat { j } \, \, \, \, \, \, \, \, \, \, \, \widehat { k }  \ \, \, 2\, \, \, \, \, \, \, \, \, \, \, \, \, \, 9\, \, \, \, \, \, \, \, \, \, \, 5\, \,  \ \, -4\, \, \, \, \, \, \, -4\, \, \, \, \, \, \, \, \, \, 1 \end{array} \right| \, \, \, \, \, \, \, \,  \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\, 29\, \hat { i } \, -\, \, 22\, \widehat { j } \, +28\, \widehat { k }  \ \Rightarrow \left| { \, \overrightarrow { r\,  } \, \times \overrightarrow { PQ }  } \right| =\sqrt { \, { { (29) }^{ 2 } }\, +{ { (-\, \, 22) }^{ 2 } }\, +{ { (28) }^{ 2 } } }  \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\sqrt { 841+484+784 } =\sqrt { 2109 }  \ \left| { \overrightarrow { r\,  } \,  } \right| =\sqrt { { 2^{ 2 } }+{ 9^{ 2 } }+{ 5^{ 2 } } }  \ \, \, \, \, \, \, \, =\sqrt { 4+81+25 } =\sqrt { 110 }  \ d=\frac { { \, \left| { \overrightarrow { r\,  } \, \times \overrightarrow { PQ }  } \right|  } }{ { \left| { \overrightarrow { r\,  } \,  } \right|  } } =\frac { { \sqrt { 2109 }  } }{ { \sqrt { 110 }  } } =\sqrt { \frac { { 2109 } }{ { 110 } }  }  \ so\, that\, the\, correct\, option\, is\, C. \end{array}$

Multiple choice maths vectors: lines in two and three dimensions distance from a point to line perpendicular distance of a point from a plane the distance from a point to a line

Perpendicular distance between the plane $  2 x-y+2 z=1  $ and origin is 

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

The perpendicular distance of a plane and origin is give as 

$\dfrac{|d|}{\sqrt {a^2+b^2+c^2}}\\dfrac{1}{\sqrt {2^2+1^2+2^2}}\\dfrac 1{\sqrt 9}=\dfrac 13$

Multiple choice maths vectors: lines in two and three dimensions distance from a point to line perpendicular distance of a point from a plane the distance from a point to a line

Find point $Q$, the foot of perpendicular drawn on line repeat $AB$, from $P\ A(1, 2, 4)\ B(3, 4,5)\ P(2, 4, 3)$.

  1. $ Q=(\dfrac{19}{9}, \dfrac{28}{9}, \dfrac{41}{9}).$
  2. $Q=(12,20,30).$
  3. $Q=(55,66,44).$
  4. $Q=(23,34,45).$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Points are $P(2,4,3), A(1,2,4), B(3,4,5).$ 
To find the foot of the perpendicular from $P$ onto $AB$.

Equation of $AB$
$x = 1+(3-1)t = 1+2t$
$y = 2+(4-2)t = 2+2t$
$z = 4+(5-4)t = 4+t$

Direction Coefficients
$ 2,  2,  1$ $Q$ lies on the $AB.$

Equation of a $3-d$ plane perpendicular to $AB$ 
$2x + 2y + z = C $

This plane passes through $P$
So,
 $2\times 2+2\times 4+3= C$

$Q$ lies on the line and the plane. 
So,
$4t+2 + 4t+4 + t+4 = 15$
So for $Q$ $t = 5/9$
Hence, $Q=(19/9, 28/9, 41/9)$

Hence, this is the answer.
Multiple choice maths vectors: lines in two and three dimensions distance from a point to line perpendicular distance of a point from a plane the distance from a point to a line

The length of the perpendicular drawn from the point $( 3 , - 1,11 )$ to the line $\dfrac { x } { 2 } = \dfrac { y - 2 } { 3 } = \dfrac { z - 3 } { 4 }  $ is:

  1. $\sqrt { 66 }$
  2. $\sqrt { 29 }$
  3. $\sqrt { 33 }$
  4. $\sqrt { 53 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The line is x/2 = (y-2)/3 = (z-3)/4 = k. Point P = (2k, 3k+2, 4k+3). Vector from P to (3, -1, 11) is (3-2k, -3-3k, 8-4k). This must be perpendicular to the line direction (2, 3, 4). 2(3-2k) + 3(-3-3k) + 4(8-4k) = 0 => 6-4k-9-9k+32-16k = 0 => 29-29k=0 => k=1. Point P = (2, 5, 7). Distance = sqrt((3-2)^2 + (-1-5)^2 + (11-7)^2) = sqrt(1 + 36 + 16) = sqrt(53).

Multiple choice maths vectors: lines in two and three dimensions distance from a point to line perpendicular distance of a point from a plane the distance from a point to a line

The length of the perpendicular drawn from $(1, 2, 3)$ to the line $\dfrac {x-6}{3}=\dfrac {y-7}{2}=\dfrac {z-7}{-2}$ is-

  1. $4$
  2. $5$
  3. $6$
  4. $7$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let us take a point on line $(3\lambda +6,2\lambda +7,-2\lambda +7)$.
Direction ratio's of line which is  perpendicular to given line 
$(3\lambda +6-1,2\lambda +7-2,-2\lambda +7-3)$
$(3\lambda +5,2\lambda +5,-2\lambda +4)$
and the direction ratio's of given line are 3,2,-2
These two lines are perpendicular, so$(3\lambda +5)*3+(2\lambda +5)*2+(-2\lambda +4)+(-2)=0$
$\lambda=-1$
So, point is $(2,4,6)$
So distance between $(3,5,9)$ and $(1,2,3)$ is $7$. (by distance formula)

Multiple choice maths vectors: lines in two and three dimensions distance from a point to line perpendicular distance of a point from a plane the distance from a point to a line

The length of the perpendicular from (1,6,3) to the line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ is 

  1. 3
  2. $\sqrt{11}$
  3. $\sqrt{13}$

  4. 5
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Line: x=k, y=2k+1, z=3k+2. Vector from (1,6,3) to (k, 2k+1, 3k+2) is (k-1, 2k-5, 3k-1). Dot product with (1, 2, 3) = 0 => k-1 + 4k-10 + 9k-3 = 0 => 14k = 14 => k=1. Point is (1, 3, 5). Distance = sqrt((1-1)^2 + (6-3)^2 + (3-5)^2) = sqrt(0 + 9 + 4) = sqrt(13).

Multiple choice maths vectors: lines in two and three dimensions distance from a point to line perpendicular distance of a point from a plane the distance from a point to a line

A line is perpendicular to the plane $x+2y+2z=0$ and passes through $(0, 1, 0)$. The perpendicular distance of this line from the origin is

  1. $\displaystyle \frac {\sqrt 5}{3}$
  2. $\displaystyle \frac {\sqrt 7}{3}$
  3. $\displaystyle \frac {2}{3}$
  4. $3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since the line is perpendicular to the plane $x+2y+2z=0$, so directon of line will be $(1,2,2)$ and it passes through $(0,1,0)$,
Therefore equation of line is $\dfrac{x}{1}=\dfrac{y-1}{2}=\dfrac{z}{2}$
So general point on the line is $(r,2r+1,2r)$
Now direction ratio of line passing through this point and origin which perpendicular to given line is $(r,2r+1,2r)$
Since they are perpendicular, so dot product will be $0$ .
$r+2\times (2r+1)+2\times 2r{=}0$
$r+4r+2+4r{=}0$
$\therefore r{=}$$\dfrac{-2}{9}$
So points on line which perpendicular from origin is $(-2/9,5/9,-4/9)$
Now applying distance formula,
${=}$ $\sqrt{{(-2/9)}^{2}+{(5/9)}^{2}+{(-4/9)}^{2}}$
${=}$ $\sqrt{4/81+25/81+16/81}$
${=}$ $\dfrac{\sqrt{5}}{3}$

Multiple choice maths vectors: lines in two and three dimensions distance from a point to line perpendicular distance of a point from a plane the distance from a point to a line

If $(a-a\prime )^2+(b-b\prime )^2+(c-c\prime )^2=p$ and $(ab\prime -a\prime b)^2+(bc\prime -b\prime c)^2+(ca\prime -c\prime a)^2=q,$ then the perpendicular distance of the line $ax+by+cz=1,$ $a\prime x+b\prime y+c\prime z=1$ from origin, is 

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

The distance of the intersection line of two planes from the origin involves the coefficients of the planes. The given expressions p and q relate to the distance formula for the line of intersection.

Multiple choice maths vectors: lines in two and three dimensions distance from a point to line perpendicular distance of a point from a plane the distance from a point to a line

Perpendiculars AP, AQ and AR are drawn to the $x-,y-$ and $z-$axes, respectively, from the point $A\left ( 1,-1,2 \right )$. The A.M. of $AP^2,$ $AQ^2$ and $AR^2$ is

  1. $4$
  2. $5$
  3. $3$
  4. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$A=(1,-1,2), P=(1,0,0), ,Q=(0,-1,0),R=(0,0,2)$
$AP^2 = [(-1)^2+2^2] = 5$,
$ AQ^2  = [1^2+2^2] = 5$,
$ AR^2 = [(-1)^2+1^2] = 2$
Hence required A.M is $=\cfrac{5+5+2}{3}=4$ 
Multiple choice maths vectors: lines in two and three dimensions distance from a point to line perpendicular distance of a point from a plane the distance from a point to a line

The length of the perpendicular drawn from $(1,2,3)$ to the line $\displaystyle \frac { x-6 }{ 3 } =\frac { y-7 }{ 2 } =\frac { z-7 }{ -2 } $ is

  1. $4$
  2. $5$
  3. $6$
  4. $7$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Direction cosines of the given line are  

$\displaystyle \frac { 3 }{ \sqrt { 17 }  } ,\frac { 2 }{ \sqrt { 17 }  } ,\frac { -2 }{ \sqrt { 17 }  } $
$\displaystyle \therefore AM=\left| \left( 6.1 \right) .\frac { 3 }{ \sqrt { 17 }  } +\left( 7-2 \right) .\frac { 2 }{ \sqrt { 17 }  } +\left( 7-3 \right) .\frac { -2 }{ \sqrt { 17 }  }  \right| =17$
$AP=\sqrt { { \left( 16-1 \right)  }^{ 2 }+{ \left( 7-2 \right)  }^{ 2 }+{ \left( 7-3 \right)  }^{ 2 } } $
$=\sqrt { 25+25+16 } =\sqrt { 66 } $
$\therefore$ length of the perpendicular is
$PM=\sqrt { { AP }^{ 2 }-{ AM }^{ 2 } } $
$=\sqrt { 66-17 } =\sqrt { 49 } =7$