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: 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 distance of the point $A(-2,3,1)$ from the line $BC$ passing through $B(-3,5,2)$ which makes equal angles with the axes is

  1. $\displaystyle \dfrac{2}{\sqrt{3}}$
  2. $\sqrt{\dfrac{14}{3}}$
  3. $\displaystyle \dfrac{16}{\sqrt{3}}$
  4. $\displaystyle \dfrac{5}{\sqrt{3}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since $ \alpha = \beta = \gamma $
$\Rightarrow l = m = n = \dfrac{1}{\sqrt{3}}$, where $l,m,n$ are direction cosines of line $PQ$
Let $M$ be a point on the line $PQ$ such that $AM \perp PQ$
So, $PM$  $ =$   Projection of $AP$ on $ PQ$
          

 $ = \mid (-2 + 3)\dfrac{1}{\sqrt{3}} + (3 - 5)\dfrac{1}{\sqrt{3}} +(1 - 2)\dfrac{1}{\sqrt{3}} \mid = \dfrac{2}{\sqrt{3}}$

and $ AP= \sqrt{(-2+3)^2 + (3-5)^2 + (1-2)^2} = \sqrt{6}$
Hence required distance is,
$AM = \sqrt{PQ^2 - QM^2} = \sqrt{\dfrac{14}{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

State the following statement is True or False

Distance of the point $P(x, y, z)$ from the plane $X  Y$ is $\sqrt {x^{2} + y^{2} + z^{2}}$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Co-ordinates of plane $XY$ is $(0,0,z)$, since $Z$ is not included. 
$\therefore $  distance of pt $P(x,y,z)$ from xy plane $(0,0,z)$ is given by
$\sqrt { { (x-0) }^{ 2 }+{ (y-0) }^{ 2 }+{ (z-z) }^{ 2 } } $

$= \sqrt { { x }^{ 2 }+{ y }^{ 2 } } $

$\therefore $ The given statement is FALSE.
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 the length of perpendicular from $ P(2, -3, 1)$ to the line $\displaystyle \frac{x- 1}{2} = \frac{y - 3}{3} = \frac{z + 2}{-1}$

  1. $5$
  2. $\displaystyle \sqrt{\dfrac{531}{14}}$
  3. $\sqrt{50}$
  4. $\sqrt{\dfrac{221}{3}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let take a point on line $(2\lambda +1,3\lambda +3,-\lambda -2)$


Direction ratio's of line which is  perpendicular to given line, 

$(2\lambda +1-2,3\lambda +3+3,-\lambda -2-1)$

$(2\lambda -1,3\lambda +6,-\lambda -3)$

And the direction ratio's of given line are $2,3,-1$.

These two lines are perpendicular, so

$(2\lambda -1)\cdot 2+(3\lambda +6)\cdot 3+(-\lambda -3)+(-1)=0$


$\lambda=\dfrac{-19}{14}$

So point is $\left (\dfrac{-24}{14},\dfrac{-15}{14},\dfrac{-9}{14}\right)$

So distance between $\left (\dfrac{-24}{14},\dfrac{-15}{14},\dfrac{-9}{14}\right) $ and $(1,3,-2)$ is $\sqrt{\dfrac{531}{14}}$...................(by distance formula) 

Multiple choice maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

If the points $(1,2,3)$ and $(2,-1,0)$ lie on the opposite sides of the plane $2x+3y-2z=k$, then

  1. $k< 1$
  2. $k> 2$
  3. $k< 1$ or $k> 2$
  4. $1< k< 2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

 Given plane equation is $2{x}+3{y}-2{z}-k=0$

$(1,2,3)$ and $(2,-1,0)$ lies on the opposite sides of the plane
$(2(1)+3(2)-2(3)-k)(2(2)+3(-1)-2(0)-k)<0$
$(2-k)(1-k)<0\implies (k-1)(k-2)<0$
$\implies 1<k<2$

Multiple choice maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

If the planes $x - cy - bz = 0,cx - y + az = 0\,$ and $bx + ay - z = 0$ pass through a stright line,then the value of ${a^2} + {b^2} + {c^2} + 2abc\,$ is:

  1. $1$
  2. $2$
  3. $3$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
If the planes $(x - cy - bz = 0), (cx - y + az = 0)$ and $(bx + ay - z = 0)$ are in same line.
$\therefore$ They must be collinear.
$\begin{vmatrix}1 & -c & -b\\ c & -1 & a\\ b & a & -1\end{vmatrix} = 0$
$\Rightarrow 1(1 - a^2) + c(-c - ab) -b(ac + b) = 0$
$\Rightarrow 1 - a^2 - c^2 - abc - abc - b^2 = 0$
$\therefore a^2 + b^2 + c^2 + 2abc = 1$
Option A is correct
Multiple choice maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

The point where the line through $A=(3, -2, 7)$ and $B= (13, 3, -8)$ meets the xy-plane

  1. $(\cfrac { 23 }{ 3 } ,\cfrac { 1 }{ 3 } ,0)$
  2. $(\cfrac { 23 }{ 6 } ,\cfrac { 1 }{ 6 } ,0)$
  3. $(\cfrac { 23 }{ 3 } ,\cfrac { 1 }{ 3 } , 1)$
  4. $(\cfrac { 23 }{ 3 } ,\cfrac { 1 }{ 3 } , 3)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Equation of line through $A(3,-2,7)$ and $B(13,3,-8)$ is:

$\cfrac { x-3 }{ 10 } =\cfrac { y+2 }{ 5 } =\cfrac { z-7 }{ -15 }$
When the line meets $x-y$ plane $\Rightarrow z=0$
$\therefore \cfrac { x-3 }{ 10 } =\cfrac { y+2 }{ 5 } =\cfrac { 7 }{ 15 } \quad \quad \Rightarrow x=\cfrac { 23 }{ 3 } ,y=\cfrac { 1 }{ 3 } \quad \quad \Rightarrow (x,y,z)=(\cfrac { 23 }{ 3 } ,\cfrac { 1 }{ 3 } ,0)$

Multiple choice maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

The ratio in which the plane $4x+5y-3z=8$ divides the line joining the points $(-2,1,5)$ and $(3,3,2)$ is

  1. $2 : 1$
  2. $1 : 2$
  3. $-2 : 1$
  4. $3 : 2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know that the ratio in which the plane $ax+by+cz+d=0$ divides the line segment joining (${x _1},{y _1},{z _1}$) and (${x _2},{y _2},{z _2}$) is

$\begin{array}{l} \dfrac { { -\left( { a{ x _{ 1 } }+b{ y _{ 1 } }+c{ z _{ 1 } }+d } \right)  } }{ { a{ x _{ 2 } }+b{ y _{ 2 } }+c{ z _{ 2 } }+d } }  \ a=4;b=5;c=-3;d=-8;{ x _{ 1 } }=-2;{ y _{ 1 } }=1;{ z _{ 1 } }=5;{ x _{ 2 } }=3;{ y _{ 2 } }=3;{ z _{ 2 } }=2 \ so,\, the\, required\, ratio=\dfrac { { -\left( { 4\left( { -2 } \right) +5\left( 1 \right) -3\left( 5 \right) -8 } \right)  } }{ { 4\left( 3 \right) +5\left( 3 \right) -3\left( 2 \right) -8 } }  \ =\dfrac { { -\left( { -8+5-15-8 } \right)  } }{ { 12+15-6-8 } }  \ =\dfrac { { 26 } }{ { 13 } }  \ =\dfrac { 2 }{ 1 } \ or\ 2:1 \end{array}$

Multiple choice maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

Let the equations of a line and a plane be $\dfrac {x+3}{2}=\dfrac {y-4}{3}=\dfrac {z+5}{2}$ and $4x-2y-z=1$, respectively, then

  1. the line is parallel to the plane.

  2. the line is perpendicular to the plane.

  3. the line lies in the plane.

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

Direction ratios of the line is $2i+3j+2k$
and normal of plane is along $4i-2j-k$
Now, $(2i+3j+2k).(4i-2j-k)=8-6-2=0$
Therefore, line is parallel to plane

Ans: A

Multiple choice maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

The ratio in which the plane $r.\left( \hat { i } -2\hat { j } +2\hat { k }  \right) =17$ divides the line joining the points $-2\hat { i } +4\hat { j } +7\hat { k } $ and $3\hat { i } -5\hat { j } +8\hat { k } $ is:

  1. $3:5$
  2. $1:10$
  3. $3:10$
  4. $1:5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the plane $r.(i-2j+3k)=17$ divide the line joining the points. 

$-2i+4j+7k$ and $2i-5j+8k$ in the ratio $t:1$ at the point $P.$

$\therefore P$ is $\displaystyle \dfrac { 3t-1 }{ t+1 } i+\dfrac { -5t+4 }{ t+1 } j+\dfrac { 8t+7 }{ t+1 } k.$

This lies on the given plane, 

$\displaystyle \therefore \dfrac { 3t-2 }{ t+1 } .1+\dfrac { -5t+4 }{ t+1 } \left( 2 \right) +\dfrac { 8t+7 }{ t+1 } \left( 3 \right) =17$

$\Rightarrow 3t-2+10t-8+24t+21=17t+17$

$\displaystyle \therefore 20t=17-21+10=6\Rightarrow =\dfrac { 6 }{ 20 } =\dfrac { 3 }{ 10 } $

$\therefore$ required ratio is $3:10$.

Multiple choice maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

Line $\vec r=\vec a+\lambda \vec b$ will not meet the plane $\vec r\cdot \vec n=q$, if-

  1. $\vec b\cdot \vec n=0, \vec a\cdot \vec n=q$
  2. $\vec b\cdot \vec n\neq 0, \vec a\cdot \vec n\neq q$
  3. $\vec b\cdot \vec n=0, \vec a\cdot \vec n\neq q$
  4. $\vec b\cdot \vec n\neq 0, \vec a\cdot \vec n=q$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given line is $\overrightarrow { r } =\overrightarrow { a } +\lambda \overrightarrow { b } $

Substitute it in plane equation $\overrightarrow { r } .\overrightarrow { n } =q$
We get $(\overrightarrow { a } +\lambda \overrightarrow { b } ).\overrightarrow { n } =q$
$\Rightarrow \overrightarrow { a } .\overrightarrow { n } +\lambda (\overrightarrow { b } .\overrightarrow { n } )=q$
If $\overrightarrow { b } .\overrightarrow { n } =0$ and $\overrightarrow { a } .\overrightarrow { n } \neq q$ then the line will not meet the plane
Therefore the correct option is $C$

Multiple choice maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

The ratio in which the plane $\vec r\cdot (\vec i-2\vec j+3\vec k)=17$ divides the line joining the points $-2\vec i+4\vec j+7\vec k$ and $3\vec i-5\vec j+8\vec k$ is-

  1. $1:5$
  2. $1:10$
  3. $3:5$
  4. $3:10$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Equation of plane in cartesian form is, $x-2y+3z-17=0$.....$(1)$
Assume this plane $(1)$ divide the line segment joining the points $(-2,4,7)$ and $(3,-5,8)$ in $m:n$ ratio
Therefore,  $\dfrac{m}{n} = \dfrac{-2-2(4)+3(7)-17}{3-2(-5)+3(8)-17}= \dfrac{-3}{10} < 0$
Hence, plane (1) divides the given line segment externally $3:10$ 

Multiple choice maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

The plane $\vec r\cdot \vec n=q$ will contain the line $\vec r=\vec a+\lambda \vec b$, if-

  1. $\vec b\cdot \vec n\neq 0, \vec a\cdot \vec n\neq q$
  2. $\vec b\cdot \vec n=0, \vec a\cdot \vec n\neq q$
  3. $\vec b\cdot \vec n=0, \vec a\cdot \vec n=q$
  4. $\vec b\cdot \vec n\neq 0, \vec a\cdot \vec n=q$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Normal of the plane $\vec { r } \cdot \vec { n } =q$ is $\vec { n } $ and direction ratio of the line $\vec { r } =\vec { a } +\lambda \vec { b } $ is $\vec { b } $
Since, line lies in the plane, normal and direction ratios should be perpendicular.
Therefore, $\vec { b } \cdot \vec { n } =0$
Also position vector $\vec { a } $ should lie on plane $\vec { r } \cdot \vec { n } =q$
Therefore, $\vec { a } \cdot \vec { n } =q$

Ans: C

Multiple choice maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

The ratio in which the line segment joining the points whose position vectors are $2\hat i-4\hat j-7\hat k$ and $-3\hat i+5\hat j-8\hat k$ is divided by the plane whose equation is $\hat r\cdot (\hat i-2\hat j+3\hat k)=13$ is-

  1. $13:12$ internally
  2. $12:25$ externally
  3. $13:25$ internally
  4. $37:25$ internally
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Equation of plane in cartesian form is, $x-2y+3z-13=0$ .....(1)$
Assume this plane (1) divide the line segment joining the points
$(2,-4,-7)$ and $(-3,5,-8)$ in $m:n$ ratio
Therefore,  $\dfrac{m}{n} = \dfrac{2-2(-4)+3(-7)-13}{-3-2(5)+3(-8)-13}= \dfrac{-12}{25} < 0$
Henc,e plane (1) divides the given line segment externally  \$12:25$.

Multiple choice maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

Which of the following lines lie on the plane $x+2y-z=0$?

  1. $x-1=y-1=1$
  2. $x-y+z=2x+y-z=0$
  3. $\vec r=2\hat i-\hat j+4\hat k+\lambda (3\hat i+\hat j+5\hat k)$

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

consider $P: x+2y-z=0$
direction ratio of normal of $P$ is $(1,2,-1)$

a) for $L _{1}: \dfrac {x-1}{1}=\dfrac {y}{-1}=\dfrac {z-5}{-1}$
Since, $(1,0,5)$ does not lie in $P$ 
Therefore, $L _{1}$ does not lie in $P$

b) for $L _{2}: x-y+z=2x+y-z=0$
direction ratio of $L _{2}$ is $(i-j+k) \times (2i+j-k)=3(j+k)$
since, $(i+2j-k).(3j+3k)=6-3=3 \neq =0$
Therefore, $L _{2}$ does not lie on $P$

c) for $L _{3}:\vec r=2\hat i-\hat j+4\hat k+\lambda (3\hat i+\hat j+5\hat k)$
Since, $(2,-1,4)$ does not lie in $P$ 
Therefore, $L _{3}$  does not lie in $P$

Ans: D

Multiple choice maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

Find the ratio in which the segment joining $(1, 2, -1)$ and $(4, -5, 2)$ is divided by the plane $2x - 3y + z = 4$

  1. $2 : 1$
  2. $3 : 2$
  3. $3 : 7$
  4. $1 : 2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using the section formula, a point dividing the segment joining (1, 2, -1) and (4, -5, 2) in the ratio k:1 has coordinates that can be substituted into the plane equation 2x - 3y + z = 4. Solving for k yields the ratio 3:7, meaning the plane divides the segment externally or internally in that ratio.