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

The line $x -2y + 4z + 4 = 0$, $x + y + z - 8 = 0$ intersects the plane $x - y + 2z + 1 = 0$ at the point

  1. $\left ( 3, 2, 3 \right )$
  2. $\left ( 5, 2, 1 \right )$
  3. $\left ( 2, 5, 1 \right )$
  4. $\left ( 3, 4, 1 \right )$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given  lines
 $x -2y + 4z + 4 = 0$    ....(1)
 $x + y + z - 8 = 0$       .....(2)
Subtracting (2) from (1), we get
$\Rightarrow y-z=4$        .....(3)
Given equation of plane $x-y+2z+1=0$
Since, the line intersects the plane, so using (3), we get 
$\Rightarrow x+z=3$      ......(4)
Hence, $y=5$, $z=1 $ and $x=2$
Hence, option C 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

$L: \displaystyle \frac{x\, +\, 1}{2}= \frac{y\, +\, 1}{3}= \frac{z\, +\, 1}{4}$
$\pi _{1}:\, x\, +\, 2y\, +\, 3z= 14,\, \pi _{2}:\, 2x\, -\, y\, +\, 3z= 27$

If the line $L$ meets the plane $\pi _{1}$ in the point $P$, and the coordinates of $P$ are $\left ( \alpha ,\, \beta ,\, \gamma  \right )$, then $\alpha ^{2}\, +\, \beta ^{2}\, +\, \gamma ^{2}$ is equal to

  1. $3$
  2. $14$
  3. $28$
  4. $29$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$L=\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z+1}{4}$     ........(i) and plane
$\pi \Rightarrow x+2y+3z=14$        .......(ii)
Given that $L$ meets plane $\pi _1$ (means intersection points) so from $eq^n$ (i)
$\Rightarrow \dfrac{x+1}{2}=\dfrac{y+1}{3}=\dfrac{x+1}{4}=K$
$\Rightarrow x= 2k-1\,\, and\,\, y=3k-1,z=4k-1  $
putting this values in ..... (ii)
So $(2k-1)+2(3k-1)+3(4k-1)=14$
$\Rightarrow 2k-1+6k-2+12k-3=14$
$20k-6=14$
$\Rightarrow 20k=20 \rightarrow k=1$
so points $x=1,y=2,z=3$ inform of $\alpha ,\beta,\gamma= \alpha =1,\beta=2,\gamma =3$
so $\alpha^2+\beta^2+\gamma^2=1^2+2^2+3^2$
$=14$
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 point where the line of intersection of the planes $ x - 2y + z = 1$ and $x + 2y - 2z = 5$, intersects the plane $2x + 2y + z + 6 = 0$

  1. $(1, -2, -4)$
  2. $(0,0,-6)$
  3. $(1,0,-8)$
  4. $(-1,-1,-2)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The line of intersection of two planes is found by solving the system of equations. Testing the point (1, -2, -4) in the first plane: 1 - 2(-2) + (-4) = 1 + 4 - 4 = 1 (Correct). In the second plane: 1 + 2(-2) - 2(-4) = 1 - 4 + 8 = 5 (Correct). In the third plane: 2(1) + 2(-2) + (-4) + 6 = 2 - 4 - 4 + 6 = 0 (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 line passing through the points $(5, 1,  a)$ and $(3, b, 1)$ crosses the $yz$-plane at the point $\left (0,\dfrac{17}{2},\dfrac{-13}{2}\right)$. Then,

  1. $a = 2, b = 8$
  2. $a = 4, b = 6$
  3. $a = 6, b = 4$
  4. $a = 8, b = 2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Equation of line passing through $(5, 1, a)$ and $(3, b, 1)$ is
$\dfrac{x-3}{5-3}= \dfrac{y-b}{1-b}= \dfrac{z-1}{a-1}$   ...(i)


Point $\left ( 0, \dfrac{17}{2}, -\dfrac{13}{2} \right )$ satisfies equation (i), we get

$-\dfrac{3}{2} = \dfrac{\dfrac{17}{2} -b}{1-b} = \dfrac{-\dfrac{13}{2}-1}{a-1}$

$\Rightarrow  a-1 = \dfrac{\left ( -\dfrac{15}{2} \right )}{\left ( -\dfrac{3}{2} \right )} = 5$
$\Rightarrow  a = 6$


Also,  $-3\left ( 1 - b \right )= 2 \left ( \dfrac{17}{2} - b\right )$

$\Rightarrow  3b - 3 = 17 - 2b$

$\Rightarrow  5b = 20   $

$  \Rightarrow  b = 4$

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

The point of intersection of line $\dfrac {x - 6}{-1} = \dfrac {y + 1}{0} = \dfrac {z + 3}{4}$ and plane $x + y - z = 3$ is

  1. $(2, 1, 0)$
  2. $(7, -1, -7)$
  3. $(1, 2, -6)$
  4. $(5, -1, 1)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The given line is $\dfrac {x - 6}{-1} = \dfrac {y + 1}{0} = \dfrac {z + 3}{4}=r$(say) ..... $(i)$

And Plane is $x + y - z = 3$ ........ $(ii)$
$\Rightarrow x=-r+6, y=-1, z=4r-3$
Then, the point $(-r + 6, - 1, 4r - 3)$ lies on the line $(i)$. 

It is given that the plane and the line intersects
Thus, the point $(-r + 6, - 1, 4r - 3)$ satisfies the plane
$\Rightarrow (-r + 6) - 1 - (4r - 3) = 3\Rightarrow r = 1$
$\therefore$ Required intersection point $= (5, -1, 1)$.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

The line passes through the points $\left ( 5,1,a \right )$ & $\left ( 3,b,1 \right )$ crosses the $yz$ plane at the point $\displaystyle \left ( 0,\frac{17}{2},-\frac{13}{2} \right )$ ,then

  1. $a= 4, b= 6$
  2. $a= 6, b= 4$
  3. $a= 8, b= 2$
  4. $a= 2, b= 8$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Equation of the line through the points $( 5,1,a  ) $ & $( 3,b,1)$ is
$\displaystyle \frac{x-5}{3-5}=\frac{y-1}{b-1}=\frac{z-a}{1-a}=\lambda $
Now it passes through $\displaystyle \left ( 0,\frac{17}{2},\frac{-13}2{} \right )$
$\displaystyle \therefore \frac{0-5}{-2}=\frac{17/2-1}{b-1}=\frac{-13/2-a}{1-a}=\lambda  :$

$ \Rightarrow \lambda =\dfrac{5}{2}$
$\displaystyle \therefore \frac{17/2-1}{b-1}=\frac{5}{2} $
$\Rightarrow b=4$
and $\displaystyle \frac{-13/2-a}{1-a}=\frac{5}{2} $
$\Rightarrow a=6$

Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

The equations of the plane through the points $(1,-1,2),(-3,2,-2)$ and perpendicular to the plane $x+2y+3z+7=0$ is  

  1. $x+16y+11z-7=0$
  2. $17x+8y-11z+13=0$
  3. $x+y+z-2=0$
  4. $x-5y-3z=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The normal vector to the plane is the cross product of the vector between the two points and the normal vector of the given plane. Points: A(1, -1, 2), B(-3, 2, -2). Vector AB = (-4, 3, -4). Normal to given plane = (1, 2, 3). Cross product = (17, 8, -11). Equation: 17(x - 1) + 8(y + 1) - 11(z - 2) = 0, which simplifies to 17x + 8y - 11z + 13 = 0.

Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

Reflection of the line $\dfrac{x-1}{-1}=\dfrac{y-2}{3}=\dfrac{z-4}{1}$ in the plane $x+y+z=7$ is:

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

The line has direction vector (-1, 3, 1) and passes through (1, 2, 4). The reflection of a line in a plane involves reflecting the direction vector and a point on the line; the normal to the plane is (1, 1, 1). Calculating the reflection of the direction vector across the plane normal yields the new direction vector (-3, -1, 1).

Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

The reflection of the point $(2, -1, 3)$ in the plane $3x-2y-z=9$ is?

  1. $\left(\dfrac{26}{7}, \dfrac{15}{7}, \dfrac{17}{7}\right)$
  2. $\left(\dfrac{26}{7}, \dfrac{-15}{7}, \dfrac{17}{7}\right)$
  3. $\left(\dfrac{16}{7}, \dfrac{26}{7}, \dfrac{-17}{7}\right)$
  4. $\left(\dfrac{1}{6}, \dfrac{2}{3}, \dfrac{3}{4}\right)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The reflection of (x1, y1, z1) in ax+by+cz+d=0 is given by (x-x1)/a = (y-y1)/b = (z-z1)/c = -2(ax1+by1+cz1+d)/(a^2+b^2+c^2). Plugging in the values gives the point (26/7, 15/7, 17/7).

Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

The image of the line $\displaystyle  \frac { x - 1 } { 3 } = \frac { y - 3 } { 1 } = \frac { z - 4 } { - 5 } $ in the plane $2 x - y + z + 3 = 0 $ is the line

  1. $

    \displaystyle\frac { x + 3 } { 3 } = \frac { y - 3 } { 1 } = \frac { z - 2 } { - 5 }

    $
  2. $\displaystyle

    \frac { x + 3 } { - 3 } = \frac { y - 5 } { - 1 } = \frac { z + 2 } { 5 }

    $
  3. $\displaystyle

    \frac { x - 3 } { 3 } = \frac { y + 5 } { 1 } = \frac { z - 2 } { - 5 }

    $
  4. $\displaystyle

    \frac { x - 3 } { - 3 } = \frac { y + 5 } { - 1 } = \frac { z - 2 } { 5 }

    $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let $A\left(1,3,4\right)$ be a point.

Let $M$ be the point on the plane.

Given equation of the plane is $2x-y+z+3=0$

Thus, the equation of the plane is

$\dfrac{x-1}{2}=\dfrac{y-3}{-1}=\dfrac{z-4}{3}=k$

Any point on the above line, $AM$ is of the form

$x=2k+1,y=-k+3,z=k+4$

Substituting the above values in the equation of the plane we have

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

$\Rightarrow\,4k+2+k-3+k+4+3=0$

$\Rightarrow\,6k+6=0$

$\Rightarrow\,k=-1$

Thus the coordinates of $M$ are

$x=2\times -1+1=-2+1=-1$

$y=-\left(-1\right)+3=1+3=4$

$z=-1+4=3$

Let $B\left({x}^{\prime},{y}^{\prime},{z}^{\prime}\right)$ be the image of $A$

Thus, $M$ is the midpoint of $AB$

$\therefore\,-1=\dfrac{1+{x}^{\prime}}{2},\,4=\dfrac{3+{y}^{\prime}}{2},\,3=\dfrac{4+{z}^{\prime}}{2}$ 

$\Rightarrow\,1+{x}^{\prime}=-2,\,\,3+{y}^{\prime}=8,\,\,4+{z}^{\prime}=6$

$\Rightarrow\,{x}^{\prime}=-2-1=-3,\,\,{y}^{\prime}=8-3=5,\,\,{z}^{\prime}=6-4=2$

$\therefore\,\dfrac{x+3}{3}=\dfrac{y-3}{1}=\dfrac{z-2}{-5}$

Hence the image of the given line.
Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

The distance between the parallel planes given by the equations, $\vec{r}.(2\hat{i}-2\hat{j}+\hat{k})+3=0$ and $\vec{r}.(4\hat{i}-4\hat{j}+2\hat{k})+5=0$ is-

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

Planes are  $2i-2j+k+3=0,4i-4j+2k+5=0,2i-2j+k+5/2=0$

distance between them is $=\cfrac{|c _1-c _2|}{\sqrt{a^2+b^2+c^2}}\=\cfrac{|3-\cfrac{5}{2}|}{\sqrt{2^2+2^2+1^2}}=\cfrac{1}{6}$

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

If $L _1$ is the line of intersection of the plane $2x-2y+3z-2=0, x-y+z+1=0$ and $L _2$ is the line of intersection of the plane   $x+2y-z-3=0, 3x-y+2z-1=0$, then the distance of origin from from the plane containing the lines $L _1$ + $L _2$ is :

  1. $\dfrac{1}{\sqrt{2}}$
  2. $\dfrac{1}{4\sqrt{2}}$
  3. $\dfrac{1}{2\sqrt{2}}$

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

The lines L1 and L2 are found by solving the systems of plane equations. The plane containing these lines is determined, and the distance from the origin is calculated using the standard point-to-plane formula.

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

The equation of plane which is passing through the point $(1,2,3)$ and which is at maximum distance from the point $(-1,0,2)$ is

  1. $2x+2y+z=9$
  2. $2x+z=5$
  3. $3x+y-z=2$

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

The plane passes through (1,2,3). The distance from (-1,0,2) is maximized when the normal vector is parallel to the vector from the plane point to the external point. Normal = (1-(-1), 2-0, 3-2) = (2,2,1). Equation: 2(x-1)+2(y-2)+1(z-3)=0 => 2x+2y+z=9.