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 the plane equation of a plane in intercept form equation of a plane in different forms different forms of equation of a plane

If from the point $P(f,g,h)$ perpendiculars $PL, PM$ be drawn to $yz$ and $zx$ planes, then the equation to the plane $OLM$ is

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

If perpendicular $PL$ and $PM$ drawn from the point $P(f,g,h)$ to the plane $yz$ and $zx$, then coordinates of $L$ and $M$ are,
$L = (0,g,h)$ and $M = (f,0,h)$
Now general equation of plane passes through origin is given by,
$x+p y+q z = 0$
Also this plane passes through $L$ and $M$
$\Rightarrow pg +qh=0 ...(1)$
and $ f+q h = 0 ...(2)$
Solving (1) and (2), we get
$p =f/g$ and $q =-f/h$
Hence, equation required plane $OLM$ is,
$\dfrac{x}{f}+\dfrac{y}{g}-\dfrac{z}{h}=0$

Multiple choice maths the plane equation of a plane in intercept form equation of a plane in different forms different forms of equation of a plane

A plane meet the co-ordinates axes in $A,B,C$ such that the centroid of triangle $ABC$ is the point $\alpha,\beta,\gamma.$ If the equation of the plane be $\displaystyle \frac{x}{\alpha}+\frac{y}{\beta}+\frac{z}{\gamma}=k$ then,$k=?$

  1. $1$
  2. $3$
  3. $2$
  4. $\alpha^{2}+\beta^{2}+\gamma^{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Equation of plane ABC in intercept form is $\displaystyle \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1  .....(i)$

So$A=(a,0,0),B=(0,b,0),C=(0,0,c)$
Also centroid of $\displaystyle \bigtriangleup ABC\  is \left(\frac{a}{3},\frac{b}{3},\frac{c}{3}\right)$
But co-ordinates of centroid is $(\alpha,\beta,\gamma)$
$\displaystyle \therefore \alpha=\frac{a}{3}$
$\therefore a=3\alpha,b=3\beta,c=3\gamma$
Putting in (i) we get
$\displaystyle \frac{x}{3\alpha}+\frac{y}{3\beta}+\frac{y}{3\gamma}=1$
$\displaystyle \frac{x}{\alpha}+\frac{y}{\beta}+\frac{z}{\gamma}=3$

Multiple choice maths the plane equation of a plane in intercept form equation of a plane in different forms different forms of equation of a plane

If a plane meets the coordinate axes in A, B and C such that the centroid of $\Delta ABC$ is $(1, 2, 4)$, then the equation of the plane is?

  1. $x+2y+4z=6$
  2. $4x+2y+z=12$
  3. $x+2y+4z=7$
  4. $4x+2y+z=7$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the required equation of the plane be $\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$

Then, it meets the coordinate axes in $A(a, 0, 0), B(0, b, 0), C(0, 0, c)$.
$\therefore$ centroid of $\Delta ABC$ is $G\left(\dfrac{a+0+0}{3}, \dfrac{0+b+0}{3}, \dfrac{0+0+c}{3}\right)$, i.e., $G\left(\dfrac{a}{3}, \dfrac{b}{3}, \dfrac{c}{3}\right)$
$\therefore \left(\dfrac{a}{3}=1, \dfrac{b}{3}=2, \dfrac{c}{3}=4\right)$
$\Rightarrow a=3, b=6, c=12$
Hence, the required equation of the plane is
$\dfrac{x}{3}+\dfrac{y}{6}+\dfrac{z}{12}=1$
$\Rightarrow 4x+2y+z=12$.

Multiple choice maths the plane equation of a plane in intercept form equation of a plane in different forms different forms of equation of a plane

The equation of a plane passing through the point $A(2, -3, 7)$ and making equal intercepts on the axes, is?

  1. $x+y+z=3$
  2. $x+y+z=6$
  3. $x+y+z=9$
  4. $x+y+z=4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the required equation of the plane be $\dfrac{x}{a}+\dfrac{y}{a}+\dfrac{z}{a}=1$, i.e., $x+y+z=a$

Since, it passes through the point $A(2, -3, 7)$, we have $2+(-3)+7=a$
$\Rightarrow a=6$
Hence, the required equation of the plane is $x+y+z=6$.

Multiple choice maths the plane equation of a plane in intercept form equation of a plane in different forms different forms of equation of a plane

A variable plane moves so that the sum of the reciprocals of its intercepts on the coordinate axes is $\dfrac{1}{2}$. Then, the plane passes through the point

  1. $(0, 0, 0)$
  2. $(1, 1, 1)$
  3. $\left(\dfrac{1}{2}, \dfrac{1}{2}, \dfrac{1}{2}\right)$
  4. $(2, 2, 2)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The equation of the plane is x/a + y/b + z/c = 1. Given 1/a + 1/b + 1/c = 1/2. If the plane passes through (2, 2, 2), then 2/a + 2/b + 2/c = 1, which simplifies to 1/a + 1/b + 1/c = 1/2. This matches the condition.

Multiple choice maths the plane equation of a plane in intercept form equation of a plane in different forms different forms of equation of a plane

The equation of the plane which makes with the coordinate axes, a triangle with centroid $(\alpha, \beta, \gamma)$ is given by?

  1. $\alpha x+\beta y+\gamma z=1$
  2. $\alpha x+\beta y+\gamma z=3$
  3. $\dfrac{x}{\alpha}+\dfrac{y}{\beta}+\dfrac{z}{\gamma}=1$
  4. $\dfrac{x}{\alpha}+\dfrac{y}{\beta}+\dfrac{z}{\gamma}=3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the equation of the plane be $\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$

Then, it meets the axes at $A(a, 0, 0), B(0, b, 0)$ adn $C(0, 0, c)$.
$\therefore$ centroid of $\Delta ABC$ is $G\left(\dfrac{a}{3}, \dfrac{b}{3}, \dfrac{c}{3}\right)$.
$\therefore$ $\left(\dfrac{a}{3}=\alpha, \dfrac{b}{3}=\beta and \dfrac{c}{3}=\gamma\right)\Rightarrow a=3\alpha, b=3\beta$ and $c=3\gamma$.
$\therefore$ the required equation of the plane is
$\dfrac{x}{3\alpha}+\dfrac{y}{3\beta}+\dfrac{z}{3\gamma}=1$
$\Rightarrow \dfrac{x}{\alpha}+\dfrac{y}{\beta}+\dfrac{z}{\gamma}=3$.

Multiple choice maths the plane equation of a plane in intercept form equation of a plane in different forms different forms of equation of a plane

From a point $P\left ( a,\, b,\, c \right )$ perpendiculars $PM$ and $PN$ are drawn to $zx$ and $xy$-planes respectively, $O$ is the origin. An equation of the plane $OMN$ is

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

If perpendicular $PM$ and $PN$ drawn from the point $P(a,b,c)$ to the plane $zx$ and $xy$, then coordinates of $M$ and $N$ are,
$M = (a,0,c)$ and $N = (a,b,0)$
Now general equation of plane passes through origin is given by,
$x+p y+q z = 0$
Also this plane passes through $M$ and $N$
$\Rightarrow a +qc=0$ ...(1)
and $ a+p b = 0$ ...(2)
Solving (1) and (2), we get
$p =-\dfrac {a}{b}$ and $q =-\dfrac {a}{c}$
Hence, equation required plane $OMN$ is,
$\dfrac{x}{a}-\dfrac{y}{b}-\dfrac{z}{c}=0$

Multiple choice maths the plane equation of a plane in intercept form equation of a plane in different forms different forms of equation of a plane

A variable plane moves so that the sum of reciprocals of its intercepts on the three coordinate axes is constant $\lambda$. It passes through a fixed point, which has coordinates

  1. $\left( \lambda ,\lambda ,\lambda \right) $
  2. $\displaystyle \left( \frac { 1 }{ \lambda } ,\frac { 1 }{ \lambda } ,\frac { 1 }{ \lambda } \right) $
  3. $\left( -\lambda ,-\lambda ,-\lambda \right) $
  4. $\displaystyle \left( -\frac { 1 }{ \lambda } ,-\frac { 1 }{ \lambda } ,-\frac { 1 }{ \lambda } \right) $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the equation of the variable plane be

$\displaystyle \frac { x }{ a } +\frac { y }{ b } +\frac { z }{ c } =1$    ...(1)
The intercepts on the coordinate axes are $a,b,c$.
The sum of reciprocals of intercepts in constant $\lambda$, therefore
$\displaystyle \frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } =\lambda \Rightarrow \frac { \left( 1/\lambda  \right)  }{ a } +\frac { \left( 1/\lambda  \right)  }{ b } +\frac { \left( 1/\lambda  \right)  }{ c } =\lambda $
$\displaystyle \therefore \left( \frac { 1 }{ \lambda  } ,\frac { 1 }{ \lambda  } ,\frac { 1 }{ \lambda  }  \right) $ lies on the plane (1)
Hence, the variable plane (1) always passes through the fixed point $\displaystyle \left( \frac { 1 }{ \lambda  } ,\frac { 1 }{ \lambda  } ,\frac { 1 }{ \lambda  }  \right) $

Multiple choice maths the plane equation of a plane in intercept form equation of a plane in different forms different forms of equation of a plane

A plane meets the coordinate axes in $A, B, C$ such that the centroid of the triangle $ABC$ is the point $(1,\, r,\, r^2)$. The plane passes through the point $(4, 8, 15)$, if $r$ is equal to

  1. $-3$
  2. $3$
  3. $5$
  4. $-5$
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation

Here $A(a,0,0)$, $B(0,b,0)$ and $C(0,0,c)$
Now, centroid $G(1,r,r^2)=\left( \dfrac { a }{ 3 } ,\dfrac { b }{ 3 } ,\dfrac { c }{ 3 }  \right) $
Therefore, $a=3$, $b=3r$ and $c=3r^2$
Now, $\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$ passes through $(4,8,15)$
Therefore, $\dfrac{4}{3}+\dfrac{8}{3r}+\dfrac{15}{3r^2}=1$

$\Rightarrow {4}r^{2}+{8}r +15=3r^{2}$
$\Rightarrow r^{2}+8r+15=0$
$\Rightarrow (r+3)(r+5)=0$
$\Rightarrow r=-3,-5$

Multiple choice maths the plane equation of a plane in intercept form equation of a plane in different forms different forms of equation of a plane

If from the point $P(f, g, h)$ perpendiculars $PL, PM$ be drawn to $yz$ and $zx$ planes then the equation to the plane $OLM$ is -

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

If perpendicular $PL$ and $PM$ drawn from the point $P(f,g,h)$ to the plane $yz$ and $zx$ then coordinates of $L$ and $M$ are,
$L = (0,g,h)$ and $M = (f,0,h)$
Now general equation of plane passes through origin is given by,
$x+p y+q z = 0$
Also this plane passes through $L$ and $M$
$\Rightarrow pg +qh=0 ...(1)$
and $ f+q h = 0 ...(2)$
Solving (1) and (2), we get
$p =\dfrac {f}{g}$ and $q =-\dfrac {f}{h}$
Hence, equation required plane $OLM$ is,
$\dfrac{x}{f}+\dfrac{y}{g}-\dfrac{z}{h} = 0$

Multiple choice maths the plane equation of a plane in intercept form equation of a plane in different forms different forms of equation of a plane

If $5, 3, 2$ are the direction ratios of a normal to the plane passing through the point $(2, 3, 1)$, then the sum of the intercepts made by the plane on the $x$ -axis and $y$ - axis is

  1. $\displaystyle \dfrac{8}{21}$
  2. $56$
  3. $\displaystyle \dfrac{56}{5}$
  4. $\displaystyle \dfrac{217}{10}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The equation of the plane will be of the form $5x + 3y+2z = d$

Since, the plane passes through $(2,3,1)$

$d = 21$

Intercepts along the axes are $\displaystyle \dfrac{21}{5} , 7 , \dfrac{21}{2} $ respectively.

Their sum of $x$-axis and $y$-axis intercepts is $\displaystyle \dfrac{56}{5} $.

Multiple choice maths the plane equation of a plane in intercept form equation of a plane in different forms different forms of equation of a plane

Equation of the plane whose intercepts are $1,2,3$ is

  1. $6x+2y+3z=1$
  2. $x+y+z=6$
  3. $6x+3y+2z=6$
  4. $6x-3y-2z=1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The equation of the plane with intercepts $a,b,c$ on the axes is,
$ \dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} $ $= 1$
Since, $a = 1 b = 2, c = 3$
The equation of the plane is,
$ \dfrac{x}{1} + \dfrac{y}{2} + \dfrac{z}{3} $ $= 1$
$\therefore 6x + 3y + 2z = 6$

Multiple choice maths the plane equation of a plane in intercept form equation of a plane in different forms different forms of equation of a plane

$5, 7$ are the intercepts of a plane on the $y$ - axis, $z$ - axis respectively. If the plane is parallel to the $x$-axis, then the equation of that plane is

  1. $5y+7z=35$
  2. $7y+5z=1$
  3. $\displaystyle \dfrac{y}{5}+\dfrac{Z}{7}=35$
  4. $7y+5z=35$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The equation of the plane with intercepts $a,b,c$ on the axes is $ \displaystyle \dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} $ $= 1$
$a =$ infinite (plane is parallel to the $x$-axis, $b = 5$, $c = 7$
$ \therefore  \dfrac{y}{5} + \dfrac{z}{7} $ $= 1$
$\therefore 7y + 5z = 35$

Multiple choice maths the plane equation of a plane in intercept form equation of a plane in different forms different forms of equation of a plane

The sum of the intercepts of the plane which bisects the line segment joining $(0,1,2)$ and $(2,3,0)$ perpendicularly is

  1. $2$
  2. $4$
  3. $6$
  4. $12$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The dr's of the normal to the plane will be $(2,2,-2)$.
Hence, the plane is , $2x + 2y - 2z = d$
The midpoint of the line segment is $(1,2,1)$.
Hence, $d = 4$.
Hence, the intercepts are $\left (  \dfrac{1}{2} , 1 ,  \dfrac{1}{2} \right )$
The sum of intercepts is $2$.