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 two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

If the equation of plane containing the line $\displaystyle \frac{-x-1}{3} = \frac{y-1}{2} = \frac{z+1}{-1}$ =1 and passing through the point (1, - 1, 0) is $ax+y+bz+c=0$, then (a+b+c) is equal to

  1. -3

  2. 3

  3. 0

  4. 2
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The line is (x+1)/-3 = (y-1)/2 = (z+1)/-1. The plane passes through this line and the point (1, -1, 0). Find the normal vector by taking the cross product of the line's direction vector (-3, 2, -1) and the vector connecting a point on the line (-1, 1, -1) to (1, -1, 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

The lengths of the intercepts on the co-ordinate axes made by the plane $5x+2y+z-13=0$ are

  1. $5, 2, 1$ unit
  2. $\dfrac{13}{5}, \dfrac{13}{2}, 13$ unit
  3. $\dfrac{5}{13}, \dfrac{2}{13}, \dfrac{1}{13}$ unit
  4. $1, 2, 5$ unit
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Solution:

Given that:
$5x+2y+z-13=0$ 
or, $\cfrac{x}{\frac{13}{5}}+\cfrac{y}{\frac{13}{2}}+\cfrac{z}{13}=1$
$\therefore$ Lengths of intercepts are $\cfrac{13}5,\cfrac{13}2$ and $13.$
Hence, B is the correct option.

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 a plane making X-intercept $4$, Y-intercept ($-6$), Z-intercept $3$ is _______.

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

Equation of a plane which cuts intercepts $4,-6,3$ on axes is
$\cfrac { x }{ 4 } +\cfrac { y }{ (-6) } +\cfrac { z }{ 3 } =1$
$\therefore$ $3x-2y+4z=12$

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

A plane $x-3y+5z=d$ passes through the point $(1,2,4)$. Intercepts on the axes are

  1. $15,-5,3$
  2. $1,-5,3$
  3. $-15,5,-3$
  4. $1,-6,20$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$(1, 2, 4)$ must satisfy this plane

$(1)(1) - (3)(2) + (5)(4) = d$ 
$\Rightarrow $ plane $\Rightarrow x - 3y + 5z = 15$
$x$ intercept $\Rightarrow x - 0 + 0 = 15 \Rightarrow 15 = x \, ml$
$y$ intercept $\Rightarrow 0 - 3y + 0 = 15 \Rightarrow y \, int = -5$
$z$ intercept $\Rightarrow 0 - 0 + 52 = 15 \Rightarrow z \, int . = 3$
$\therefore A = (15, -5 , 3)$

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 the point $P(a, b, c)$, let perpendiculars $PL$ and $PM$ be drawn to $YOZ$ and $ZOX$ planes, respectively. Then the equation of the plane $OLM$ is-

  1. <p class="MsoNormal">$\displaystyle \dfrac {x}{a}+\dfrac {y}{b}+\dfrac {z}{c}=0$</p>
  2. <p class="MsoNormal">$\displaystyle \dfrac {x}{a}+\dfrac {y}{b}-\dfrac {z}{c}=0$</p>
  3. <p class="MsoNormal">$\displaystyle \dfrac {x}{a}-\dfrac {y}{b}-\dfrac {z}{c}=0$</p>
  4. <p class="MsoNormal">$\displaystyle \dfrac {x}{a}-\dfrac {y}{b}+\dfrac {z}{c}=0$</p>
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

If the intercepts made on the axes by the plane which bisects the line joining the points $(1, 2, 3)$ and $(-3, 4, 5)$ at right angles are $(a,0,0), (0,b,0)$ and $(0,0,c)$ then $(a,b,c)$ is 

  1. $\left (-\dfrac {9}{2}, 9, 9\right)$
  2. $\left (\dfrac {1}{2}, 1, 1\right)$
  3. $\left (1, -\dfrac {1}{2}, 1\right)$
  4. $\left (1, \dfrac {1}{2}, 1\right)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given points are (1,2,3) and (-3,4,5)
Mid point of this segment is, $(-1,3,4) = M$(say)
and direction ratio are, $(4,-2,-2)$
Therefore, normal vector perpendicular to required plane is $\vec{n} = 4\hat{i}-2\hat{j}-2\hat{k}$
Since required plane is bisecting given points perpendicularly, so point $M$ will lie in the plane.
Therefore equation of plane is given by,
$((x+1)\hat{i}+(y-3)\hat{j}+(z-4)\hat{k} ) \cdot \vec{n} = 0$
$\Rightarrow ((x+1)\hat{i}+(y-3)\hat{j}+(z-4)\hat{k} ) \cdot (4\hat{i}-2\hat{j}-2\hat{k}) = 0$
$\Rightarrow 2x-y-z+9=0$
Hence, intercepts made on the axes are $\left(-\cfrac{9}{2}, 9, 9\right)$

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 makes intercept $3$ and $4$ with $x$ and $z$ axes and parallel to y-axis is 

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

Consider the intercept $3$unit and $4$ unit with $x$ and $z$axes,

Now, equation of a plane which cut intercept $a,b$ and $c$ from $x-$axis, $y-$axis and $z-$axis is,

$\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$


But given that, it is parallel to $y-$axis ,

So, $b=0$

$\dfrac{x}{a}+\dfrac{z}{c}=1$


Given that, $a=3$ unit and $c=4$ unit.

So,

$ \dfrac{x}{3}+\dfrac{z}{4}=1 $

$ 4x+3y=12 $

So,

$ \dfrac{x}{3}+\dfrac{z}{4}=1 $

$4x+3z=12 $


Hence, this is the answer. 

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 $PL$ and $PM$ be drawn to $yz$ and $zx$ planes, then equation to the plane $OLM$ is

  1. $\displaystyle \frac{x}{f} + \frac{y}{g} - \frac{z}{h} =0 $
  2. $\displaystyle \frac{x}{f} + \frac{y}{g} + \frac{z}{h} =0 $
  3. $\displaystyle \frac{x}{f} - \frac{y}{g} + \frac{z}{h} =0 $
  4. $-\displaystyle \frac{x}{f} + \frac{y}{g} + \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

$p =\dfrac fg$ 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 the plane $x-3y+5z=d$, passes through the point $(1, 2, 4)$, then the intercept on x, y, z axes are?

  1. $15, -5, 3$
  2. $1, -5, 3$
  3. $-15, 5, -3$
  4. $1, -6, 20$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the plane x - 3y + 5z = d to pass through (1, 2, 4), d = 1 - 3(2) + 5(4) = 1 - 6 + 20 = 15. The intercept form is x/15 + y/(-5) + z/3 = 1. Thus, intercepts are 15, -5, 3.