Tag: the plane

Quizzes (3) Questions (79)

Questions Related to the plane

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

Two system of rectangular axes have the same origin. If a plane cuts them at distances, $a$, $b$, $c$ and ${a} _{1}$,${b} _{1}$ , ${c} _{1}$ from the origin, then

  1. $\dfrac { 1 }{ { a }^{ 2 } } +\dfrac { 1 }{ { b }^{ 2 } } +\dfrac { 1 }{ { c }^{ 2 } } =\dfrac { 1 }{ { a } _{ 1 }^{ 2 } } +\dfrac { 1 }{ { b } _{ 1 }^{ 2 } } +\dfrac { 1 }{ { c } _{ 1 }^{ 2 } }$
  2. $\dfrac { 1 }{ { a }^{ 2 } } -\dfrac { 1 }{ { b }^{ 2 } } +\dfrac { 1 }{ { c }^{ 2 } } =\dfrac { 1 }{ { a } _{ 1 }^{ 2 } } -\dfrac { 1 }{ { b } _{ 1 }^{ 2 } } +\dfrac { 1 }{ { c } _{ 1 }^{ 2 } }$
  3. ${ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }={ a } _{ 1 }^{ 2 }+{ b } _{ 1 }^{ 2 }+{ c } _{ 1 }^{ 2 }$
  4. ${ a }^{ 2 }-{ b }^{ 2 }+{ c }^{ 2 }={ a } _{ 1 }^{ 2 }-{ b } _{ 1 }^{ 2 }+{ c } _{ 1 }^{ 2 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let the equation of the plane be
$\dfrac { x }{ a } +\dfrac { y }{ b } +\dfrac { z }{ c } =1$  and $\dfrac { x }{ a _1 } +\dfrac { y }{ b _1 } +\dfrac { z }{ c _1 } =1$
$ax+by+cz+d=0\quad perpendicular\quad distance\quad from\quad origin\quad is\quad \dfrac { \left| d \right|  }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } }  } $
as they have the same origin their perpendicular distance is constant.
$\dfrac { 1 }{ \sqrt { \dfrac { 1 }{ { a }^{ 2 } } +\dfrac { 1 }{ { b }^{ 2 } } +\dfrac { 1 }{ { c }^{ 2 } }  }  } =\dfrac { 1 }{ \sqrt { \dfrac { 1 }{ { a }^{ 2 } } +\dfrac { 1 }{ { b }^{ 2 } } +\dfrac { 1 }{ { c }^{ 2 } }  }  }$
$\dfrac { 1 }{ { a _1}^{ 2 } } +\dfrac { 1 }{ { b _1 }^{ 2 } } +\dfrac { 1 }{ { c _1 }^{ 2 } } =\dfrac { 1 }{ { a }^{ 2 } } +\dfrac { 1 }{ { b }^{ 2 } } +\dfrac { 1 }{ { c }^{ 2 } } $


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.