Tag: equation of a plane in intercept form

Questions Related to equation of a plane in intercept form

The intercepts of the plane $2x-3y+5z-30=0$ are 

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

  1. $15,-10,6$

  2. $5,10,6$

  3. $1/8,-1/6,1/4$

  4. $3,-4,6$

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Correct Option: A
Explanation:

We have,

$\begin{array}{l} 2x-3y+5z=30 \ \frac { x }{ { 15 } } +\frac { y }{ { -10 } } +\frac { z }{ 6 } =1 \end{array}$
Intercept we $(15,-10,6)$
Then,
OPtion $A$ is correct answer.

If $A=(3,1,-2) , B=(-1,0,1)$ and $l, m$ are the projections of AB on the y-axis, zx plane respectively then $3l^2-m+1$=

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

  1. 0

  2. 1

  3. 11

  4. 27

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Correct Option: A

A plane $
\pi
 $ makes intercept 3 and 4 respectively on z-axis and x-axis. If $
\pi
 $ is parallel to y-axis, then its equation is


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

  1. 3x+4z=12

  2. 3z+4x=12

  3. 3y+4z=12

  4. 3z+4y=12

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Correct Option: A
Explanation:
$X-$intercept $a=4$
and $Z-$intercept $c=3$
Required equation $\dfrac{x}{4}+\dfrac{z}{3}=1$
or $3x+4z=12$

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

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

  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

Show answer
Correct Option: B
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.

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

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

  1. $3x-4y+6z=12$

  2. $3x-2y+4z=12$

  3. $4x-6y+3z=1$

  4. $4x-3y+2z=12$

Show answer
Correct Option: B
Explanation:

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

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

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

  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. ${ a }^{ 2 }-{ b }^{ 2 }+{ c }^{ 2 }={ a } _{ 1 }^{ 2 }-{ b } _{ 1 }^{ 2 }+{ c } _{ 1 }^{ 2 }$

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Correct Option: A
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 } } $


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

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

  1. $15,-5,3$

  2. $1,-5,3$

  3. $-15,5,-3$

  4. $1,-6,20$

Show answer
Correct Option: A
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 $\Rightarrow 0 - 0 + 52 = 15 \Rightarrow z \, int . = 3$
$\therefore A = (15, -5 , 3)$

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-

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

  1. $\displaystyle \dfrac {x}{a}+\dfrac {y}{b}+\dfrac {z}{c}=0$

  2. $\displaystyle \dfrac {x}{a}+\dfrac {y}{b}-\dfrac {z}{c}=0$

  3. $\displaystyle \dfrac {x}{a}-\dfrac {y}{b}-\dfrac {z}{c}=0$

  4. $\displaystyle \dfrac {x}{a}-\dfrac {y}{b}+\dfrac {z}{c}=0$

Show answer
Correct Option: B
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,
$\dfrac{x}{a}+\dfrac{y}{b}-\dfrac{z}{c}=0$

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 

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

  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. $\left (1, \dfrac {1}{2}, 1\right)$

Show answer
Correct Option: A
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 $\left(-\cfrac{9}{2}, 9, 9\right)$

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

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

  1. $3x+4z=12$

  2. $4x+3z=12$

  3. $3y+4z=12$

  4. $4y+3x=12$

Show answer
Correct Option: B
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.