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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

lf a plane meets the coordinate axes at $A,B,C$ , then equation of plane is such that centroid of triangle $ABC$ is $\left (\displaystyle \dfrac{1}{3}\dfrac{2} {3},\dfrac{4}{3}\right)$

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

Assume equation of plane is, $ax+by+cz=d$
Now this plane intersect axes at $A,B$ and $C$,
$\Rightarrow A = \left (\dfrac{d}{a}, 0, 0\right), B =  \left (0, \dfrac{d}{b}, 0\right)$ and $ C =\left (0, 0, \dfrac{d}{c}\right)$
So the centroid of triangle $ABC$ is, $\left (\dfrac{d}{3a}, \dfrac{d}{3b}, \dfrac{d}{3c}\right)$
Comparing this with given value $ a=d, b = 2d$ and $ c = 4d$
Hence equation of the plane is 

$4x+2y+z = 4$

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 a point $P(a,b,c)$ perpendicular $PA$ and $PB$ are drawn to $yz$ and $zx$ planes, find the equation of the plane $OAB$:

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

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

The coordinates of $A$ and $B$ are $(0,b,c)$ and $(a,0,c)$ respectively.

The equation of the plane passing through $O(0,0,0),A(0,b,0)$ and $B(a,0,c)$ is given by 
$\displaystyle \begin{vmatrix} x-0 & y-0 & z-0 \ 0-0 & b-0 & c-0 \ a-0 & 0-0 & c-0 \end{vmatrix}=0\Rightarrow bcx+acy-abz=0$
$\displaystyle \Rightarrow \frac { x }{ a } +\frac { y }{ b } -\frac { z }{ c } =0$

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

The equation of the plane which is parallel to y-axis and cuts off intercepts of length 2 and 3 from x-axis and z-axis is :

  1. $ 3x + 2z = 1$
  2. $ 3x+ 2z = 6 $
  3. $ 2x+ 3z = 6 $
  4. $ 3x+ 2z = 0 $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation of plane parallel to $y-axis$ is, $ax+bz+1=0$      ----- ( 1 )


Here,  $x=2$ and $z=3$


Substituting $x=2$ and $z=0$ in equation ( 1 ),

$\Rightarrow$  $2a+0+1=0$

$\Rightarrow$  $x=\dfrac{-1}{2}$

Substituting $x=0$ and $z=3$ in equation ( 1 ),

$\Rightarrow$  $0+3b+1=0$

$\Rightarrow$  $b=\dfrac{-1}{3}$

Substituting value of $a$ and $b$ equation (  1 ) we get,

$\Rightarrow$  $\dfrac{-1}{2}x-\dfrac{1}{3}z+1=0$

$\Rightarrow$  $3x+2z=6$

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

The expression of $x+y+z=1$ in form of $x\cos { \alpha  } +y\cos { \beta  } +z\cos { \gamma  } =p$ is _______.

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

$\cfrac { x }{ \sqrt { 3 }  } +\cfrac { y }{ \sqrt { 3 }  } +\cfrac { z }{ \sqrt { 3 }  } =1$
$\quad \rightarrow P=\cfrac { \left| -1 \right|  }{ \sqrt { 1+1+1 }  } =\cfrac { 1 }{ \sqrt { 3 }  } $
$\therefore x+y+z=1$
$\therefore \cfrac { x }{ \sqrt { 3 }  } +\cfrac { y }{ \sqrt { 3 }  } +\cfrac { z }{ \sqrt { 3 }  } =\cfrac { 1 }{ \sqrt { 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

The sum of Y and Z intercepts of the plane $3x+4y-6z=12$ is ___________.

  1. $10$
  2. $4$
  3. $1$
  4. $5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$3x+4y-6z=12$
$\therefore\dfrac{3x}{12}+\dfrac{4y}{12}+\dfrac{(-6)z}{12}=1$
$\therefore \dfrac{x}{4}+\dfrac{y}{3}+\dfrac{z}{(-2)}=1$
$\therefore$ y intercept $b=3$ and z intercept $c=-2$
$\therefore$ y intercept $+$ z intercept $=3+(-2)=1$

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 plane $ax+by+cz=1$ meets the coordinate axes in $A, B$ and $C$. The centroid of the triangle is:

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

The plane $ax + by + cz = 1$ meets the coordinate axis in $A,B,C$, then the coordinates will be,

$A\left( {a,0,0} \right)$, $B\left( {0,b,0} \right)$ and $C\left( {0,0,c} \right)$

The equation of the plane in intercept form is,

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

The intercepts that the plane make on the axis is $\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}$

Let C denotes the centroid, then,

$C = \left( {\dfrac{{\dfrac{1}{a} + 0 + 0}}{3},\dfrac{{0 + \dfrac{1}{b} + 0}}{3},\dfrac{{0 + 0 + \dfrac{1}{c}}}{3}} \right)$

Therefore, the coordinates of the centroid will be $\left( {\dfrac{1}{{3a}},\dfrac{1}{{3b}},\dfrac{1}{{3c}}} \right)$.

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

If a plane passes through a fixed point $\left ( 2, 3, 4 \right )$ and meets the axes of reference in $A$, $B$ and $C$, the point of intersection of the planes through $A$, $B$, $C$ parallel to the coordinate planes can be

  1. $\left ( 6, 9, 12 \right )$
  2. $\left ( 4, 12, 16 \right )$
  3. $\left ( 1, 1, -1 \right )$
  4. $\left ( 2, 3, -4 \right )$
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

Let us say a plane P $ax+by+cz=k$ passes through $\left( 2,3,4 \right) $ so $2a+3b+4c=k \quad -(1)$

$A\left( \dfrac { k }{ a } ,0,0 \right) ,\quad B\left( 0,\dfrac { k }{ b } ,0 \right) ,\quad C\left( 0,0,\dfrac { k }{ c }  \right) $

Points of intersection will be $\left< \dfrac { k }{ a } ,\dfrac { k }{ b } ,\dfrac { k }{ c }  \right> $

Let $\dfrac { k }{ a } =x\quad \dfrac { k }{ b } =y\quad \dfrac { k }{ c } =z$ so in $(1)$

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

$\dfrac { 2 }{ x } +\dfrac { 3 }{ y } +\dfrac { 4 }{ z } =1\quad -(1)$

$(a)$ if $(x,y,z) = (6,9,12)$

$\dfrac { 2 }{ 6 } +\dfrac { 3 }{ 9 } +\dfrac { 4 }{ 12 } =\dfrac { 1 }{ 3 } +\dfrac { 1 }{ 3 } +\dfrac { 1 }{ 3 } =1$ Hence true.

$(b)$ $\left< 4,12,16 \right> $

$\dfrac { 2 }{ 4 } +\dfrac { 3 }{ 12 } +\dfrac { 4 }{ 16 } =\dfrac { 1 }{ 2 } +\dfrac { 1 }{ 4 } +\dfrac { 1 }{ 4 } =1$ Hence correct

$(c)$ $\left< 1,1,-1 \right> $

$\dfrac { 2 }{ 1 } +\dfrac { 3 }{ 1 } +\dfrac { 4 }{ -1 } =1$ Hence this is also correct.

$(d)$ $\left< 2,3,-4 \right> $

$\dfrac { 2 }{ 2 } +\dfrac { 3 }{ 3 } +\dfrac { 4 }{ -4 } =2-1=1$ This is also correct.