Tag: equation of a plane in intercept form

Questions Related to equation of a plane in intercept form

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

  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 } } $

Show answer
Correct Option: D
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 }  } $

The sum of Y and Z intercepts of the plane $3x+4y-6z=12$ 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. $10$

  2. $4$

  3. $1$

  4. $5$

Show answer
Correct Option: C
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 $=3+(-2)=1$

The plane $ax+by+cz=1$ meets the coordinate axes in $A, B$ and $C$. The centroid of the triangle 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. $(3a, 3b, 3c)$

  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. $\left( \dfrac { 1 }{ 3a } ,\dfrac { 1 }{ 3b } ,\dfrac { 1 }{ 3c } \right)$

Show answer
Correct Option: D
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 $\left( {\dfrac{1}{{3a}},\dfrac{1}{{3b}},\dfrac{1}{{3c}}} \right)$.

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

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. $\left ( 6, 9, 12 \right )$

  2. $\left ( 4, 12, 16 \right )$

  3. $\left ( 1, 1, -1 \right )$

  4. $\left ( 2, 3, -4 \right )$

Show answer
Correct Option: A,B,C,D
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.