Plane Intercepts - Class XII

We have,

Solution:

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

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

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,
$\dfrac{x}{a}+\dfrac{y}{b}-\dfrac{z}{c}=0$

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 $\left(-\cfrac{9}{2}, 9, 9\right)$

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. 

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,

$\dfrac{x}{f}+\dfrac{y}{g}-\dfrac{z}{h} = 0$

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.

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
$p =f/g$ and $q =-f/h$
Hence, equation required plane $OLM$ is,
$\dfrac{x}{f}+\dfrac{y}{g}-\dfrac{z}{h}=0$

Equation of plane ABC in intercept form is $\displaystyle \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1  .....(i)$

Let the required equation of the plane be $\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$

Let the required equation of the plane be $\dfrac{x}{a}+\dfrac{y}{a}+\dfrac{z}{a}=1$, i.e., $x+y+z=a$

The equation of the plane is x/a + y/b + z/c = 1. Given 1/a + 1/b + 1/c = 1/2. If the plane passes through (2, 2, 2), then 2/a + 2/b + 2/c = 1, which simplifies to 1/a + 1/b + 1/c = 1/2. This matches the condition.

Let the equation of the plane be $\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$

The given plane is $2x-3y+4z=12$

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

Let the equation of the variable plane be

Here $A(a,0,0)$, $B(0,b,0)$ and $C(0,0,c)$
Now, centroid $G(1,r,r^2)=\left( \dfrac { a }{ 3 } ,\dfrac { b }{ 3 } ,\dfrac { c }{ 3 }  \right) $
Therefore, $a=3$, $b=3r$ and $c=3r^2$
Now, $\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$ passes through $(4,8,15)$
Therefore, $\dfrac{4}{3}+\dfrac{8}{3r}+\dfrac{15}{3r^2}=1$

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
$p =\dfrac {f}{g}$ and $q =-\dfrac {f}{h}$
Hence, equation required plane $OLM$ is,
$\dfrac{x}{f}+\dfrac{y}{g}-\dfrac{z}{h} = 0$

The equation of the plane will be of the form $5x + 3y+2z = d$

Since, the plane passes through $(2,3,1)$

$d = 21$

Intercepts along the axes are $\displaystyle \dfrac{21}{5} , 7 , \dfrac{21}{2} $ respectively.

Their sum of $x$-axis and $y$-axis intercepts is $\displaystyle \dfrac{56}{5} $.

The equation of the plane with intercepts $a,b,c$ on the axes is,
$ \dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} $ $= 1$
Since, $a = 1 b = 2, c = 3$
The equation of the plane is,
$ \dfrac{x}{1} + \dfrac{y}{2} + \dfrac{z}{3} $ $= 1$
$\therefore 6x + 3y + 2z = 6$

The equation of the plane with intercepts $a,b,c$ on the axes is $ \displaystyle \dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} $ $= 1$
$a =$ infinite (plane is parallel to the $x$-axis, $b = 5$, $c = 7$
$ \therefore  \dfrac{y}{5} + \dfrac{z}{7} $ $= 1$
$\therefore 7y + 5z = 35$

The dr's of the normal to the plane will be $(2,2,-2)$.
Hence, the plane is , $2x + 2y - 2z = d$
The midpoint of the line segment is $(1,2,1)$.
Hence, $d = 4$.
Hence, the intercepts are $\left (  \dfrac{1}{2} , 1 ,  \dfrac{1}{2} \right )$
The sum of intercepts is $2$.

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 

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

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

$\cfrac { x }{ \sqrt { 3 }  } +\cfrac { y }{ \sqrt { 3 }  } +\cfrac { z }{ \sqrt { 3 }  } =1$
$\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 }  } $

$3x+4y-6z=12$
$\therefore\dfrac{3x}{12}+\dfrac{4y}{12}+\dfrac{(-6)z}{12}=1$
$\therefore \dfrac{x}{4}+\dfrac{y}{3}+\dfrac{z}{(-2)}=1$
$\therefore$ 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 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 $\left( {\dfrac{1}{{3a}},\dfrac{1}{{3b}},\dfrac{1}{{3c}}} \right)$.

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.