Tag: angle between two planes

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Multiple choice maths the plane angle between planes angle between two planes problems involving equation of plane

Consider the planes $3x-6y+2z+5=0$ and $4x-12y+3z=3$. The plane $67x-162y+47z+44=0$ bisects the angle between the given planes which-

  1. Contains origin

  2. Is acute

  3. Is obtuse

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

For $3x-6y+2z+5=0$ and $-4x+12y-3z+3=0$ bisector are
$\displaystyle \frac { 3x-6y+2z+5 }{ \sqrt { 9+36+4 }  } =\pm \frac { -4x+12y-3z+3 }{ \sqrt { 16+144+9 }  } $
The plane which bisects the angle between the plane that contains the origin
$13\left( 3x-6y+2z+5 \right) =7\left( -4x+12y-3z+3 \right) \ \Rightarrow 67x-162y+47z+44=0$
Further $3\times \left( -4 \right) +\left( -6 \right) \times 12+2\times \left( -3 \right) <0$
Hence, the origin lies in the acute angle.

Multiple choice maths the plane angle between planes angle between two planes problems involving equation of plane

Angle between planes $2x-y+z$ $=$ $6$ and $x+y+2z$ $=$ $7,$ is -

  1. $\dfrac { \pi }{ 4 } $
  2. $\dfrac { \pi }{ 2 } $
  3. $\dfrac { \pi }{ 3 } $
  4. $\dfrac {- \pi }{ 4 } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Plane $1$: $2x-y+z=6$
normal vector is $\bar{n _1}=2\hat{i}-\hat{j}+\hat{k}$
Plane $2$: $x+y+2z=7$
normal vector is $\bar{n _2}=\hat{i}+\hat{j}+2\hat{k}$
Angle between planes is same as the angle between their normal.
$\Rightarrow \cos\theta =\dfrac{\bar{n _1}\cdot\bar{n _2}}{|\bar{n _1}||\bar{n _2}|}$
$=\dfrac{(2\hat{i}-\hat{j}+\hat{k})\cdot(\hat{i}+\hat{j}+2\hat{k})}{(\sqrt{4+1+1})\sqrt{1+1+4}}$
$=\left|\dfrac{2-1+2}{\sqrt{6}\cdot \sqrt{6}}\right|$
$=\dfrac{3}{6}$
$=\dfrac{1}{2}$
$\Rightarrow \cos\theta =\dfrac{1}{2}$
$\Rightarrow \theta =\dfrac{2}{3}$.
Multiple choice maths the plane angle between planes angle between two planes problems involving equation of plane

The equation of the plane bisecting the angle between the planes $\displaystyle 3x +4y = 4$ and $\displaystyle 6x - 2y + 3z + 5 = 0$ that contains the origin, is

  1. $\displaystyle 9x - 38y + 15z + 43 = 0$
  2. $\displaystyle 51x + 18y + 15z = 3$
  3. $\displaystyle 9x + 2y + 3z + 1 = 0$
  4. $\displaystyle 17x + 9y + 15z = 26$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Equation of given planes can be written as
 $3x+4y=4 , 6x−2y+3z+5=0$

formula is
  $\dfrac {a _1x+b _1y+c _1z+d _1}{\sqrt {a _1^2+b _1^2+c _1^2}}$ = +  or  - $\dfrac {a _2x+b _2y+c _2z+d _2}{\sqrt {a _2^2+b _2^2+c _2^2}}$

by substituting the values in the given formula we will get 

$\dfrac {3x+4y+0z+-41}{\sqrt {3^2+4^2+0^2}}$ = + or - $\dfrac {6x+-2y+3z+5}{\sqrt {6^2+(-2)^2+3^2}}$

$\Rightarrow$ $21x+28y-28 = +\  or\  - 30x-10y+160+25$

so when adding the above equation we will get $51x + 18y + 160z - 3 = 0$

is the plane bisecting the angle containing the origin, and when subtracting we will get $9x - 38y + 160z + 53 = 0$ is the other bisecting plane.

Hence the plane $51x + 18y + 160z - 3 = 0\  or\  51x + 18y + 160z  = 3$ bisects the acute angle and therefore origin lies in the acute angle.

Multiple choice maths the plane angle between planes angle between two planes problems involving equation of plane

The equation of the plane bisecting the obtuse angle between the planes $\displaystyle x+y+z= 1$ and $\displaystyle x+2y-4z= 5$ is

  1. $\displaystyle \left ( \sqrt{7}-1 \right )x+\left ( \sqrt{7}-2 \right )y+\left ( \sqrt{7}+4 \right )z+5-\sqrt{7}= 0$
  2. $\displaystyle \left ( \sqrt{7}+1 \right )x+\left ( \sqrt{7}+2 \right )y+\left ( \sqrt{7}+4 \right )z+5-\sqrt{7}= 0$
  3. $\displaystyle \left ( \sqrt{7}+1 \right )x+\left ( \sqrt{7}+2 \right )y+\left ( \sqrt{7}-4 \right )z=\sqrt{7}$

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

Given planes are  $ x+y+z-1=0.....(1)$ and $x+2y-4z-5=0.........(2)$
Therefore equation of planes bisecting these planes are
$\dfrac{x+y+z-1}{\sqrt{3}}=\pm\dfrac{x+2y-4z-5}{\sqrt{21}}$

$\Rightarrow x+y+z-1=\pm\dfrac{x+2y-4z-5}{\sqrt{7}}$

$\Rightarrow (\sqrt{7}-1) x+(\sqrt{7}-2)y+(\sqrt{7}+4)z = \sqrt{7}+5 ...(3)$ and $(\sqrt{7}+1) x+(\sqrt{7}+2)y+(\sqrt{7}-4)z = \sqrt{7}-5  ....(4)$
If $\theta$ is the angle between $(1)$ and $(3)$, then

$  \cos\theta = \dfrac{(\sqrt{7}-1).1+(\sqrt{7}-2).1+(\sqrt{7}+4).1}{(\sqrt{(\sqrt{7}-1)^2+(\sqrt{7}-2)^2+(\sqrt{7}+4)^2}).(\sqrt{3})}= \dfrac{3\sqrt{7}+2}{(\sqrt{40+2\sqrt{7}}).(\sqrt{3})}> \dfrac{1}{2}$

$\Rightarrow \theta > 45^\circ$
Hence, plane $(1)$ bisects the obtuse angle between the given planes.
Therefore equation of plane bisecting acute angle  between given plane is
$(\sqrt{7}-1) x+(\sqrt{7}-2)y+(\sqrt{7}+4)z = \sqrt{7}+5 $

Hence, option 'D' is correct.

Multiple choice maths the plane angle between planes angle between two planes problems involving equation of plane

Let two planes $p _{1}:2x-y+z=2$, and $p _{2}:x+2y-z=3$ are given. The equation of the bisector of angle of the planes  $P _{1}$ and $P _{2}$ which does not contains origin, is

  1. $x-3y+2z+1=0$
  2. $x+3y=5$
  3. $x+3y+2z+2=0$
  4. $3x+y=5$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given planes are $p _{1}:2x-y+z=2$ and $p _{2}:x+2y-z=3$

Normals to the planes
$N _1:\dfrac{1}{\sqrt{6}}(2,-1,1)$
$N _2:\dfrac{1}{\sqrt{6}}(1,2,-1)$

Let $N$ be the normal vector of angle bisector
$N=  N _1+N _2$ or $ N _1-N _2$
$N = (3,1,0)$ or $(1,-3,2)$

The equation of plane is
$P = P _1+ \lambda P _2$
$P= 2x-y+z-2 + \lambda (x+2y-z -3) $

If $N = (3,1,0)$, then $\lambda = 1$,
Equation of Plane $=  P = 3x+y- 5$
It does not pass through origin.

Hence, option D is correct.