Tag: the plane

Questions Related to the plane

Find the equation of the bisector planes of the angles between the planes $2x - y + 2z + 3 = 0$ and $3x - 2y + 6z + 8 = 0$.

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

  1. $ 5x-y-4z-22=0$

  2. $ 23x-13y+32z+26 = 0 $

  3. $ 19x-y-4z+26 = 0 $

  4. none of these

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Correct Option: A,B
Explanation:

Equation of the planes is $2x-4y+2z+3=0$ and $3x-2y+6z+8=0$
Then equation of the plane bisection the angles between them are
$\displaystyle \frac { 2x-4y+2z+3 }{ \sqrt { 4+16+4+9 }  } =\pm \frac { 3x-2y+6z+8 }{ \sqrt { 9+4+36+64 }  } $
$\displaystyle \Rightarrow \frac { 2x-4y+2z+3 }{ \sqrt { 33 }  } =\pm \frac { 3x-2y+6z+8 }{ \sqrt { 113 }  } $
$\Rightarrow 5x-y-4z-22=0$ and $23x-13y+32z+26=0$

The angle between two planes is equal to

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

  1. the angle between the tangents to them from any point

  2. the angle between the normals to them from any point

  3. the angle between the lines parallel to the planes from any point

  4. None of the above

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Correct Option: D
Explanation:

The angle between two intersecting planes is equal to the acute angle determined by the normal vectors of the two planes.

This is different from the angle between the normals to the planes from any point.

lf the planes $ x+2y-z+5=0,\ 2x-ky+4z+3=0$ are perpendicular, then $ {k} $ is

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

  1. $1$

  2. $-1$

  3. $0$

  4. $2$

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Correct Option: B
Explanation:

I am using the constant $\lambda$ instead of $k$ to avoid the confusion. 

The normals to the planes are given by $i+2j-k$ and $2i-\lambda j+4k$, respectively. 
Since, they are perpendicular dot product between normals are zero. 
Thus, $(i+2j-k).(2i-\lambda+4k)=0$. 
$\Rightarrow2-2\lambda-4=0 \Rightarrow \lambda=-1$

In the space the equation $by+ cz+ d= 0$ represents a plane perpendicular to the plane:

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

  1. $YOZ$

  2. $ZOX$

  3. $XOY$

  4. $Z= k$

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Correct Option: A
Explanation:

Consider $P : bx+cz+d=0$
a) Equation of $YOZ$ plane is $x=0$
Since, $(i).(bj+ck)=0$
Therefore, $P$ is perpendicular to $YOZ$

b)  Equation of $ZOX$ plane is $y=0$
Since, $(j).(bj+ck)=b \neq 0$
Therefore, $P$ is not perpendicular to $ZOX$

c)  Equation of $XOY$ plane is $z=0$
Since, $(k).(bj+ck)=c \neq 0$
Therefore, $P$ is not perpendicular to $XOY$

d)  Consider. $z=k$
Since, $(k).(bj+ck)=c \neq 0$
Therefore, $P$ is not perpendicular to $z=k$

Ans: A

If the planes $ 2x-y+ \lambda z- 5=0$ and $x+4y+2z- 7= 0$ are perpendicular, then $\lambda=$

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

  1. $1$

  2. $-1$

  3. $2$

  4. $-2$

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Correct Option: A
Explanation:

Since, the planes $2x-y+\lambda z-5=0$ & $x+4y+2z-7=0$ are perpendicular to each other
Therefore, $\left( 2i-j+\lambda k \right) .\left( i+4j+2k \right) =0$
$\Rightarrow 2-4+2\lambda =0$
$\Rightarrow \lambda =1$

Ans: A

If the planes $\vec{r}. (2\widehat{i}- \widehat{j}+ 2\widehat{k})= 4$ and $\vec{r}. (3\widehat{i}+ 2\widehat{j}+\lambda\widehat{k})= 3$ are perpendicular, then $\lambda =$

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

  1. $2$

  2. $-2$

  3. $3$

  4. $-3$

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Correct Option: B
Explanation:

Since, the planes $\vec { r } .(2\widehat { i } -\widehat { j } +2\widehat { k } )=4$ & $\vec{r}. (3\widehat{i}+ 2\widehat{j}+\lambda\widehat{k})= 3\ $ are perpendicular to each other
Therefore, $\left( 2i-j+2 k  \right) .\left( 3i+2j+\lambda k \right) =0$
$\Rightarrow 6-2+2\lambda =0$
$\Rightarrow \lambda =-2$

Ans: B

The angle between the planes, $\vec{r}.(2\widehat{i}- \widehat{j}+\widehat {k})=6$ and $\vec{r}.(\widehat{i}+ \widehat{j}+2\widehat {k})=5$ , is:

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

  1. $\dfrac{\pi}{3}$

  2. $\dfrac{2\pi}{3}$

  3. $\dfrac{\pi}{6}$

  4. $\dfrac{5\pi}{6}$

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Correct Option: A
Explanation:

Angle between $2x-y+z=6$ & $x+y+2z=5$ is
$\theta =\cos ^{ -1

}{ \left[ \dfrac { \left( 2i-j+k \right) .\left( i+j+2k \right)  }{

\sqrt { \left( { 2 }^{ 2 }+{ 1 }^{ 2 }+{ 1 }^{ 2 } \right) \left( 1^{ 2

}+{ 1 }^{ 2 }+2^{ 2 } \right)  }  }  \right]  } =\dfrac { \pi  }{ 3 } $

Ans: B

The angle between the planes $ 3x-6y+2z+5=0 $ 7 $ 4x-12y+3z=3 $.Which is bisected by the plane
$ 67x-162y+47z+44 = 0 $is the angle which-

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

  1. contains origin

  2. is acute

  3. is obtuse

  4. is right angle

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Correct Option: A

A plane$ P _{1}$ has the equation $2x-y+z=4$ and the plane $P _{2}$ has the equation $x+ny+2z=11.$ If the angle between $P _{1}$ and $P _{2}$ is $\pi /3$ then the value (s) of '$n$' is (are)

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

  1. $7/2$

  2. $17,-1$

  3. $-17,1$

  4. $-7/2$

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Correct Option: C
Explanation:
A plane$ P _{1}$ has the equation $2x-y+z=4$ and the plane $P _{2}$ has the equation $x+ny+2z=11.$  
The direction vector of the normal of the first plane is $2i-j+k$ and second plane is $i+nj+2k$.
The angle between $P _{1}$ and $P _{2}$ is $\pi /3$
$\cos { \dfrac { \pi  }{ 3 }  } =\dfrac { \left( 2i-j+k \right) .\left( i+nj+2k \right)  }{ \sqrt { \left( { 2 }^{ 2 }+{ 1 }^{ 2 }+{ 1 }^{ 2 } \right) \left( { 1 }^{ 2 }+{ n }^{ 2 }+{ 2 }^{ 2 } \right)  }  } $
$\Rightarrow \dfrac { 1 }{ 2 } =\dfrac { 4-n }{ \sqrt { 6\left( 5+{ n }^{ 2 } \right)  }  } $
$\Rightarrow \cos^2(\dfrac{\pi}{3})=\dfrac{(4-n)^2}{6(5+n^2)} $
$\Rightarrow n^2+16n-17=0\Rightarrow (n+17)(n-1)=0$
$\Rightarrow n=-17,1$

The angle between the planes $\displaystyle x + y + z = 0$ and $\displaystyle 3x - 4y + 5z = 0$ is

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

  1. $\displaystyle \cos ^{-1}\left ( \frac{1}{5} \sqrt{\frac{2}{5}} \right )$

  2. $\displaystyle \frac{\pi }{2}$

  3. $\displaystyle \frac{\pi }{3}$

  4. $\displaystyle \cos ^{-1}\left ( \frac{2}{5} \sqrt{\frac{2}{3}} \right )$

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Correct Option: D
Explanation:

The angle between $x+y+z=0$ & $3x-4y+5z=0$ is
$\theta =\cos ^{ -1

}{ \left[ \dfrac { \left(\vec  i+\vec j+\vec k \right) .\left( 3\vec i-4\vec j+5\vec k \right)  }{

\sqrt { \left( { 1 }^{ 2 }+{ 1 }^{ 2 }+{ 1 }^{ 2 } \right) \left( 3^{ 2

}+{ 4 }^{ 2 }+5^{ 2 } \right)  }  }  \right]  } $

$= \cos ^{ -1 } \left( \dfrac { 2 }{ 5 } \sqrt { \dfrac { 2 }{ 3 }  }  \right) $

Ans: D