Tag: cartesian equation of plane

Questions Related to cartesian equation of plane

The Cartesian equation of the plane $\vec r=(1+\lambda-\mu)\hat i+(2-\lambda)\hat j+(3-2\lambda+2\mu)\hat k$ is-

maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

  1. $2x+y=5$

  2. $2x-y=5$

  3. $2x+z=5$

  4. $2x-z=5$

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Correct Option: C
Explanation:

Given, $\vec{r} = (1+\lambda-\mu)\hat{i}+(2-\lambda)\hat{j}+(3-2\lambda+2\mu)\hat{k}$
$\Rightarrow x\hat{i}+y\hat{j}+z\hat{k} = (1+\lambda-\mu)\hat{i}+(2-\lambda)\hat{j}+(3-2\lambda+2\mu)\hat{k}$
Comparing coefficient, we get
$ 1+\lambda-\mu = x, 2-\lambda=y, 3-2\lambda+2\mu=z$
$\Rightarrow\lambda = 2-y, \mu=1+\lambda - x = 3-y-x$
Eliminating $\mu$ and $\lambda$, we get
$2x+z=5$ which is required equation of plane in cartesian form.

The equation of a plane which passes through the point of intersection of lines $\dfrac {x-1}{3}=\dfrac {y-2}{1}=\dfrac {z-3}{2}$, and $\dfrac {x-3}{1}=\dfrac {y-1}{2}=\dfrac {z-2}{3}$ and at greatest distance from point $(0, 0, 0)$ is-

maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

  1. $4x+3y+5z=25$

  2. $4x+3y+5z=50$

  3. $3x+4y+5z=49$

  4. $x+7y-5z=2$

Show answer
Correct Option: B
Explanation:

Any point on the first line is $P\left( 3\lambda +1,\lambda +2,2\lambda +3 \right) $
and on the second line is $Q\left( u+3,2u+1,3u+2 \right) $
$P$ and $Q$ represent the same point if $\lambda =u=1$
And the point of intersection of the given line is $P\left( 4,3,5 \right) $
The plane given in (a),(b),(c) and (d) all pass through $P$.
The plane at greatest distance is one which is at a distance equalt to $OP$ from the origin.
So, the distance of the plane from origin is $\sqrt { { 4 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 } } =\sqrt { 50 } $
The equation of plane is $4x+3y+5z=50$

Let $A (1, 1, 1), B(2, 3, 5)$ and $C(-1, 0, 2)$ be three points, then equation of a plane parallel to the plane $ABC$ and at the distance $2$ is

maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

  1. $2x-3y+z-2\sqrt {14}=0$

  2. $2x-3y+z-\sqrt {14}=0$

  3. $2x-3y+z+2=0$

  4. $2x-3y+z-2=0$

Show answer
Correct Option: A
Explanation:

$\vec{AB}=i+2j+4k$
$\vec{BC}=3i+3j+3k$

Hence, $\vec{AB}\times\vec{BC}=3(2i-3j+k)$
Now the unit normal of the plane of $ABC$ will be 
$=\dfrac{2i-3j+k}{\sqrt{14}}$
The required plane is parallel to the plane $ABC$.
Hence, its unit normal will be parallel to the normal of $ABC$.
Therefore, the equation of the required plane is 
$r.(\dfrac{2i-3j+k}{\sqrt{14}})=d$
$2x-3y+z=d\sqrt{14}$
Now $d$ is $2$.
Hence, the equation is $2x-3y+z=2\sqrt{14}$.

The plane which passes through the point $(3, 2, 0)$ and the line $\dfrac {x-3}{1}=\dfrac {y-6}{5}=\dfrac {z-4}{4}$ is:

maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

  1. $x-y+z=1$

  2. $x+y+z=5$

  3. $x+2y-z=1$

  4. $2x-y+z=5$

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Correct Option: A

Equation of the plane passing through the points $(2, 2, 1)$ and $(9, 3, 6)$, and perpendicular to the plane $2x+6y+6z-1=0$ is-

maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

  1. $3x+4y+5z=9$

  2. $3x+4y-5z=9$

  3. $3x-4y+5z=9$

  4. None of the above.

Show answer
Correct Option: B
Explanation:

Equation of plane passes through $(2,2,1)$ is given by,
$a(x-2)+b(y-2)+c(z-1) = 0.......(A)$
Given it also passes through $(9,3,6)$
$\Rightarrow 7a+b+5c=0 ......(1)$
and this plane is perpendicular to plane $2x+6y+6z-1=0$
$\Rightarrow 2a+6a+6c=0 .....(2)$
Solving (1) and (2), we get $ a= \dfrac{-3a}{5}, b = \dfrac{-4a}{5}$
Putting these values in (A) our required plane is,
$3x+4y-5z=9$

The cartesian equation of the plane $\overrightarrow { r } =\left( 1+\lambda -\mu  \right) i+\left( 2-\lambda  \right) j+\left( 3-2\lambda +2\mu  \right) k$ is:

maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

  1. $2x+y=5$

  2. $2x-y=5$

  3. $2x+z=5$

  4. $2x-z=5$

Show answer
Correct Option: C
Explanation:

We have $\overrightarrow { r } =\left( 1+\lambda -\mu  \right) i+\left( 2-\lambda  \right) j+\left( 3-2\lambda +2\mu  \right) k$

$\Rightarrow \overrightarrow { r } =\left( i+2j+k \right) +\lambda \left( i=j-2k \right) +\mu \left( -i+2k \right) $
which is a plane through $\overrightarrow { a } =i+2j+3k$ and parallel to the vectors
$\overrightarrow { b } =i-j-2k$ and $\overrightarrow { c } =-i+2k$
Therefore, it is perpendicular to the vector
$\overrightarrow { n } =\overrightarrow { b } \times \overrightarrow { c } =-2i-k$
Hence, its vector equation is 
$\left( \overrightarrow { r } -\overrightarrow { a }  \right) .\overrightarrow { n } =0\Rightarrow \overrightarrow { r } .\overrightarrow { n } =\overrightarrow { a } .\overrightarrow { n } \ \Rightarrow \overrightarrow { r } .\left( -2i-k \right) =-2-3\Rightarrow \overrightarrow { r } \left( 2i+k \right) =5$
So, the cartesian equation is
$\left( xi+yj+zk \right) .\left( 2i+k \right) =5\Rightarrow 2x+z=5$

If $lx+my+nz=p$ is equation of plane in normal form, then :

maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

  1. $l^2+m^2+n^2=1$

  2. l, m , n are d.c's of a normal to the plane

  3. p > 0

  4. All of these

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Correct Option: A

The equation of the plane through the points $(2,3,1)$ and $(4,-5,3)$ and parallel to $x$-axis is

maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

  1. $x-z-1=0$

  2. $4x+y-11=0$

  3. $y+4z-7=0$

  4. None of these

Show answer
Correct Option: C
Explanation:
The line segment passing through $(2,3,1)$ and $(4,-5,3)$ is given by $2i-8j+2k$. 
Hence, the normal to the plane is given by $(2i-8j+2k) \times i=2j+8k$. 
Hence, the equation of plane is given by $2y+8z+d=0$. 
Since it passes through $(2,3,1)$ we get $6+8+d=0 \Rightarrow d=-14$. 
Thus, the equation of plane is given by $y+4z-7=0$.  

Equation of the plane passing through the point $(1, 1, 1)$ and perpendicular to each of the planes $x+ 2y+ 3z= 7$ and $2x- 3y +4z= 0$, is

maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

  1. $17x- 2y +7z= 12$

  2. $17x+ 2y -7z= 12$

  3. $17x+ 2y +7z= 12$

  4. $17x- 2y -7z= 12$

Show answer
Correct Option: B
Explanation:

Let $ax+by+cz=1$ be the desired plane.
Since, it is perpendicular to $x+2y+3z=7$ & $2x-3y+4z=0$
Therefore, $a+2b+3c=0$        .... (1)
and $2a-3b+4c=0$       ...(2)
and it passes through $(1,1,1)$
Therefore, $a+b+c=1$      ...(3)
Solving $(1),(2)$ and $(3)$ simultaneously, we get

$a=\dfrac {17}{12}$, $b=\dfrac {1}{6}$, $c=-\dfrac {7}{12}$
Therefore, desired plane is $17x+2y-7z=12$

Ans: B

The cartesian form of the plane 
$ { r } =(s-2t)\hat { i+(3-t)\hat { j+(2s+t)\hat { k }  }  } $ is 

maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

  1. $ 2 x-5 y-z-15=0$

  2. $2 x-5 y+z-15=0$

  3. $2 x-5 y-z+15+0$

  4. $2 x+5 y-z+15=0$

Show answer
Correct Option: C
Explanation:

Since $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$ is the proof any vector $(x,y,z)$on the plane. The given equation can be written as.

$x\hat{i}+y\hat{j}++z\hat{k}=(s—2t)\hat{i}+(3-t)\hat{j}+(2s+t)\hat{k}\\x=(s-2t)\quad y=(3-t)\quad z=2s+t$

Similarly We get $x-2y=s-6$ and $y+z=3+2s$

Now eliminating $s$ we get $2(x-2y)-(y+z)=-15$

$2x-5y-z+15=0$ is the required form of the equation.