Tag: general form of the equation of a plane

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Questions Related to general form of the equation of a plane

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

The cartesian equation of the plane which is at a distance of 10 unite from the original and perpendicular to the vector i + 2j -2k is 

  1. x+2y+2z = 30

  2. x - 2y - 2z =30

  3. x - 2y + 2z = 30

  4. x+2y-2z = 30
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The equation of a plane at distance d from origin with normal vector n is r dot (n/|n|) = d. Here n = i + 2j - 2k, so |n| = sqrt(1+4+4) = 3. The equation is x + 2y - 2z = 10 * 3 = 30.

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

The normal form of $2x-2y+z=5$ is

  1. $12x-4y+3z=39$
  2. <p class="MsoNormal">$\displaystyle \dfrac{-6}{7}x+\dfrac{2}{7}y+\dfrac{3}{7}z=1$</p>
  3. <p class="MsoNormal">$\displaystyle \dfrac{12}{13}x-\dfrac{-4}{13}y+\dfrac{3}{13}z=3$</p>
  4. <p class="MsoNormal">$\displaystyle \dfrac{2}{3}x-\dfrac{2}{3}y+\dfrac{1}{3}z=\dfrac{5}{3}$</p>
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The dr's of the normal to the plane are $(2,-2,1)$.
The dc's will be $\left ( \dfrac{2}{3} , \dfrac{-2}{3} , \dfrac{1}{3} \right)$
Hence, the equation of the plane in the normal form will be,
$ \dfrac{2x}{3} - \dfrac{2y}{3} + \dfrac{z}{3} $ = $ \dfrac{5}{3} $

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

If a line is given by  $\dfrac{x-2}3 = \dfrac{y+10}5 = \dfrac{z+6}2$,  then which of the following points lies on this line?

  1. $(5,11,0)$
  2. $(3,10,0)$
  3. $(11,5,0)$
  4. $(0,5,11)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Putting $x=11$ and $y=5$ and $z=0$ in the given equation of line, we get
$\begin{array}{l} \dfrac { { x-2 } }{ 3 } =\dfrac { { y+10 } }{ 5 } =\dfrac { { z+6 } }{ 2 }  \\ \dfrac { { 11-2 } }{ 3 } =\dfrac { { 5+10 } }{ 5 } =\dfrac { { 0+6 } }{ 2 }  \\ \dfrac { 9 }{ 3 } =\dfrac { { 15 } }{ 3 } =\dfrac { 6 }{ 2 }  \\ 3=3=3 \end{array}$
therefore $(11,5,0)$ satisfies given equation of line 
hence  $(11,5,0)$ lies on the given line

Option $C$ is the correct answer.
Multiple choice maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

Find the equation of the plane through the points $(1, 0, -1), (3, 2, 2)$ and parallel to the line $\dfrac{x-1}{1}=\dfrac{y-1}{-2}=\dfrac{z-2}{3}$.

  1. $4x-y-2z=6$
  2. $4x-y-2z=-6$
  3. $4x-y+2z=6$
  4. $4x+y-2z=6$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$L:\cfrac { x-1 }{ 1 } =\cfrac { y-1 }{ -2 } =\cfrac { z-2 }{ 3 } \ P(1,0,-1),Q(3,2,2)$

$ \therefore$ Direction of  $\overrightarrow { PQ } =2\hat { i } +2\hat { j } +3\hat { k } $
Plane is parallel to line $L$ & contains $\overrightarrow { PQ } $. (normal to plane $\bot$ to $PQ$ & $L$)
$\therefore \overrightarrow { n } =\left| \begin{matrix} \hat { i }  & \hat { j }  & \hat { k }  \ 2 & 2 & 3 \ 1 & -2 & 3 \end{matrix} \right| =(12)\hat { i } -3\hat { j } +(-6)\hat { k } $
direction ratios of normal are $(4,-1,-2)$
$\therefore$ Equation plane passing through $(1,0,-1)$ and having directions of normal $(4,-1,-2)$.
$4x-y-2z=6$

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

The equation of the plane passing through the straight line $\dfrac{x-1}{2}=\dfrac{y+1}{-1}=\dfrac{z-3}{4}$ and perpendicular to plane $x+2y +z=12$ is:

  1. $9x+2y-5z+8 =0$
  2. $9x +2y -5z +10=0$
  3. $9x-2y +5z +6=0$
  4. $9x -2y -5z+4=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Let the DR of the normal of the plane be $<a, b, c>$
Since it passes through a line.
$\therefore$ Normal of the plane must be perpendicular to the line
$\therefore a + 2b + c = 0$ ...... $(1)$ ($\perp$ to another plane)
$2a - b + 4c = 0$ ...... $(2)$
On solving :
$\dfrac{a}{8 + 1} = \dfrac{-b}{4 - 2} = \dfrac{c}{-1 - 4}$
$\Rightarrow \dfrac{a}{9} = \dfrac{b}{-2} = \dfrac{c}{-5}$
$a = 9, b = -2, c = -5$
Any point on the straight line
$(2\alpha + 1, -\alpha - 1, 4\alpha + 3)$
putting $\alpha = 1$
$(3, -2, 7)$
$\therefore$ Equation of the line plane with normal DR $(9, -2, -5)$ and passing through $(3, -2, 7)$
$\therefore 9(x - 3) - 2(y + 2) - 5(z - 7)=0$
$\Rightarrow 9x - 27 - 2y - 4 - 5z + 35 = 0$
$\Rightarrow 9x - 2y - 5z + 4 = 0$

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

Equation of the plane containing the straight lines $\dfrac{x}{2} = \dfrac{y}{3} = \dfrac{z}{4}$ and perpendicular to the plane containing the straight lines  $\dfrac{x}{3} = \dfrac{y}{4} = \dfrac{z}{2}$ and $\dfrac{x}{4} = \dfrac{y}{2} = \dfrac{z}{3}$

  1. $x + 2y - 2z = 0$
  2. $3x + 2y - 2z = 0$
  3. $x - 2y + z = 0$
  4. $5x + 2y - 4z = 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Vector normal to plane $P _1$ is $\overrightarrow{n _1}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\3&4&2\4&2&3\end{vmatrix}$


                                                       $=\hat{i}(12-4)-\hat{j}(9-8)+\hat{k}(6-8)$
                                                       $=8\hat{i}-\hat{j}-10\hat{k}$
Plane $P _2$ is perpendicular to this plane and it leaking the line $\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{2}{4}$

$\overrightarrow{n _2}$ is normal to

$\overrightarrow{n _2}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\8&-1&-10\2&3&4\end{vmatrix}$

       $=\hat{i}(-4+30)-\hat{j}(32+20)+\hat{k}(24+2)$

       $=26\hat{i}-52\hat{j}+26\hat{k}$

This plane $T _2$ having normal $n _2$ passing through $(0,0,0)$ is $26x-52y+26z=0$ $i.e.$ $x-2y+z=0$

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

Let a,b,c be any real numbers.Suppose that there are real numbers x,y,z not all zero such that $x=cy+bz , y=az+cx$ and $z=bx+ay$, then ${a^2} + {b^2} + {c^2} + 2abc $ is equal to

  1. 2

  2. -1

  3. 0

  4. 1
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$a, b, c$ real numbers

$x, y, z$ real numbers not all zero

$x=cy+bz\rightarrow x-cy-bz=0 --- (1)$

$y=az+cx\rightarrow cx+y-az=0---(2)$

$z=bx+cy\rightarrow -bx-ay+z=0 --- (3)$

The system of equation have trivial solution then

$\begin{vmatrix} 1 & -c & -b \\ c & -1 & a \\ b & a & -1 \end{vmatrix}=0$

$\Rightarrow 1(1-a^{2})+c(-c-ab)-b(a+b)=0$

$\Rightarrow a^{2}+b^{2}+c^{2}+2abc=1$

$D$ is correct
Multiple choice maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

The direction cosines of the normal to the plane $x+2y-3z+4=0$ are

  1. $\cfrac { -1 }{ \sqrt { 14 } } ,\cfrac { -2 }{ \sqrt { 14 } } ,\cfrac { 3 }{ \sqrt { 14 } } $
  2. $\cfrac { 1 }{ \sqrt { 14 } } ,\cfrac { 2 }{ \sqrt { 14 } } ,\cfrac { 3 }{ \sqrt { 14 } } $
  3. $\cfrac { -1 }{ \sqrt { 14 } } ,\cfrac { 2 }{ \sqrt { 14 } } ,\cfrac { 3 }{ \sqrt { 14 } } $
  4. $\cfrac { 1 }{ \sqrt { 14 } } ,\cfrac { -2 }{ \sqrt { 14 } } ,\cfrac { -3 }{ \sqrt { 14 } } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Clearly, the normal to the plane has DR's $\equiv(1, 2, -3)$

$\therefore $ DCs  are $\equiv \pm \left(\dfrac{1}{\sqrt{1^2 + 2^2 + (-3)^2}} , \dfrac{2}{\sqrt{14}} , \dfrac{-3}{\sqrt{14}}\right)$
$= \pm \left(\dfrac{1}{\sqrt{14}} , \dfrac{2}{\sqrt{14}} , \dfrac{-3}{\sqrt{14}} \right)$
$\therefore A$