Tag: the plane

Quizzes (3) Questions (79)

Questions Related to the plane

Multiple choice maths the plane equation of a plane in intercept form equation of a plane in different forms different forms of equation of a plane

The expression of $x+y+z=1$ in form of $x\cos { \alpha  } +y\cos { \beta  } +z\cos { \gamma  } =p$ is _______.

  1. $x+y+z=1$
  2. $\cfrac { x }{ 2\sqrt { 3 } } +\cfrac { y }{ 2\sqrt { 3 } } +\cfrac { z }{ 2\sqrt { 3 } } =\cfrac { 1 }{ \sqrt { 3 } } $
  3. $\cfrac { x }{ \sqrt { 3 } } +\cfrac { y }{ \sqrt { 3 } } +\cfrac { z }{ \sqrt { 3 } } =1$
  4. $\cfrac { x }{ \sqrt { 3 } } +\cfrac { y }{ \sqrt { 3 } } +\cfrac { z }{ \sqrt { 3 } } =\cfrac { 1 }{ \sqrt { 3 } } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

Multiple choice maths the plane equation of a plane in intercept form equation of a plane in different forms different forms of equation of a plane

The sum of Y and Z intercepts of the plane $3x+4y-6z=12$ is ___________.

  1. $10$
  2. $4$
  3. $1$
  4. $5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$3x+4y-6z=12$
$\therefore\dfrac{3x}{12}+\dfrac{4y}{12}+\dfrac{(-6)z}{12}=1$
$\therefore \dfrac{x}{4}+\dfrac{y}{3}+\dfrac{z}{(-2)}=1$
$\therefore$ y intercept $b=3$ and z intercept $c=-2$
$\therefore$ y intercept $+$ z intercept $=3+(-2)=1$

Multiple choice maths the plane equation of a plane in intercept form equation of a plane in different forms different forms of equation of a plane

The plane $ax+by+cz=1$ meets the coordinate axes in $A, B$ and $C$. The centroid of the triangle is:

  1. $(3a, 3b, 3c)$
  2. $\left( \dfrac { a }{ 3 } ,\dfrac { b }{ 3 } ,\dfrac { c }{ 3 } \right)$
  3. $\left( \dfrac { 3 }{ a } ,\dfrac { 3 }{ b }, \dfrac { 3 }{ c } \right)$
  4. $\left( \dfrac { 1 }{ 3a } ,\dfrac { 1 }{ 3b } ,\dfrac { 1 }{ 3c } \right)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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 $\left( {\dfrac{1}{{3a}},\dfrac{1}{{3b}},\dfrac{1}{{3c}}} \right)$.

Multiple choice maths the plane equation of a plane in intercept form equation of a plane in different forms different forms of equation of a plane

If a plane passes through a fixed point $\left ( 2, 3, 4 \right )$ and meets the axes of reference in $A$, $B$ and $C$, the point of intersection of the planes through $A$, $B$, $C$ parallel to the coordinate planes can be

  1. $\left ( 6, 9, 12 \right )$
  2. $\left ( 4, 12, 16 \right )$
  3. $\left ( 1, 1, -1 \right )$
  4. $\left ( 2, 3, -4 \right )$
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

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.

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

Find the planes bisecting the acute angle between the planes $x-y+2x+1=0$ and $2x+y+z+2=0$

  1. $x+z-1=0$
  2. $x+z+1=0$
  3. $x-z-1=0$

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

Given planes are $x-y+2{z}+1=0,2{x}+y+z+2=0$

The plane bisecting the acute angle between the planes will be $\dfrac{x-y+2{z}+1}{\sqrt{1+1+4}}=-\dfrac{2{x}+y+z+2}{\sqrt{4+1+1}}$
$\implies x+z+1=0$

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

The planes $x-3y+4z-1=0$ and $kx-4y+3z-5=0$ are perpendicular then value of $k$ is

  1. $24$
  2. $-24$
  3. $12$
  4. $0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let direction ratios of the perpendicular to the plane 
$x-3y+4z-1=4$ are $a _{2}=1, b _{1}=-3, c _{1}=4$
and that of planes will be perpendicular if 
$a _{1}a _{2}+b _{1}b _{2}+c _{1}c _{2}=0$
$k+(-3) \times (-4)+4 \times{3}=0$
$k+12+12=0$
$k=-24$








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

The equation of the plane which bisects the angle between the planes $3x-6y+2z+5=0$ and $4x-12y+3z-3=0$ which contains the origin is ?

  1. $33x-13y+32z+45=0$
  2. $x-3y+z-5=0$
  3. $33x+13y+32z+45=0$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given ,

The required equation of plane bisects the given two planes.
$\begin{array}{l} \therefore \frac { { 3x-6y+2z+5 } }{ { \sqrt { { 3^{ 2 } }+{ { \left( { -6 } \right)  }^{ 2 } }+{ { \left( 2 \right)  }^{ 2 } } }  } } =\pm \frac { { 4x-12y+3z-3 } }{ { \sqrt { { 4^{ 2 } }+{ { \left( { -12 } \right)  }^{ 2 } }+{ 3^{ 2 } } }  } }  \ \frac { { 3x-6y+2z+5 } }{ { \sqrt { 49 }  } } =\pm \frac { { 4x-12y+3z-3 } }{ { \sqrt { 169 }  } }  \ \frac { { 3x-6y+2z+5 } }{ 7 } =\pm \frac { { 4x-12y+3z-3 } }{ { 13 } }  \ 39x-78y+26z+65=\pm 28x-84y+21z-21 \end{array}$
Now, solving for the positive value, we get
$39x - 78y + 26z + 65$.........(i)
or,  $11x + 6y + 5z + 36 = 0$
And for negative value, we get
$\begin{array}{l} 39x-78y+26z+65=-\left( { 28x-84y+21z-21 } \right)  \ or,\, \, 67x-162y+47z+44=0.......\left( { ii } \right)  \end{array}$
$\because $None of the answer matches with the given equation.
Hence,
Option $D$ is correct  in this case.

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

The corner of a square OPQR is folded up so that the plane OPQ is perpendicular to the plane OQR, the angle between OP and QR is 

  1. $\dfrac { \pi }{ 2 } $
  2. $\dfrac { \pi }{ 3 } $
  3. $\dfrac { \pi }{ 4 } $
  4. $\dfrac { \pi }{ 6 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When the corner of a square is folded such that two planes are perpendicular, the geometry dictates that the angle between the original lines OP and QR becomes 90 degrees.

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

The angle between the plane passing through the points $A(0,\ 0,\ 0),\ B(1,\ 1,\ 1),\ C(3,\ 2,\ 1)$ & the plane passing through $A(0,\ 0,\ 0),\ B(1,\ 1,\ 1),\ D(3,\ 1,\ 2)$ is

  1. $90^{o}$
  2. $45^{o}$
  3. $120^{o}$
  4. $30^{o}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\begin{array}{l} { \pi _{ 1 } }=ax+by+cz=0 \ a+b+c=0 \ 3a+2b+c=0 \ \frac { a }{ { -1 } } =\frac { { -b } }{ { 1-3 } } =\frac { c }{ { 2-3 } }  \ \frac { a }{ { -1 } } =\frac { b }{ 2 } =\frac { c }{ { -1 } }  \ -x+2y-z=0 \ { \pi _{ 1 } }:\, x-2y+z=0 \ and, \ { \pi _{ 2 } }=ax+by+cz=0 \ a+b+c=0 \ 3a+b+2c=0 \ \frac { a }{ 1 } =\frac { { -b } }{ { 2-3 } } =\frac { c }{ { 1-3 } }  \ \frac { a }{ 1 } =\frac { b }{ 1 } =\frac { c }{ { -2 } }  \ { \pi _{ 2 } }:\, x+y-2z=0 \ Now, \ \cos  \theta =\frac { { \left( { 1-2-2 } \right)  } }{ { \sqrt { 6 } \sqrt { 6 }  } } =\frac { { -3 } }{ 6 } =\frac { { -1 } }{ 2 }  \ \therefore \theta ={ 120^{ \circ  } } \ Hence,\, the\, option\, C\, is\, the\, correct\, answer. \end{array}$

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

The angle between the planes
$\vec{r}(\hat{i}+2\hat{j}+\hat{k})=4$ and $\vec{r}(\hat{-i}+\hat{j}+2\hat{k})=9$

  1. $30^{\mathrm{o}}$
  2. $60^{\mathrm{o}}$
  3. $45^{\mathrm{o}}$
  4. $90^{0}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Angle between two planes is the angle between their normal vectors.

For the first plane, normal vector is $\vec{n _0}=(1,2,1)$
For second plane, normal vector is $\vec{n _1}=(-1,1,2)$
Let $\theta$ be the angle between the planes, it is also the angle between their normals.
$\implies \cos \theta = \dfrac{{n} _{1}.{n} _{2}}{|n _1||n _2|} $

$\implies \cos \theta = \dfrac{(1,2,1)\cdot (-1,1,2)}{\sqrt{1^2+2^2+1^2}\sqrt{(-1)^2+1^2+2^2}} $

$\implies \cos \theta = \dfrac{-1+2+2}{6}=\dfrac{1}{2}$
$\implies \theta $ = $ {60}^{o}$