Tag: three dimensional geometry

Questions Related to three dimensional geometry

Consider the planes  $3x - 6y - 2z = 15$  and  $2x + y - 2z = 5$.  Which of the following vectors is parallel to the line of intersection of given plane

line of intersection of two planes line and a plane vectors, lines and planes three dimensional geometry maths

  1. $13i + 2j + 15k$

  2. $14i + 2j + 13k$

  3. $13i + 3j + 15k$

  4. $14i + 2j + 15k$

Show answer
Correct Option: D
Explanation:

Consider the problem 

Let 
$\begin{array}{l} 3x-6y{ { - } }2z=15 \ 2x+y-2z=5 \end{array}$

For $z=0$
we get 
$x=3,\;y=-1$
Direction ratios of the plane 

$3,-6,-2$ and $2,1,-2$
and 
direction ratios of intersected line
$14,2,15$
Therefore,

$\dfrac{{x - 3}}{{14}} = \dfrac{{y + 1}}{2} = \dfrac{{z - 0}}{2} = \lambda $

of planes So, vectors which parallel to the intersection plane 
$14\hat i + 2\hat j + 15\hat k$

Hence option $D$ is the correct answer.

The equations of the line of intersection of the planes $\displaystyle x + y + z = 2$ and $\displaystyle 3x - y + 2z = 5$ in symmetric form are

line of intersection of two planes line and a plane vectors, lines and planes three dimensional geometry maths

  1. $\displaystyle \dfrac{x - \dfrac{7}{4}}{4} = \dfrac{y - \dfrac{1}{4}}{-1} = \dfrac{z}{-3}$

  2. $\displaystyle \dfrac{x}{3} = \dfrac{y + \dfrac{1}{3}}{1} = \dfrac{z - \dfrac{7}{4}}{-4}$

  3. $\displaystyle \frac{x}{1} = \frac{3y + 1}{1} = \frac{3z - 7}{-4}$

  4. none of these

Show answer
Correct Option: B
Explanation:

Given, $\vec{n _{1}}=\hat i+\hat j+\hat k$
$\vec{n _{2}}=3\hat i-\hat j+2\hat k$

Therefore the direction vector of the line is parallel to 
$\vec{n _{1}}\times \vec{n _{2}}$
$=(\hat i+\hat j+\hat k)\times(3\hat i-\hat j+2\hat k)$
$=3\hat i+\hat j-4\hat k$
Hence, the equation of the line will be of the form
$\dfrac{x-\alpha}{3}=\dfrac{y-\beta}{1}=\dfrac{z-\gamma}{-4}$

Consider the planes $\displaystyle 3x-6y-2z=15$ and $\displaystyle 2x+y-2z=5.$ 


Assertion: The parametric equations of the line of intersection of the given planes are $\displaystyle x=3+14t, y=1+2t, z=15t.$ because  

Reason: The vector $\displaystyle 14\hat{i}+2\hat{j}+15\hat{k}$ is parallel to the line of intersection of given planes.

line of intersection of two planes line and a plane vectors, lines and planes three dimensional geometry maths

  1. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

  2. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

  3. Assertion is correct but Reason is incorrect

  4. Both Assertion and Reason are incorrect

Show answer
Correct Option: D
Explanation:

$3x-6y-2z=15$

$2x+y-2z=5$
For $z=0$ we get $x=3,y=-1$
Direction vectors of plane are $<3 -6 -2 >$ and $<2, 1 ,-2>$
Then the dr's of line of intersection of planes  is $<14, 2 ,15>$
$\therefore \dfrac{x-3}{14}=\dfrac{y+1}{2}=\dfrac{z-0}{15}=\lambda$
$\Longrightarrow x=14\lambda+3y=2\lambda-1z=15\lambda$
hence Both Assertion and Reason are incorrect 

The line of intersection of the planes $\displaystyle \bar r (3\hat i - \hat j + \hat k) = 1$ and $\displaystyle \bar r (\hat i + 4\hat j - 2\hat k) = 2$ is parallel to the vector

line of intersection of two planes line and a plane vectors, lines and planes three dimensional geometry maths

  1. $\displaystyle -2\hat i + 7\hat j + 13\hat k$

  2. $\displaystyle 2\hat i - 7\hat j - 13\hat k$

  3. $\displaystyle 2\hat i + 7\hat j + 13\hat k$

  4. $\displaystyle 2\hat i + 2\hat j + 13\hat k$

Show answer
Correct Option: A
Explanation:
We know that $\vec{r} = {x}\hat{i} +{y}\hat{j}+{z}\hat{k}$
Hence
The equations of the planes can be written as $3x - y + z - 1 = 0$ and $x + 4y - 2z - 2 = 0$
We find the coordinates of a point lying on their intersection line.
Let $z = t$
$\therefore 3x - y = 1 - t \ ...(1)$
$x + 4y = 2t + 2 \ ...(2)$
Multiplying equation (1) by 4 and adding that to equation (2), we get 
$13x = 4 - 4t + 2t + 2 = 6 - 2t$
$\therefore x = \cfrac{6 - 2t}{13}$
Substituting this in equation (1), we get 
$y = 3x - 1 + t = \cfrac{18 - 6t - 13 + 13t}{13} = \cfrac{5 + 7t}{13}$
Thus, the point can be written as $\left ( \cfrac{6 - 2t}{13}, \cfrac{5 + 7t}{13}, t \right )$
i.e. $\left ( \cfrac{6}{13}, \cfrac{5}{13}, 0 \right ) + \left ( \cfrac{-2}{13}, \cfrac{7}{13}, 1 \right ) t$
$\Rightarrow$ the line of intersection is parallel to $-2\hat{i} + 7\hat{j} + 13\hat{k}$

Consider three planes$P _1: x-y+z=1$$P _2: x+y-z=-1$$P _3: x-3y+3z=2$Let $L _1, L _2, L _3$ be the lines of intersection of the planes ${P} _{2}$ and ${P} _{3},\ {P} _{3}$ and ${P} _{1}$, and ${P} _{1}$ and ${P} _{2}$, respectively.
STATEMENT-$1$ : At least two of the lines ${L} _{1},\ {L} _{2}$ and ${L} _{3}$ are non-parallel.
and 
STATEMENT -$2$ : The three planes do not have a common point.

line of intersection of two planes line and a plane vectors, lines and planes three dimensional geometry maths

  1. Statement-1 is True, Statement -2 is True; Statement-2 is a correct explanation for Statement-1

  2. Statement -1 is True, Statement -2 is True; Statement-2 is NOT a correct explanation for Statement-1

  3. Statement -1 is True, Statement -2 is False

  4. Statement -1 is False, Statement -2 is True

Show answer
Correct Option: D
Explanation:

Given three planes are 
${ P } _{ 1 }:x-y+z=1$   ...$(1)$
${ P } _{ 1 }:x+y-z=-1$    ....$(2)$
and ${ P } _{ 1 }:x-3y+3z=2$    ...$(3)$
Solving Eqs. $(1)$ and $(2)$, we have 
$x=0,z=1+y$
which does not satisfy Eq. $(3)$
As, $x-3y+3z=0-3y+3\left( 1+y \right) =3\left( \neq 2 \right) $
$\therefore$ Statement II is true. 
Nest,since we know that direction ratio's of line of intersection of planes ${ a } _{ 1 }x+{ b } _{ 1 }y+{ c } _{ 1 }z+{ d } _{ 1 }=0$ and
${ a } _{ 2 }x+{ b } _{ 2 }y+{ c } _{ 2 }z+{ d } _{ 2 }=0$
$b _{ 1 }{ c } _{ 2 }-{ b } _{ 2 }{ c } _{ 1 },{ c } _{ 1 }{ a } _{ 2 }-{ a } _{ 1 }{ c } _{ 2 },{ a } _{ 1 }{ b } _{ 1 }$
Using above result, we get
Direction ratio's of lines ${ L } _{ 1 },{ L } _{ 2 }$ and ${ L } _{ 3 }$ are $o,2,2;0,-4,-4;0,-2,-2$ respectively. 
$\Rightarrow $ All the three lines ${ L } _{ 1 },{ L } _{ 2 }$ and ${ L } _{ 3 }$ are parallel pairwise. 
$\therefore $ Statement I is false. 

Let L be the line of intersection of the planes $2x + 3y + z = 1$ and $x + 3y + 2z = 2$. If L makes an angle $\alpha$ with the positive x-axis, then $\cos \alpha$ equals

line of intersection of two planes line and a plane vectors, lines and planes three dimensional geometry maths

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

  2. $\dfrac{1}{2}$

  3. $1$

  4. $\dfrac{1}{\sqrt{2}}$

Show answer
Correct Option: A
Explanation:
We have 
$2x+3y+z=1$
$\Longrightarrow 2x+3y=1-z$      ...(i)
$x+3y+2z=2$
$\Longrightarrow x+3y=2-2z$    ...(ii)
$(i)-(ii)$
$\Longrightarrow 2x+3y-x-3y=1-z-2+2z$
$\Longrightarrow x=z-1$
$\Longrightarrow z=\cfrac{x+1}{1}$
Putting values of $z$ in (ii), we get,
$z-1+3y=2-2z$
$\Longrightarrow 3y=2-2z-z+1$
$3y=-3(z-1)$
$\Longrightarrow y=-(z-1)=-z+1$
$\therefore z=\cfrac{y-1}{1}$
Hence, we have $\cfrac{x+1}{1}=\cfrac{y-1}{1}=\cfrac{z}{1}$
thus $\cos\alpha=\cfrac{a}{(\sqrt {(a^2+b^2+c^2)})}$
$\therefore \cos\alpha=\cfrac{1}{\sqrt3}$

Find the angle between the line of intersection of the planes $\overrightarrow { r } .\left( i+2j+3k \right) =0$ and $\overrightarrow { r } .\left( 3i+2j+3k \right) =0$ with coordinate axes

line of intersection of two planes line and a plane vectors, lines and planes three dimensional geometry maths

  1. with $x$-axis $\displaystyle \dfrac { \pi  }{ 2 } $

  2. with $y$-axis $\displaystyle \cos ^{ -1 }{ \left( \dfrac { 3 }{ \sqrt { 13 }  }  \right)  } $

  3. with $y$-axis $\displaystyle \cos ^{ -1 }{ \left( \dfrac { 2 }{ \sqrt { 13 }  }  \right)  } $

  4. all of these

Show answer
Correct Option: A,B
Explanation:

Direction ratios: $\displaystyle \begin{vmatrix} i & j & k \ 1 & 2 & 3 \ 3 & 2 & 3 \end{vmatrix}=\left( 0,6,4 \right) $ or $(-,3,-2)$

Therefore angle with $x$-axis: $\displaystyle \dfrac { \pi  }{ 2 } $
with $y$- axis $\displaystyle \cos ^{ -1 }{ \left( \dfrac { 3 }{ \sqrt { 13 }  }  \right)  } $
with $z$-axis $\displaystyle \cos ^{ -1 }{ \left( \dfrac { -2 }{ \sqrt { 13 }  }  \right)  } $

Statement-I: The point $A(3,1,6)$ is the mirror image of the point $B(1,3,4)$ in the plane $x-y+z=5$.
Statement-2: The plane $x-y+z=5$ bisects the line segment joining $A(3,1,6)$ and $B(1,3,4)$.

maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

  1. (1 ) StatementI is true. Statement-1 is true: Statement--2 is a correct explanation for Statement-1.

  2. (2) StatementI is true, Statement-2 is true: Statement-9 is not a correct explanation for statement-1.

  3. (3) Statement--I is true, Statement-2 is false.

  4. (4) StatementI is false. Statement-2 is true.

Show answer
Correct Option: A
Explanation:

Mid-point of AB=$\begin{array}{l} = \left( {\dfrac{{3 + 1}}{2},\dfrac{{1 + 3}}{2},\dfrac{{4 + 6}}{2}} \right)\ = (2,2,5)\end{array}$
lies on the plane as it satisfies the equation of the plane
and DR s of AB $ = (2, - 2,2)$
DR s of normal to the plane $= (1, - 1,1)$
AB is the perpendicular bisector.
Hence, A is the image of 2 
 

If the points $(1,2,3)$ and $(2,-1,0)$ lie on the opposite sides of the plane $2x+3y-2z=k$, then

maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

  1. $k< 1$

  2. $k> 2$

  3. $k< 1$ or $k> 2$

  4. $1< k< 2$

Show answer
Correct Option: D
Explanation:

 Given plane equation is $2{x}+3{y}-2{z}-k=0$

$(1,2,3)$ and $(2,-1,0)$ lies on the opposite sides of the plane
$(2(1)+3(2)-2(3)-k)(2(2)+3(-1)-2(0)-k)<0$
$(2-k)(1-k)<0\implies (k-1)(k-2)<0$
$\implies 1<k<2$

If the planes $x - cy - bz = 0,cx - y + az = 0\,$ and $bx + ay - z = 0$ pass through a stright line,then the value of ${a^2} + {b^2} + {c^2} + 2abc\,$ is:

maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

  1. $1$

  2. $2$

  3. $3$

  4. none of these

Show answer
Correct Option: A
Explanation:
If the planes $(x - cy - bz = 0), (cx - y + az = 0)$ and $(bx + ay - z = 0)$ are in same line.
$\therefore$ They must be collinear.
$\begin{vmatrix}1 & -c & -b\\ c & -1 & a\\ b & a & -1\end{vmatrix} = 0$
$\Rightarrow 1(1 - a^2) + c(-c - ab) -b(ac + b) = 0$
$\Rightarrow 1 - a^2 - c^2 - abc - abc - b^2 = 0$
$\therefore a^2 + b^2 + c^2 + 2abc = 1$
Option A is correct