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Line of intersection of two planes - class-XII

Description: line of intersection of two planes
Number of Questions: 27
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Tags: three dimensional geometry maths vectors, lines and planes
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The line of intersection of the planes $\overrightarrow { r } .\left( 3\hat { i } -\hat { j } +\hat { k }  \right) =1$ and $\overrightarrow { r } .\left( \hat { i } +4\hat { j } -2\hat { k }  \right) =2$ is parallel to vector

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

  2. $2\hat { i } +7\hat { j } -13\hat { k } $

  3. $-2\hat { i } -7\hat { j } +13\hat { k } $

  4. $2\hat { i } +7\hat { j } +13\hat { k } $


Correct Option: A

There are two different planes, one passing though the x-axis and the other passing through y-axis. The angle between the planes is $\cfrac{\pi}{4}$. Then locus of a point on the line of intersection of the planes in.

  1. $(x^2+y^2+z^2)x^2=y^2z^2$

  2. $(x^2+y^2+z^2)z^2=x^2y^2$

  3. $(x^2+y^2+z^2)y^2=x^2z^2$

  4. None of these


Correct Option: A

The line of intersection of the planes 
$r.\left( {3\hat i - \hat j + \hat k} \right) = 1$ and $r.\left( {\hat i + 4\hat j - 2\hat k} \right) = 2$ is parallel to the vector

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

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

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

  4. $2\hat i + 7\hat j + 13\hat k$


Correct Option: A
Explanation:

$\overrightarrow { r } .(3\hat { i } -\hat { j } +\hat { k } )=1$ and $\overrightarrow { r } .(\hat { i } +4\hat { j } -2\hat { k } )=2$

direction vector of normal to plane are,
${ \overrightarrow { b }  } _{ 1 }=3\hat { i } -\hat { j } +\hat { k } ,\quad { \overrightarrow { b }  } _{ 2 }=\hat { i } +4\hat { j } -2\hat { k } $
Let $lmn$ be directions of line of intersection.
$3l-m+n=0$
$l+4m-2n=0$
$\therefore \cfrac { l }{ -2 } =\cfrac { -m }{ -7 } =\cfrac { n }{ 13 } \quad \Rightarrow (l,m,n)=(-2,7,13)$
$\therefore$ vector parallel to line of intersection of planes is,
$\overrightarrow { r } =-2\hat { i } +7\hat { j } +13\hat { k } $

A unit vector parallel to the intersection of the planes $\vec r\cdot (\hat i-\hat j+\hat k)=5$ and $\vec r\cdot (2\hat i+\hat j-3\hat k)=4$ can be

  1. $\dfrac {2\hat i+5\hat j+3\hat k}{\sqrt {38}}$

  2. $\dfrac {2\hat i-5\hat j+3\hat k}{\sqrt {38}}$

  3. $\dfrac {-2\hat i-5\hat j-3\hat k}{\sqrt {38}}$

  4. $\dfrac {-2\hat i+5\hat j-3\hat k}{\sqrt {38}}$


Correct Option: A,C
Explanation:

Vector parallel to intersection of planes 


$\vec p= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 1 \ 2 & 1 & -3 \end{vmatrix}$


$= \hat{i} (3 - 1) - \hat{j} (-3 - 2) + \hat{k} (1 + 2)$

$= 2 \hat{i} + 5 \hat{j} + 3 \hat{k}$

$|\vec p|=\sqrt{2^2 + 5^2 + 3^2}=\sqrt{38}$

unit vector parallel to intersection of planes

$= \pm \dfrac{(2 \hat{i} + 5 \hat{j} + 3 \hat{k} )}{\sqrt{2^2 + 5^2 + 3^2}}$

$= \pm \dfrac{(2 \hat{i} + 5 \hat{j} + 3 \hat{k})}{\sqrt{38}}$

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:

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

  2. $1$

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

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


Correct Option: D
Explanation:

The direction vector of the line of the intersection of the planes $2x+3y+z=1$ and $x+3y+2x=2$ is given by

$n _{1}\times n _{2}$
$=(2i+3j+k)\times (i+3j+2k)$
$=3i-3j+3k$
Hence the unit vector along the direction of the line will be 
$-\dfrac{i-j+k}{\sqrt{3}}$
Thus 
$cos\alpha=\dfrac{1}{\sqrt{3}}$ where $\alpha$ is the angle that the line makes with x axis.

A non-zero vector $\vec{a}$ is parallel to the line of intersection of the plane determined  by the vectors $\hat{i},\hat{i}+\hat{j}$ and the plane determined by the vectors $\hat { i } -\hat { j } ,\hat { i } -\hat { k }$. The angle between $\vec{a}$ and $\hat { i } -2\hat { j } +2\hat { k } $ is

  1. $\pi/3$

  2. $\pi/4$

  3. $\pi/6$

  4. $none\ of\ these$


Correct Option: A

The planes $bx-ay=n,cy-bz=1,az-cx=m$ intersect in a line if

  1. $al+bm+cn=0$

  2. $al-bm+cn=0$

  3. $al-bm-cn+1=0$

  4. $al+bm+cn=1$


Correct Option: A
Explanation:
Since the planes

$bx-ay=n$,    $\Rightarrow x=\dfrac { n+ay }{ b } \quad \longrightarrow \left( 1 \right) $

$cy-bz=l$,    $\Rightarrow z=\dfrac { cy-l }{ b } \quad \longrightarrow \left( 2 \right) $
and, $az-cx=m$
intersect in a line, eliminating $x,y,z$ we will get the desired condition.
Substitute $(1)$ and $(2)$ in $(3)$
$a\left( \dfrac { cy-l }{ b }  \right) -c\left( \dfrac { n+ay }{ b }  \right) =m$
$\Rightarrow acy-al-cn-acy=mb$
$\Rightarrow al+cn+bm=0$

$\Rightarrow al+bm+cn=0$

Answer : Option B.

Let $L$ be the line of intersection of the planes $2x+3y+z=1$ and $x+3y+2z=2$.

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

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

  3. $1$

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


Correct Option: A

The equation of plane through the line of intersection of the planes $2x+3y+4z-7=0, x+y+z-1=0$ and perpendicular to the plane $x-5y+3z-6=0$ is

  1. $x+2y+3z=6$

  2. $x-2y+z=6$

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

  4. $x+2y+6z=3$


Correct Option: A

The direction cosines of a line parallel to the planes $\displaystyle 3x + 4y + z = 0$ and $\displaystyle x - 2y - 3z = 5$ are

  1. $\displaystyle \left ( -1, : 1, : -1 \right )$

  2. $\displaystyle \left ( -\frac{1}{\sqrt{3}}, : -\frac{1}{\sqrt{3}}, : \frac{1}{\sqrt{3}} \right )$

  3. $\displaystyle \left ( -\frac{1}{\sqrt{3}}, : \frac{1}{\sqrt{3}}, : \frac{-1}{\sqrt{3}} \right )$

  4. no line possible


Correct Option: C
Explanation:

Given equations of the planes are $\displaystyle 3x + 4y + z = 0$ and $\displaystyle x - 2y - 3z = 5$
required line is parallel to the given palnes,i.e perpendicular to the normals to the planes whose direction ratios are
$(3,4,1)$ and $(1,-2,-3)$ respectively
let $(a,b,c)$ be direction ratios of the line.
$\Rightarrow 3a+4b+c=0 and a-2b-3c=0$
$\Rightarrow a=-b=c$
$\therefore$ direction cosines of the line are $\displaystyle \left ( -\frac{1}{\sqrt{3}}, : \frac{1}{\sqrt{3}}, : -\frac{1}{\sqrt{3}} \right )$ or $\displaystyle \left ( \frac{1}{\sqrt{3}}, : -\frac{1}{\sqrt{3}}, : \frac{1}{\sqrt{3}} \right )$   

If $\displaystyle \left ( 3, : \lambda, : \mu \right )$ is a point on the line then $\displaystyle 2x + y + z = 0 = x - 2y + z -1$ then

  1. $\displaystyle \lambda = \frac{-8}{3}, : \mu = - \frac{1}{3}$

  2. $\displaystyle \lambda = \frac{-1}{3}, : \mu = - \frac{8}{3}$

  3. $\displaystyle \lambda = \dfrac{-4}{3} : \mu = \dfrac{-14}{3}$

  4. $\displaystyle \lambda = -5, : \mu = -1$


Correct Option: C
Explanation:

Since the point lies on the both planes we have, 

$6+\lambda+\mu=0$       ...(1)
$3-2\lambda+\mu-1=0$           ....(2) 
Subtracting equation 2 from 1 we get $\lambda=\cfrac{-4}{3}$. 

And substituting on equation 1 we get $\mu=\cfrac{-14}{3}$.

The variable plane $\displaystyle \left ( 2 \lambda + 1 \right )x + \left ( 3 - \lambda \right )y + z = 4$ always passes through the line 

  1. $\displaystyle \frac{x}{0} = \frac{y}{0} = \frac{x + 4}{1}$

  2. $\displaystyle \frac{x}{1} = \frac{y}{2} = \frac{z}{-3}$

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

  4. none of these


Correct Option: C
Explanation:

Given equation of plane is $ (2 \lambda+1)x+(3- \lambda)y+z=4$


Here $\lambda$ is a variable, we have to eliminate it.


The given equation can be written as $ \lambda(2x-y)+x+3y+z=4$

Now observe that $\lambda$ will be eliminated from the equation, if $2x-y=0$

$\Rightarrow 2x=y$

$\Rightarrow x=\dfrac{y}{2}$

The given equation becomes $x+3y+z=4$

$\Rightarrow 7x+z=4$

$\Rightarrow x=\dfrac{z-4}{-7}$

So, the given plane always passes through $ \dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z-4}{-7}$ irrespective of $\lambda$

Therefore, option $C$ is correct.

The equation of the plane which contains the origin and the line of intersection of the planes $\vec r.\vec a=\vec p$ and $\vec r.\vec b=\vec q$ is

  1. $\vec r.\left( \vec p\vec a-\vec q\vec b \right) =0$

  2. $\vec r.\left(\vec  p\vec a+\vec q\vec b \right) =0$

  3. $\vec r.\left(\vec  q\vec a+\vec p\vec b \right) =0$

  4. $\vec r.\left( \vec q\vec a-\vec p\vec b \right) =0$


Correct Option: D
Explanation:

Any plane through the inetersection of $\vec r.\vec a=\vec p$ and $\vec r.\vec b=\vec q$ is

$r.\left( \vec a-\lambda \vec b \right) =\vec p-\lambda \vec q$   ...(1)
Since it passes through the origin,
$\displaystyle \therefore 0.\left( \vec a-\lambda \vec b \right) =\vec p-\lambda \vec q$
$\Rightarrow \vec p-\lambda \vec q=0$
$\Rightarrow \lambda =\dfrac { \vec p }{ \vec q } $
Putting this value of $\lambda$ in (1), we get
$\displaystyle \vec r.\left( \vec a-\frac { \vec p }{\vec  q } \vec b \right) =\vec p-\frac {\vec  p }{ \vec q } \vec q=0\Rightarrow \vec r.\left( \vec a\vec q-\vec p\vec b \right) =0$
This is the required equation.

The distance of the point $(1, -2, 3)$ from the plane $x-y+z=5$ measured parallel to the line. $\frac { x }{ 2 } =\frac { y }{ 3 } =\frac { z }{ -6 } ,\quad is:$

  1. 1

  2. 6/7

  3. 7/6

  4. 1/6


Correct Option: B
Explanation:
Given,

Let $P=(1,-2,3)$

given plane $x-y+z=5$

to find the distance of point $P=(1,-2,3)$ from the plane $x-y+z=5$ measured along the parallel line to

$\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{-6}$

equation of line passing through $(1,-2,3)$and having DR's $(2,3,-6)$

let $\dfrac{x-1}{2}=\dfrac{y+2}{3}=\dfrac{z-3}{-6}=\lambda $

$x=2\lambda+1,y=3\lambda-2,z=-6\lambda+3$

$\Rightarrow 2\lambda+1-3\lambda+2-6\lambda+3=2+3-6$

$-7\lambda =5-6$

$\therefore \lambda =\dfrac{1}{7}$

Therefore the coordinates of Q are

$\left ( \dfrac{2}{7}+1,\dfrac{3}{7}-2,-\dfrac{6}{7}+3 \right )$

$=\left ( \dfrac{9}{7},-\dfrac{11}{7},\dfrac{15}{7} \right )$

$PQ=\sqrt{\left ( \dfrac{9}{7}-1 \right )^2+\left ( -\dfrac{11}{7}+2 \right )^2+\left ( \dfrac{15}{7}-3 \right )^2}$

$=\sqrt{\dfrac{4}{49}+\dfrac{9}{49}+\dfrac{36}{49}}$

$=\sqrt{\dfrac{49}{49}}$

$=1$

Which of the following does not represent a straight line?

  1. $ax+by+cz+d=0,ax+b'y+cz+d=0(b\neq b')$

  2. $ax+by+cz+d=0,a'x+by+cz+d=0(a\neq a')$

  3. $ax+by+cz+d=0,ax+by+cz+d'=0(d\neq d')$

  4. $ax+by+cz+d=0,ax+by+c'z+d=0(c\neq c')$


Correct Option: C
Explanation:

A. $ax+by+cz+d=0,ax+b'y+cz+d=0(b\neq b')$
Both the planes are different and are not parallel so they will definitely intersect on a line. Thus option A represents a line.
Similarly B and D represents a line. 
But C does not represents line. since $ax+by+cz+d=0,ax+by+cz+d'=0(d\neq d')$ represents two parallel planes which never intersects. 
Hence, option 'C' is correct choice.

Consider a plane $x+2y+3z=15$ and a line $\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-2}{4}$ then find the distance of origin from point of intersection of line and plane.

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

  2. $\dfrac{9}{2}$

  3. $\dfrac{5}{2}$

  4. $4$


Correct Option: B
Explanation:

Let $\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-2}{4}=\lambda$ $\Rightarrow x=2\lambda +1, y=3\lambda -1, z=4\lambda +2$
Now substitution in $x+2y+3z=15$
$\Rightarrow (2\lambda +1)+2(3\lambda -1)+3(4\lambda +2)=15$
$\Rightarrow 2\lambda +1+6\lambda -2+12\lambda +6=15$ $\Rightarrow 20\lambda +5=15$ $\Rightarrow \lambda =\dfrac{1}{2}$
Hence point of intersection is $(2, \dfrac{1}{2}, 4)$
Hence distance from origin is $\sqrt{4+\dfrac{1}{4}+16}=\sqrt{\dfrac{81}{4}}=\dfrac{9}{2}$.

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 

  1. $1$

  2. $\displaystyle \frac{1}{\sqrt{2}}$

  3. $\displaystyle \frac{1}{\sqrt{3}}$

  4. $\displaystyle \frac{1}{2}$


Correct Option: C
Explanation:

Given planes are $2x+3y+z=1$ and $x+3y+2z=2$


Direction ratios of line $L$ through their intersection is given by the cross product of direction cosines of plane.


Let the direction ratios of line are represented by $\vec r$, then

$\vec { r } =(2i+3j+k)\times (i+3j+2k)$

$ \vec { r } =\left| \begin{matrix} i & j & k \\ 2 & 3 & 1 \\ 1 & 3 & 2 \end{matrix} \right| $

$ \vec { r } =i(6-3)-j(4-1)+k(6-3)$

$ \vec { r } =3i-3j+3k$

Direction ratio of $x$ axis is $\vec a =i$

$\vec { r } .\vec { a } =\left| \vec { r }  \right| \left| \vec { a }  \right| \cos { \alpha  } $

$ (3i-3j+3k).(i)=(3\sqrt { 3 } )(1)\cos { \alpha  } $

$ 3-0+0=3\sqrt { 3 } \cos { \alpha  } $

$ \cos { \alpha  } =\dfrac { 3 }{ 3\sqrt { 3 }  } =\dfrac { 1 }{ \sqrt { 3 }  } $

So, option C is correct.

The vector equation of the line of intersection of the planes $r.(i+2j+3k)=0$ and $r.(3i+2j+k)=0$ is

  1. $r=\lambda (i+2j+k)$

  2. $r=\lambda (i-2j+k)$

  3. $r=\lambda (i+2j-3k)$

  4. None of these


Correct Option: B
Explanation:

The line of intersection of the planes $r.(i+2j+3k)=0$ and $r.(3i+2j+k)=0$ is parallel to the vector 

$\left( i+2j+3k \right) \times \left( 3i+2j+k \right) =-4i+8j-4k$
Since both the planes pass through the origin, therefore their line of intersection will also pass through the origin.
Thus, the required line passes through the origin and is parallel to the vector$-4i+8j-4k$
Hence, its equation is
$r=0+\lambda '\left( -4i+8j-4k \right) \Rightarrow r=\lambda \left( i-2j+k \right) $ where $\lambda =-4\lambda'$

The direction ratios of the line $x-y+z-5=0=x-3y-6$ are 

  1. $3,1,-2$

  2. $2,-4,1$

  3. $\displaystyle \dfrac { 3 }{ \sqrt { 14 }  } ,\dfrac { 1 }{ \sqrt { 14 }  } ,\dfrac { -2 }{ \sqrt { 14 }  } $

  4. $\displaystyle \dfrac { 2 }{ \sqrt { 14 }  } ,\dfrac { -4 }{ \sqrt { 14 }  } ,\dfrac { 1 }{ \sqrt { 14 }  } $


Correct Option: A
Explanation:

If $l,m,n$ are the d.c's of the line, then

$1.l-1.m+1.n=0$

and $1/l-3.m+0.n=0$

$\displaystyle \therefore \dfrac { l }{ 0+3 } +\dfrac { m }{ 1-0 } =\dfrac { n }{ -3+1 } $

Hence, the dr's of the line are $3,1,-2$.

The line of intersection of the planes $\overrightarrow { r } .\left( 3i-j+k \right) =1$ and $\overrightarrow { r } .\left( i+4j-2k \right) =2$ is parallel to the vector:

  1. $2i+7j+13k$

  2. $-2i-7j+13k$

  3. $2i+7j-13k$

  4. $-2i+7j+13k$


Correct Option: A
Explanation:

The line of intersection of the planes $\overrightarrow { r } .\left( 3i-k+k \right) =1$ and $\overrightarrow { r } .\left( i+4j-2k \right) =2$ is perpendicular to each, if the normal vector $\overrightarrow { { n } _{ 1 } } =3i-j+k$ and $\overrightarrow { { n } _{ 2 } } =i+4j-2k$.

$\therefore$ It is parallel to the vector, 
$\overrightarrow { { n } _{ 1 } } \times \overrightarrow { { n } _{ 2 } } =\left( 3i-j+k \right) \times \left( i+4j-2k \right) =2i+7j+13k$

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

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

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

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

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


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

  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


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.

  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


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

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


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.

  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


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

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

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

  3. $1$

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


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

  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


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)  } $

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