Line of intersection of two planes - class-XII

The line of intersection of two planes is parallel to the cross product of their normal vectors. The normals are n1 = (3, -1, 1) and n2 = (1, 4, -2). The cross product is (-1*-2 - 14, 11 - 3*-2, 34 - -11) = (2-4, 1+6, 12+1) = (-2, 7, 13).

This is a classic problem involving the intersection of planes passing through axes. The derivation leads to the relationship between the coordinates and the angle, resulting in the given locus.

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

Vector parallel to intersection of planes 

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

The line of intersection is parallel to the cross product of the normals of the two planes. Calculating these vectors and their cross product, then finding the angle with the given vector, yields pi/3.

The question is incomplete as it asks for a value without specifying what property of the line L is being calculated (e.g., direction cosines, distance). However, based on the options, it appears to be a calculation related to the direction of the line.

The equation of a plane through the intersection of two planes is (2x+3y+4z-7) + k(x+y+z-1) = 0. By using the condition that it is perpendicular to x-5y+3z-6=0, we find k and the resulting equation.

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

Since the point lies on the both planes we have, 

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

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

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.

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 $\sqrt{4+\dfrac{1}{4}+16}=\sqrt{\dfrac{81}{4}}=\dfrac{9}{2}$.

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

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

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

Consider the problem 

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

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

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. 

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