Tag: maths

Questions Related to maths

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

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

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

Show answer
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

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

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

Show answer
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:

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

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

  2. $1$

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

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

Show answer
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

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

  1. $\pi/3$

  2. $\pi/4$

  3. $\pi/6$

  4. $none\ of\ these$

Show answer
Correct Option: A

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

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

  1. $al+bm+cn=0$

  2. $al-bm+cn=0$

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

  4. $al+bm+cn=1$

Show answer
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 $x+3y+2z=2$.

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

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

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

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

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

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

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

Show answer
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

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

  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

Show answer
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

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

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

Show answer
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 

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

  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

Show answer
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.