Tag: maths

Questions Related to maths

Consider the planes $3x-6y+2z+5=0$ and $4x-12y+3z=3$. The plane $67x-162y+47z+44=0$ bisects the angle between the given planes which-

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

  1. Contains origin

  2. Is acute

  3. Is obtuse

  4. None of these

Show answer
Correct Option: A,B
Explanation:

For $3x-6y+2z+5=0$ and $-4x+12y-3z+3=0$ bisector are
$\displaystyle \frac { 3x-6y+2z+5 }{ \sqrt { 9+36+4 }  } =\pm \frac { -4x+12y-3z+3 }{ \sqrt { 16+144+9 }  } $
The plane which bisects the angle between the plane that contains the origin
$13\left( 3x-6y+2z+5 \right) =7\left( -4x+12y-3z+3 \right) \ \Rightarrow 67x-162y+47z+44=0$
Further $3\times \left( -4 \right) +\left( -6 \right) \times 12+2\times \left( -3 \right) <0$
Hence, the origin lies in the acute angle.

Angle between planes $2x-y+z$ $=$ $6$ and $x+y+2z$ $=$ $7,$ is -

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

  1. $\dfrac { \pi }{ 4 } $

  2. $\dfrac { \pi }{ 2 } $

  3. $\dfrac { \pi }{ 3 } $

  4. $\dfrac {- \pi }{ 4 } $

Show answer
Correct Option: C
Explanation:
Plane $1$: $2x-y+z=6$
normal vector is $\bar{n _1}=2\hat{i}-\hat{j}+\hat{k}$
Plane $2$: $x+y+2z=7$
normal vector is $\bar{n _2}=\hat{i}+\hat{j}+2\hat{k}$
Angle between planes is same as the angle between their normal.
$\Rightarrow \cos\theta =\dfrac{\bar{n _1}\cdot\bar{n _2}}{|\bar{n _1}||\bar{n _2}|}$
$=\dfrac{(2\hat{i}-\hat{j}+\hat{k})\cdot(\hat{i}+\hat{j}+2\hat{k})}{(\sqrt{4+1+1})\sqrt{1+1+4}}$
$=\left|\dfrac{2-1+2}{\sqrt{6}\cdot \sqrt{6}}\right|$
$=\dfrac{3}{6}$
$=\dfrac{1}{2}$
$\Rightarrow \cos\theta =\dfrac{1}{2}$
$\Rightarrow \theta =\dfrac{2}{3}$.

The equation of the plane bisecting the angle between the planes $\displaystyle 3x +4y = 4$ and $\displaystyle 6x - 2y + 3z + 5 = 0$ that contains the origin, is

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

  1. $\displaystyle 9x - 38y + 15z + 43 = 0$

  2. $\displaystyle 51x + 18y + 15z = 3$

  3. $\displaystyle 9x + 2y + 3z + 1 = 0$

  4. $\displaystyle 17x + 9y + 15z = 26$

Show answer
Correct Option: B
Explanation:
Equation of given planes can be written as
 $3x+4y=4 , 6x−2y+3z+5=0$

formula is
  $\dfrac {a _1x+b _1y+c _1z+d _1}{\sqrt {a _1^2+b _1^2+c _1^2}}$ = +  or  - $\dfrac {a _2x+b _2y+c _2z+d _2}{\sqrt {a _2^2+b _2^2+c _2^2}}$

by substituting the values in the given formula we will get 

$\dfrac {3x+4y+0z+-41}{\sqrt {3^2+4^2+0^2}}$ = + or - $\dfrac {6x+-2y+3z+5}{\sqrt {6^2+(-2)^2+3^2}}$

$\Rightarrow$ $21x+28y-28 = +\  or\  - 30x-10y+160+25$

so when adding the above equation we will get $51x + 18y + 160z - 3 = 0$

is the plane bisecting the angle containing the origin, and when subtracting we will get $9x - 38y + 160z + 53 = 0$ is the other bisecting plane.

Hence the plane $51x + 18y + 160z - 3 = 0\  or\  51x + 18y + 160z  = 3$ bisects the acute angle and therefore origin lies in the acute angle.

The equation of the plane bisecting the obtuse angle between the planes $\displaystyle x+y+z= 1$ and $\displaystyle x+2y-4z= 5$ is

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

  1. $\displaystyle \left ( \sqrt{7}-1 \right )x+\left ( \sqrt{7}-2 \right )y+\left ( \sqrt{7}+4 \right )z+5-\sqrt{7}= 0$

  2. $\displaystyle \left ( \sqrt{7}+1 \right )x+\left ( \sqrt{7}+2 \right )y+\left ( \sqrt{7}+4 \right )z+5-\sqrt{7}= 0$

  3. $\displaystyle \left ( \sqrt{7}+1 \right )x+\left ( \sqrt{7}+2 \right )y+\left ( \sqrt{7}-4 \right )z=\sqrt{7}$

  4. None of these

Show answer
Correct Option: D
Explanation:

Given planes are  $ x+y+z-1=0.....(1)$ and $x+2y-4z-5=0.........(2)$
Therefore equation of planes bisecting these planes are
$\dfrac{x+y+z-1}{\sqrt{3}}=\pm\dfrac{x+2y-4z-5}{\sqrt{21}}$

$\Rightarrow x+y+z-1=\pm\dfrac{x+2y-4z-5}{\sqrt{7}}$

$\Rightarrow (\sqrt{7}-1) x+(\sqrt{7}-2)y+(\sqrt{7}+4)z = \sqrt{7}+5 ...(3)$ and $(\sqrt{7}+1) x+(\sqrt{7}+2)y+(\sqrt{7}-4)z = \sqrt{7}-5  ....(4)$
If $\theta$ is the angle between $(1)$ and $(3)$, then

$  \cos\theta = \dfrac{(\sqrt{7}-1).1+(\sqrt{7}-2).1+(\sqrt{7}+4).1}{(\sqrt{(\sqrt{7}-1)^2+(\sqrt{7}-2)^2+(\sqrt{7}+4)^2}).(\sqrt{3})}= \dfrac{3\sqrt{7}+2}{(\sqrt{40+2\sqrt{7}}).(\sqrt{3})}> \dfrac{1}{2}$

$\Rightarrow \theta > 45^\circ$
Hence, plane $(1)$ bisects the obtuse angle between the given planes.
Therefore equation of plane bisecting acute angle  between given plane is
$(\sqrt{7}-1) x+(\sqrt{7}-2)y+(\sqrt{7}+4)z = \sqrt{7}+5 $

Hence, option 'D' is correct.

Let two planes $p _{1}:2x-y+z=2$, and $p _{2}:x+2y-z=3$ are given. The equation of the bisector of angle of the planes  $P _{1}$ and $P _{2}$ which does not contains origin, is

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

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

  2. $x+3y=5$

  3. $x+3y+2z+2=0$

  4. $3x+y=5$

Show answer
Correct Option: D
Explanation:
Given planes are $p _{1}:2x-y+z=2$ and $p _{2}:x+2y-z=3$

Normals to the planes
$N _1:\dfrac{1}{\sqrt{6}}(2,-1,1)$
$N _2:\dfrac{1}{\sqrt{6}}(1,2,-1)$

Let $N$ be the normal vector of angle bisector
$N=  N _1+N _2$ or $ N _1-N _2$
$N = (3,1,0)$ or $(1,-3,2)$

The equation of plane is
$P = P _1+ \lambda P _2$
$P= 2x-y+z-2 + \lambda (x+2y-z -3) $

If $N = (3,1,0)$, then $\lambda = 1$,
Equation of Plane $=  P = 3x+y- 5$
It does not pass through origin.

Hence, option D is correct.

The dist.of a point P on the ellipse $\cfrac{{{x^2}}}{{12}} + \cfrac{{{y^2}}}{4} = 1$ from centre is $\sqrt 6 $ then the eccentric angle of P is 

position of point wrt ellipse ellipse maths

  1. $\cfrac{\pi }{2}$

  2. $\cfrac{\pi }{6}$

  3. $\cfrac{\pi }{4}$

  4. $\cfrac{\pi }{3}$

Show answer
Correct Option: D
Explanation:
Let the point be $P = (a\cos\theta , b\sin\theta )$
 
Distance of point $P$ from centre is $\sqrt 6$
$\therefore \sqrt { { a }^{ 2 }\cos^{ 2 }\theta + b^{ 2 }\sin^{ 2 }\theta} = \sqrt { 6 }$
$\Rightarrow { a }^{ 2 }\cos^{ 2 }\theta + b^{ 2 }\sin^{ 2 }\theta = 6$
$\Rightarrow 12\cos^{ 2 }\theta + 4\sin^{ 2 }\theta = 6 \cos^{ 2 }\theta + \sin^{ 2 }\theta = 1$
$\Rightarrow 8\cos^{ 2 }\theta = 2$
$\Rightarrow \cos\theta = \cfrac { 1 }{ 2 }$
Hence, $\theta = \cfrac { \pi }{ 3 }$

Point $(1,2)$ lies _____ the ellipse $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$.

position of point wrt ellipse ellipse maths

  1. inside

  2. outside

  3. on

  4. None of the above

Show answer
Correct Option: A
Explanation:

The region (disk) bounded by the ellipse is given by the equation:

$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}\leq 1$ centered at $(h,k)$ ..... $(1)$
Point $(x,y)$ lies inside the ellipse if it satisfies $(1)$
Take $(x,y)=(1,2)$
Consider, $\dfrac{(x-0)^2}{16}+\dfrac{(y-0)^2}{9}$

                 $=\dfrac{1^{2}}{16}+\dfrac{2^{2}}{9}=\dfrac{73}{144}<1$
Hence, $(1,2)$ lies inside the given ellipse.

Eccentric angle of a point on the ellipse $x^{2}+3y^{2}=6$ at a distance $2$ units. from the centre of the ellipse is

position of point wrt ellipse ellipse maths

  1. $2\pi/3$

  2. $\pi/3$

  3. $4\pi/3$

  4. $none\ of\ these$

Show answer
Correct Option: B

Let the equation of the ellipse be $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$. Let $f(x,y) = \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - 1$. To determine whether the point $(x _1,y _1)$ lies inside the ellipse, the necessary condition is:

position of point wrt ellipse ellipse maths

  1. $f(x _1,y _1) < 0$

  2. $f(x _1,y _1) > 0$

  3. $f(x _1,y _1) = 0$

  4. None of these

Show answer
Correct Option: A
Explanation:

$f(x,y) = \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - 1$ ........ $(1)$ 

The region (disk) bounded by the ellipse is given by the equation:
$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}\leq 1$ centered at $(h,k)$.

The given equation of ellipse is $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2} = 1$ centered at origin i.e. $(0,0)$.
The region bounded by this ellipse is 
$\dfrac{(x)^2}{a^2}+\dfrac{(y)^2}{b^2}\leq 1$ ...... $(1)$
The point $x _{1},y _{1}$ lies inside the given ellipse if it satisfies $(1)$
i.e. if $\dfrac{(x _{1})^2}{a^2}+\dfrac{(y _{1})^2}{b^2}\leq 1$ ...... $(1)$
if $\dfrac{(x _{1})^2}{a^2}+\dfrac{(y _{1})^2}{b^2} - 1<0$ ...... $(1)$
$\implies$ $f(x _{1}, y _{1})<0$  ....... From $(1)$
Hence, option A is correct.

The locus of a point whose distance form the point $(3,0)$ is $3/5$ times its distance from the line $x=p$ is an ellipse with centre at the origin. The value of $p$ is 

position of point wrt ellipse ellipse maths

  1. $5$

  2. $7$

  3. $\dfrac{25}{3}$

  4. $\dfrac{25}{9}$

Show answer
Correct Option: C