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

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-

  1. Contains origin

  2. Is acute

  3. Is obtuse

  4. None of these
Reveal answer Fill a bubble to check yourself
A,B Correct answer
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.

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

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

  1. $\dfrac { \pi }{ 4 } $
  2. $\dfrac { \pi }{ 2 } $
  3. $\dfrac { \pi }{ 3 } $
  4. $\dfrac {- \pi }{ 4 } $
Reveal answer Fill a bubble to check yourself
C Correct answer
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.
$\Rightarrow \cos\theta =\dfrac{\bar{n _1}\cdot\bar{n _2}}{|\bar{n _1}||\bar{n _2}|}$
$=\dfrac{(2\hat{i}-\hat{j}+\hat{k})\cdot(\hat{i}+\hat{j}+2\hat{k})}{(\sqrt{4+1+1})\sqrt{1+1+4}}$
$=\left|\dfrac{2-1+2}{\sqrt{6}\cdot \sqrt{6}}\right|$
$=\dfrac{3}{6}$
$=\dfrac{1}{2}$
$\Rightarrow \cos\theta =\dfrac{1}{2}$
$\Rightarrow \theta =\dfrac{2}{3}$.
Multiple choice maths the plane angle between planes angle between two planes problems involving equation of plane

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

  1. $\displaystyle 9x - 38y + 15z + 43 = 0$
  2. $\displaystyle 51x + 18y + 15z = 3$
  3. $\displaystyle 9x + 2y + 3z + 1 = 0$
  4. $\displaystyle 17x + 9y + 15z = 26$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  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
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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

  1. $x-3y+2z+1=0$
  2. $x+3y=5$
  3. $x+3y+2z+2=0$
  4. $3x+y=5$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.
Multiple choice position of point wrt ellipse ellipse maths

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 

  1. $\cfrac{\pi }{2}$
  2. $\cfrac{\pi }{6}$
  3. $\cfrac{\pi }{4}$
  4. $\cfrac{\pi }{3}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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, $\theta = \cfrac { \pi }{ 3 }$
Multiple choice position of point wrt ellipse ellipse maths

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

  1. inside

  2. outside

  3. on

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice position of point wrt ellipse ellipse maths

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

  1. $2\pi/3$
  2. $\pi/3$
  3. $4\pi/3$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The given ellipse equation is x^2 + 3y^2 = 6, which can be rewritten in standard form as x^2/6 + y^2/2 = 1. A point on the ellipse can be parameterized as (sqrt(6) cos(theta), sqrt(2) sin(theta)). The distance from the centre (0,0) is given as 2 units, so (sqrt(6) cos(theta))^2 + (sqrt(2) sin(theta))^2 = 2^2, which simplifies to 6 cos^2(theta) + 2 sin^2(theta) = 4. Solving this yields 4 cos^2(theta) + 2(cos^2(theta) + sin^2(theta)) = 4, so 4 cos^2(theta) + 2 = 4, meaning cos^2(theta) = 1/2. Thus cos(theta) = plus or minus 1/sqrt(2), making theta equal to pi/4, 3pi/4, etc., which are not in the options. Therefore, none of these is correct.

Multiple choice position of point wrt ellipse ellipse maths

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:

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice position of point wrt ellipse ellipse maths

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 

  1. $5$
  2. $7$
  3. $\dfrac{25}{3}$
  4. $\dfrac{25}{9}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The definition of an ellipse is the locus of a point whose distance from a focus is e times its distance from the directrix. Here e = 3/5 and the focus is (3,0). The directrix is x=p, so the equation is sqrt((x-3)^2 + y^2) = (3/5)|x-p|. Squaring both sides and simplifying to the form x^2/a^2 + y^2/b^2 = 1, we find the center is at (ae, 0). Since the center is at the origin, the focus must be at (ae, 0), implying ae = 3. With e = 3/5, a = 5. The directrix is x = a/e = 5/(3/5) = 25/3.