Tag: maths

Quizzes (518) Questions (9324)

Questions Related to maths

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

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. $\vec r.\left( \vec q\vec a-\vec p\vec b \right) =0$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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

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

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

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

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

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

$=1$
Multiple choice line of intersection of two planes line and a plane vectors, lines and planes three dimensional geometry maths

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. $ax+by+cz+d=0,ax+by+c'z+d=0(c\neq c')$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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. $4$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $\sqrt{4+\dfrac{1}{4}+16}=\sqrt{\dfrac{81}{4}}=\dfrac{9}{2}$.

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

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. $\displaystyle \frac{1}{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $\lambda =-4\lambda'$

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

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

  1. $3,1,-2$
  2. $2,-4,1$
  3. <p class="MsoNormal">$\displaystyle \dfrac { 3 }{ \sqrt { 14 } } ,\dfrac { 1 }{ \sqrt { 14 } } ,\dfrac { -2 }{ \sqrt { 14 } } $</p>
  4. <p class="MsoNormal">$\displaystyle \dfrac { 2 }{ \sqrt { 14 } } ,\dfrac { -4 }{ \sqrt { 14 } } ,\dfrac { 1 }{ \sqrt { 14 } } $</p>
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $3,1,-2$.

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

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. $-2i+7j+13k$
Reveal answer Fill a bubble to check yourself
A Correct answer
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, 
$\overrightarrow { { n } _{ 1 } } \times \overrightarrow { { n } _{ 2 } } =\left( 3i-j+k \right) \times \left( i+4j-2k \right) =2i+7j+13k$

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

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. $14i + 2j + 15k$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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. <p class="MsoNormal">$\displaystyle \dfrac{x - \dfrac{7}{4}}{4} = \dfrac{y - \dfrac{1}{4}}{-1} = \dfrac{z}{-3}$</p>
  2. <p class="MsoNormal">$\displaystyle \dfrac{x}{3} = \dfrac{y + \dfrac{1}{3}}{1} = \dfrac{z - \dfrac{7}{4}}{-4}$</p>
  3. $\displaystyle \frac{x}{1} = \frac{3y + 1}{1} = \frac{3z - 7}{-4}$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
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
$\dfrac{x-\alpha}{3}=\dfrac{y-\beta}{1}=\dfrac{z-\gamma}{-4}$