Tag: three dimensional geometry

Quizzes (19) Questions (122)

Questions Related to three dimensional geometry

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 $\overrightarrow { r } .\left( 3\hat { i } -\hat { j } +\hat { k }  \right) =1$ and $\overrightarrow { r } .\left( \hat { i } +4\hat { j } -2\hat { k }  \right) =2$ is parallel to vector

  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 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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 - 1*4, 1*1 - 3*-2, 3*4 - -1*1) = (2-4, 1+6, 12+1) = (-2, 7, 13).

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

There are two different planes, one passing though the x-axis and the other passing through y-axis. The angle between the planes is $\cfrac{\pi}{4}$. Then locus of a point on the line of intersection of the planes in.

  1. $(x^2+y^2+z^2)x^2=y^2z^2$
  2. $(x^2+y^2+z^2)z^2=x^2y^2$
  3. $(x^2+y^2+z^2)y^2=x^2z^2$

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

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.

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 
$r.\left( {3\hat i - \hat j + \hat k} \right) = 1$ and $r.\left( {\hat i + 4\hat j - 2\hat k} \right) = 2$ is parallel to the vector

  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. $2\hat i + 7\hat j + 13\hat k$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 } $

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

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

  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. $\dfrac {-2\hat i+5\hat j-3\hat k}{\sqrt {38}}$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
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

$= \pm \dfrac{(2 \hat{i} + 5 \hat{j} + 3 \hat{k} )}{\sqrt{2^2 + 5^2 + 3^2}}$

$= \pm \dfrac{(2 \hat{i} + 5 \hat{j} + 3 \hat{k})}{\sqrt{38}}$

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. $\dfrac {1}{2}$
  2. $1$
  3. $\dfrac {1}{\sqrt 2}$
  4. $\dfrac {1}{\sqrt 3}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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

  1. $\pi/3$
  2. $\pi/4$
  3. $\pi/6$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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.

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

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

  1. $al+bm+cn=0$
  2. $al-bm+cn=0$
  3. $al-bm-cn+1=0$
  4. $al+bm+cn=1$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.
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 $x+3y+2z=2$.

  1. $\dfrac{1}{\sqrt{3}}$
  2. $\dfrac{1}{2}$
  3. $1$
  4. $\dfrac{1}{\sqrt{2}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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.

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

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

  1. $x+2y+3z=6$
  2. $x-2y+z=6$
  3. $2x+y+z=5$
  4. $x+2y+6z=3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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.

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

The direction cosines of a line parallel to the planes $\displaystyle 3x + 4y + z = 0$ and $\displaystyle x - 2y - 3z = 5$ are

  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
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $\displaystyle \left ( \frac{1}{\sqrt{3}}, : -\frac{1}{\sqrt{3}}, : \frac{1}{\sqrt{3}} \right )$