Tag: introduction to three dimensional geometry

Quizzes (6) Questions (119)

Questions Related to introduction to three dimensional geometry

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

The distance between the points $(\cos \, \theta , \, \sin \, \theta) $ and $ (\sin \, \theta - \cos \, \theta)$ is 

  1. $\sqrt{3}$

  2. $\sqrt{2}$

  3. $2$

  4. $1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Distance between the point $(\cos\theta, \sin\theta)$ and $(\sin\theta, -\cos\theta)$ is :
$ = \sqrt{(\sin\theta - \cos\theta)^2 + (-\cos\theta - \sin\theta)^2}$
$ = \sqrt{1 - 2 \sin\theta \cos\theta + 1 + 2 \sin\theta \cos\theta}$
$(\cos^2\theta + \sin^2\theta = 1)$
$= \sqrt{2}$
Option B is the correct answer
Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

If the distance between a point P and the point (1, 1, 1) on the line $\frac{{x\, - \,1}}{3}\, = \,\frac{{y - \,1}}{4}\, = \,\frac{{z\, - 1}}{{12}}$ is 13, then the coordinates of P are

  1. (3, 4, 12)

  2. $\left( {\frac{3}{{13}},\,\frac{4}{{13}},\,\frac{{12}}{{13}}} \right)$

  3. (4, 5, 12)

  4. (40, 53, 157)
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\begin{array}{l} According\, to\, question: \ we\, have\, the\, line\, equ=\frac { { x-1 } }{ 3 } =\frac { { y-1 } }{ 4 } =\frac { { z-1 } }{ { 12 } } \, and\, point\, (P)\, is\, 13. \ Now, \ line\, of\, equ: \ \Rightarrow \frac { { x-1 } }{ 3 } =\frac { { y-1 } }{ 4 } =\frac { { z-1 } }{ { 12 } } =13 \ Now\, find\, x,y\, & \, z\, \, coordinates: \ \, \Rightarrow \frac { { x-1 } }{ 3 } =13 \ \, \, \, \, \, \, \, \therefore \, \, \, \, x=39+1=40 \ \Rightarrow \frac { { y-1 } }{ 4 } =13 \ \, \, \, \, \, \, \, \therefore \, \, y=\, 52+1=53 \ \Rightarrow \frac { { z-1 } }{ { 12 } } =13 \ \, \, \, \, \, \, \, \, \therefore \, \, z=156+1=157 \ Now,\, we\, get\, \, new\, coordinates\, of\, P:\, \, \, \left( { 40,53,157 } \right) \, \,  \ so\, that\, \, the\, correct\, option\, is\, D. \end{array}$

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

If the lines $\frac{x - 0}{1} =\frac{y+1}{2}=\frac{z-1}{-1}$ and $\frac{x+1}{k}=\frac{y-3}{-2}=\frac{z-2}{1}$ are at right angles, then the value of k is

  1. $5$

  2. $0$

  3. $3$

  4. $-1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If $({ l } _{ 1 }{ m } _{ 1 }{ n } _{ 1 })$ and  $({ l } _{ 2 }{ m } _{ 2 }{ n } _{ 2 })$ are directions of two $\bot$ lines then,

${ l } _{ 1 }{ l } _{ 2 }+{ m } _{ 1 }{ m } _{ 2 }+n _{ 1 }{ n } _{ 2 }=0\ k-4-1=0\ k=5$

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

If the point $(x, y)$ is equidistant from the points $(a + b, b - a)$ and $(a - b , a + b)$, then  $bx = ay$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If a point (x,y) is equidistant from A(a,b) and B(c,d) then,

$x = \frac{{a + c}}{2},y = \frac{{b + d}}{2}$
So,
$\begin{array}{l}x = \frac{{\left( {a + b} \right) + \left( {a - b} \right)}}{2},y = \frac{{\left( {b - a} \right) + \left( {a + b} \right)}}{2}\x = \frac{{2a}}{2},y = \frac{{2b}}{2}\x = a,y = b\end{array}$

putting values in bx=ay
Solving LHS,
b(a)=ab
Solving RHS,
a(b)=ab
As LHS=RHS the equation bx=ay is true.

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

The area of triangle whose vertices are $(1, 2, 3), (2, 5, -1)$ and $(-1, 1, 2)$ is

  1. $150\ sq. units$

  2. $145\ sq. units$

  3. $\sqrt {155}/2\ sq. units$

  4. $155/2\ sq. units$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the vertices of triangle are 


$A(1,2,3),B(2,5,-1)$ and $C(-1,1,2)$ 

Then, 

$\begin{array}{l} \overrightarrow { AB } =\overrightarrow { OB } -\overrightarrow { OA } =\hat { i } +3\hat { j } -4\hat { k }  \  \ \overrightarrow { AC } =\overrightarrow { OC } -\overrightarrow { OA } =-2\hat { i } -\hat { j } -\hat { k }  \end{array}$

Then, 

$\begin{array}{l} \overrightarrow { AB } \times \overrightarrow { AC } =\left| { \begin{array} { *{ 20 }{ c } }{ \hat { i }  } & { \hat { j }  } & { \hat { k }  } \ 1 & 3 & { -4 } \ { -2 } & { -1 } & { -1 } \end{array} } \right|  \  \ =-7\hat { i } +9\hat { j } +5\hat { k }  \  \ \left| { \overrightarrow { AB } \times \overrightarrow { AC }  } \right| =\sqrt { { { \left( { -7 } \right)  }^{ 2 } }+{ { \left( 9 \right)  }^{ 2 } }+{ { \left( 5 \right)  }^{ 2 } } }  \  \ =\sqrt { 49+8+25 }  \  \ =\sqrt { 155 }  \end{array}$

Area of triangle $ABC=\frac{1}{2}|\vec {AB} \times \vec {AC}|$

$=\frac{1}{2} \times \sqrt {155}$

$=\frac{\sqrt {155}}{2}$ sq. units 

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

The points $(10,7,0)$, $(6,6-1)$ and $(6,9,-4)$ form a 

  1. Right -angled triangle

  2. Isosceles triangle

  3. Both $(1)$ & $(2)$

  4. Equilateral triangle
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$P(10, 7, 0),\ Q(6, 6, -1),\ R(6, 9, -4)$
$PQ=\sqrt {4^2+(-1)^2 +(-1)^2}$
$=\sqrt {16+1+1}=3\sqrt 3$
$QR=\sqrt {0^2+3^2+-3^2}$
$PR=\sqrt {4^2+2^2+(-4)^2}$
$=\sqrt {16+4+16}=6$
$PQ^2+QR^2=(3\sqrt 2)^2+(3\sqrt 2)^2=6^2=PR^2$
$\therefore \ \triangle PQR$ is a right angle $D$ of $Q,\ PQ=QR$
$\triangle PQR$ is a isoscels traingle
$(C)$ Both $(1)$ & $(2)$


Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

The shortest distance of the point $(1,2,3)$ from ${x}^{2}+{y}^{2}=0$ is 

  1. $5$

  2. $\sqrt{5}$

  3. $2$

  4. $\sqrt{14}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

${x}^{2}+{y}^{2}=0$

$\therefore$ $(x,y)\equiv (0,0)$
Let  point be $\equiv (0,0,z)$
$d\equiv \sqrt { { (1-0) }^{ 2 }+{ (2-0) }^{ 2 }+{ (z-3) }^{ 2 } } =\sqrt { 1+4+{ (z-3) }^{ 2 } } \Rightarrow { (z-3) }^{ 2 }\ge 0$
${ d } _{ min }=\sqrt { 5 } $

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

The distance between two points $(1,1)$ and $\left( {\dfrac{{2{t^2}}}{{1 + {t^2}}},\dfrac{{{{\left( {1 - t} \right)}^2}}}{{1 + {t^2}}}} \right)$ is 

  1. 4t

  2. 3t

  3. 1

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Two point are $A(1,1)$ and $B\left(\dfrac{2t^{2}}{1+t^{2}}, \dfrac{(1+t)^{2}}{1+t^{1}}\right)$

$AB=\sqrt{\left(1-\dfrac{2t^{2}}{1+t^{2}}\right)^{2}+\left(1-\dfrac{(1-t^{2})}{1+t^{2}}\right)^{2}}$

$=\sqrt{\left(\dfrac{1+t^{2}-2t^{2}}{1+t^{2}}\right)^{2}+\left(\dfrac{1+t^{2}-1-t^{2}+2t}{1+t^{2}}\right)^{2}}$

$=\sqrt{\left(\dfrac{1-t^{2}}{1+t^{2}}\right)^{2}+\left(\dfrac{RT}{1+t^{2}}\right)^{2}}$

$=\sqrt{\dfrac{1+t^{4}-2t^{2}+4t^{2}}{(1+t^{2})^{2}}}$

$=\sqrt{\dfrac{(1+t^{2})^{2}}{(1+t^{2})^{2}}}$

$=1$