Tag: introduction to three dimensional geometry

Questions Related to introduction to three dimensional geometry

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

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

  1. $\sqrt{3}$

  2. $\sqrt{2}$

  3. $2$

  4. $1$

Show answer
Correct Option: B
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

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

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

  1. (3, 4, 12)

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

  3. (4, 5, 12)

  4. (40, 53, 157)

Show answer
Correct Option: D
Explanation:

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

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

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

  1. $5$

  2. $0$

  3. $3$

  4. $-1$

Show answer
Correct Option: A
Explanation:

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

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

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

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

  1. True

  2. False

Show answer
Correct Option: A
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.

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

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

  1. $150\ sq. units$

  2. $145\ sq. units$

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

  4. $155/2\ sq. units$

Show answer
Correct Option: C
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 

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

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

  1. Right -angled triangle

  2. Isosceles triangle

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

  4. Equilateral triangle

Show answer
Correct Option: C
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 $(1)$ & $(2)$


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

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

  1. $5$

  2. $\sqrt{5}$

  3. $2$

  4. $\sqrt{14}$

Show answer
Correct Option: B
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 } $

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 

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

  1. 4t

  2. 3t

  3. 1

  4. none of these

Show answer
Correct Option: C
Explanation:
Two point are $A(1,1)$ and $B\left(\dfrac{2t^{2}}{1+t^{2}}, \dfrac{(1+t)^{2}}{1+t^{1}}\right)$

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

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

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

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

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

$=1$

The plane passing through the point $\left(-2,-2,2\right)$ and containing the line joining the points $\left(1,1,1\right)$ and $\left(1,-1,2\right)$ makes intercepts on the coordinates axes, the sum whose lengths is ?

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

  1. $3$

  2. $4$

  3. $6$

  4. $12$

Show answer
Correct Option: A

The points $A(1,2,3); B-(-1,-2,-1); C(2,3,2)$ and $D(4,7,6)$ form 

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

  1. Square

  2. Rectangle

  3. Parallelogram

  4. Rhombus

Show answer
Correct Option: A