Tag: introduction to three dimensional geometry

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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

Calculate the distance between the points $(-3,6,7)$ and $(2,-1,4)$ in $3D$ space.

  1. $4.36$

  2. $5.92$

  3. $7.91$

  4. $9.11$

  5. $22.25$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know the distance formula:

$d=\sqrt{(x _2-x _1)^2+(y _2-y _1)^2+(z _2-z _1)^2}$

The coordinates are $(-3, 6, 7)$ and$ (2, -1, 4)$

$d=\sqrt{(2-(-3))^2+((-1)-6)^2+(4-7)^2}$

$d=\sqrt{(5)^2+((-7)^2+(-3)^2}$

$d=\sqrt{25+49+9}$

$d=\sqrt{83}$

$d = 9.11$

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

A point on the line $\displaystyle \frac{{x + 2}}{1} = \frac{{y - 3}}{{ - 4}} = \frac{{z - 1}}{{2\sqrt 2 }}$ at a distance 6 from the point (2, 3, 1) is

  1. $(4-21, 1+12\sqrt{2})$

  2. $\left( {\frac{{ - 4}}{5},\frac{{ - 9}}{5},1} \right)$

  3. $\left( {\frac{{ - 16}}{5},\frac{{39}}{5},\frac{{5 - 12\sqrt 2 }}{5}} \right)$

  4. $\left( {\frac{{ - 16}}{5}, - 21,1 + 12\sqrt 2 } \right)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$(\lambda -2, -4\lambda+3, 2\sqrt 2\lambda+1)\leftrightarrow (-2, 3, 1)$
$(\lambda-2+2)^2+(-4\lambda+3-3)^2+2\sqrt 2\lambda+1-1)^2=36$
$\lambda^2+16\lambda^2+8\lambda^2=36$
$\lambda=\pm \frac {6}{5}$
$\therefore point=(\frac {-4}{5}, \frac {-9}{5}, 1)$

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 between z-axis and the line 
$x+y+2z-3=0=2x+3y+4z-4$, is _____________

  1. $1$

  2. $2$

  3. $4$

  4. $3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$x+y+2z-3=0=2x+3y+4z-4$ at $z-axis, x=y=0$

$x+y=3-2z$ and $2x+3y=4(1-z)$
solving $x$ and $y$ in function $z$,
$2x+3(3-2z-x)=4(1-z)$
$\implies 2x+9-6z-3x=4-4z$
$\implies x=9-6z-4+4z$
$\implies x=5-2z\quad equation (2)$
$\implies (5-2z)+y=3-2z$
$\implies y=3-2z-5+2z$
$y=-2\quad equation (2)$
So, $\sqrt {x^2+y^2}=\sqrt {(-2)^2+(5-2z)^2}$
at, $z=\cfrac {5}{2}$, this would be minimum.

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

In a $\triangle {ABC}$, side $AB$ has the equation $2x+3y=29$ and the side $AC$ has the equation $x+2y=16$. If the mid point of $BC$ is $(5,6)$, then the equation of $BC$ is

  1. $2x+y=7$

  2. $x+y=1$

  3. $2x-y=17$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let co-ordinates of $B$ be $(x _1, y _1)$ & $C$ be $(x _2, y _2)$

$\therefore$ $(5,6)$ is the mid point,

so, $\dfrac{x _1 + x _2}{2} = 5, \dfrac{y _1 + y _2}{2} = 6$

$\Rightarrow x _1 + x _2 = 10, y _1 + y _2 = 12$

$B(x _1, y _1)$ lies on the line $2x + 3y = 29$

$\therefore 2x _1 + 3y _1 = 29$  ----(1)

$C(x _2, y _2)$ lies on the line $x + 2y = 16,$

$\therefore x _2 + 2y _2 = 16$  ----(2)

$\therefore$ putting $x _1, y _1$ in the form of $x _2, y _2$ in (1)

$2(10 - x _2) + 3(12 - y _2) = 29$  {$x _1 = 10 - x _2, y _1 = 12 - y _2$}

$\Rightarrow 20 - 2x _2 + 36 - 3y _2 = 29$

$\Rightarrow 2x _2 + 3y _2 = 27$  ----(3)

on subtracting $(3)$ and $(2)$ $\times$ $2$

$-y _2 = -5$

$y _2 = 5$

Putting $y _2 \,  in (2)$

$x _2 + 2(5) = 16$

$x _2 = 6$

$x _1 = 10 - x _2$

      $= 4$

$y _1 = 12 - 5 = 7$

Equation :

$\dfrac{x - x _1}{x _2 - x _1} = \dfrac{y - y _1}{y _2 - y _1}$

$\Rightarrow \dfrac{x - 4}{2} = \dfrac{y - 7}{-2}$

$\Rightarrow -x + 4 = y - 7$

$\Rightarrow x + y = 11$
Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

A swimmer can swim $2$ km in $15$ minutes in a lake and in a river he can swim a distance of $4$ km in $20$ minutes along the stream. If a paper boat is put in the river, then the distance covered by it in $\displaystyle $2$ \, \frac{1}{2}$2 hours will be 

  1. $18$ km

  2. $12$ km

  3. $8$ km

  4. $10$ km
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Speed of the man in still water $\cfrac{2}{15/60}$ = $8$ km/hr
Speed of the man in downstream = $\cfrac{4}{20/60}$ = $12$ km/hr
Speed of stream = $4$ km/hr
Distance covered by paper boat in $2$ $\frac{1}{2}$ hours = $\cfrac{5}{2} \, \times$  $4$ = $10$ km 

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

The point equidistant from the point $O(0, 0, 0), A(a, 0, 0), B(0, b, 0)$ and $C(0, 0, c)$ has the coordinates

  1. $(a, b, c)$

  2. $(a/2, b/2, c/2)$

  3. $(a/3, b/3, c/3)$

  4. $(a/4, b/4, c/4)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$P(x, y, z)$
$PO=\sqrt{x^2+y^2+z^2}$
$PA=\sqrt{(a-x)^2+(-y)^2+(-z)^2}=\sqrt{(a-z)^2+y^2+z^2}$
$PB=\sqrt{x^2+(b-y)^2+z^2}$
$PC=\sqrt{x^2+y^2+(C-z)^2}$
$\Rightarrow PO=PA$
$\sqrt{x^2+y^2+z^2}=\sqrt{(a-x)^2+y^2+z^2}$
$x^2+y^2+z^2=(a-x)^2+y^2+z^2$
$x^2=(a-x)^2$
$x=a-x$
$\Rightarrow x=\dfrac{a}{2}$
$y=b/2$
$z=c/2$.
Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

The distances of the point $P(1,2,3)$ from the coordinates axes are:

  1. $\sqrt {13} ,\sqrt {10} ,\sqrt 5 $

  2. $\sqrt {11} ,\sqrt {10} ,\sqrt 5 $

  3. $\sqrt {13} ,\sqrt {20} ,\sqrt {15} $

  4. $\sqrt {23} ,\sqrt {10} ,\sqrt 5 $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The point: $(1, 2, 3)$
$\therefore$ Co-ordinates axes are $(1, 0, 0), (0, 2, 0), (0, 0, 3)$
Distance of $(1, 2, 3)$ from $(1, 0, 0)$
$= \sqrt{(1 - 1)^2 + (0 - 2)^2 + (0 - 3)^2)}$
$= \sqrt{4 + 9} = \sqrt{13}$
Distance of $(1, 2, 3)$ from $(0, 2, 0)$
$= \sqrt{(0 - 1)^2 + (2 - 2)^2 + (0 - 3)^2)}$
$= \sqrt{1 + 9} = \sqrt{10}$
Distance of $(1, 2, 3)$ from $(0, 0, 3)$
$= \sqrt{(0 - 1)^2 + (0 - 2)^2 + (3 - 3)^2)}$
$= \sqrt{5}$
Hence, the distances of the point $P(1,2,3)$ from the coordinates axes$= \sqrt{13}, \sqrt{10}, \sqrt{5}$
Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

The values of a for which $(8, -7, a), (5, 2, 4)$ and $(6, -1, 2)$ are collinear, is given by?

  1. $2$

  2. $-2$

  3. $-1$

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

$\begin{matrix} then\, a=? \ The\, equation\, lineAB\, is\,  \ \Rightarrow \dfrac { { x-8 } }{ { 5-8 } } =\dfrac { { y+7 } }{ { 2+7 } } =\dfrac { { z-a } }{ { 4-a } } ....\left( 1 \right)  \ po{ int }\, c\, lies\, in\, the\, line\, AB \ po{ int }\left( { 6,-1,2 } \right) satisfy\, eq\left( 1 \right)  \ \Rightarrow \dfrac { { -2 } }{ { -3 } } =\dfrac { { -1+7 } }{ 9 } =\dfrac { { 2-a } }{ { 4-a } }  \ \Rightarrow \dfrac { 2 }{ 3 } =\dfrac { { 2-a } }{ { 4-a } }  \ \Rightarrow 8-2a=6-3a \ \Rightarrow 3a-2a=6-8 \ a=-2 \  \end{matrix}$

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

The locus of a point P which moves such that $PA^2-PB^2=2k^2$ where A and B are $(3, 4, 5)$ and $(-1, 3, -7)$ respectively is 

  1. $8x+2y+24z-9+2k^2=0$

  2. $8x+2y+24z-2k^2=0$

  3. $8x+2y+24z+9+2k^2=0$

  4. $8x-2y+24z-2k^2=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$P{A}^{2}-P{B}^{2}=2{k}^{2}$
$\Rightarrow \left[{\left(x-3\right)}^{2}+{\left(y-4\right)}^{2}+{\left(z-5\right)}^{2}\right]-\left[{\left(x+1\right)}^{2}+{\left(y-3\right)}^{2}+{\left(z+7\right)}^{2}\right]=2{k}^{2}$
$\Rightarrow \left(x-3+x+1\right)\left(x-3-x-1\right)+\left(y-4+y-3\right)\left(y-4-y+3\right)+\left(z-5+z+7\right)\left(z-5-z-7\right)=2{k}^{2}$
$\Rightarrow \left(2x-2\right)\left(-4\right)+\left(2y-7\right)\left(-1\right)+\left(2z+1\right)\left(-12\right) =2{k}^{2}$
$\Rightarrow -8x+8-2y+7-24z-24=2{k}^{2}$
$\Rightarrow -8x-2y-24z-9=2{k}^{2}$
$\Rightarrow 8x+2y+24z+9+2{k}^{2}=0$