Tag: introduction to three dimensional geometry

Questions Related to introduction to three dimensional geometry

The distance of the point (1,−2,4)(1,−2,4) from the plane passing through the point (1,2,2)(1,2,2) and perpendicular to the planes x−y+2z=3x−y+2z=3 and 2x−2y+z+12=0,2x−2y+z+12=0, is :

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

  1. $2\sqrt{2}$

  2. $4$

  3. $\sqrt2$

  4. $23$

Show answer
Correct Option: A

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

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

  1. $4.36$

  2. $5.92$

  3. $7.91$

  4. $9.11$

  5. $22.25$

Show answer
Correct Option: D
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)$

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

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

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

$d=\sqrt{83}$

$d = 9.11$

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

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

  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. $\left( {\frac{{ - 16}}{5}, - 21,1 + 12\sqrt 2 } \right)$

Show answer
Correct Option: B
Explanation:

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

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

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

  1. $1$

  2. $2$

  3. $4$

  4. $3$

Show answer
Correct Option: A
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.

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

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

  1. $2x+y=7$

  2. $x+y=1$

  3. $2x-y=17$

  4. None of these

Show answer
Correct Option: B
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 :

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

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

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

$\Rightarrow x + y = 11$

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 

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

  1. $18$ km

  2. $12$ km

  3. $8$ km

  4. $10$ km

Show answer
Correct Option: D
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 

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

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

  1. $(a, b, c)$

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

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

  4. $(a/4, b/4, c/4)$

Show answer
Correct Option: B
Explanation:
$P(x, y, z)$
$PO=\sqrt{x^2+y^2+z^2}$
$PA=\sqrt{(a-x)^2+(-y)^2+(-z)^2}=\sqrt{(a-z)^2+y^2+z^2}$
$PB=\sqrt{x^2+(b-y)^2+z^2}$
$PC=\sqrt{x^2+y^2+(C-z)^2}$
$\Rightarrow PO=PA$
$\sqrt{x^2+y^2+z^2}=\sqrt{(a-x)^2+y^2+z^2}$
$x^2+y^2+z^2=(a-x)^2+y^2+z^2$
$x^2=(a-x)^2$
$x=a-x$
$\Rightarrow x=\dfrac{a}{2}$
$y=b/2$
$z=c/2$.

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

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

  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 $

Show answer
Correct Option: A
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$= \sqrt{13}, \sqrt{10}, \sqrt{5}$

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

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

  1. $2$

  2. $-2$

  3. $-1$

  4. $1$

Show answer
Correct Option: A
Explanation:

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

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 

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

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

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

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

  4. $8x-2y+24z-2k^2=0$

Show answer
Correct Option: C
Explanation:
$P{A}^{2}-P{B}^{2}=2{k}^{2}$
$\Rightarrow \left[{\left(x-3\right)}^{2}+{\left(y-4\right)}^{2}+{\left(z-5\right)}^{2}\right]-\left[{\left(x+1\right)}^{2}+{\left(y-3\right)}^{2}+{\left(z+7\right)}^{2}\right]=2{k}^{2}$
$\Rightarrow \left(x-3+x+1\right)\left(x-3-x-1\right)+\left(y-4+y-3\right)\left(y-4-y+3\right)+\left(z-5+z+7\right)\left(z-5-z-7\right)=2{k}^{2}$
$\Rightarrow \left(2x-2\right)\left(-4\right)+\left(2y-7\right)\left(-1\right)+\left(2z+1\right)\left(-12\right) =2{k}^{2}$
$\Rightarrow -8x+8-2y+7-24z-24=2{k}^{2}$
$\Rightarrow -8x-2y-24z-9=2{k}^{2}$
$\Rightarrow 8x+2y+24z+9+2{k}^{2}=0$