Tag: distance between two points in 3d

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

$P(0,5,6),Q(1,4,7),R(2,3,7)$ and $S(3,5,16)$ are four points in the space. The point nearest to the origin $O(0,0,0)$ is

  1. $P$

  2. $Q$

  3. $R$

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

The $4$ points are as given.
We calculate their individual distance from the origin.
$OP =$ $ {({5}^{2} + {6}^{2})}^{0.5} $ = $ {(61)}^{0.5} $
$OQ =$ $ {({1}^{2} + {4}^{2} + {7}^{2} )}^{0.5} $ = $ {(66)}^{0.5} $
$OR =$ $ {({2}^{2} + {3}^{2} + {7}^{2} )}^{0.5} $ = $ {(62)}^{0.5} $
$OS =$ $ {({3}^{2} + {5}^{2} + {16}^{2} )}^{0.5} $= $ {(290)}^{0.5} $
Hence, $P$ is the nearest to the origin.

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

The name of the figure formed by the points $(-1, -3, 4), (5, -1,1), (7, -4, 7)$ and $(1, -6, 10)$ is a

  1. square

  2. rhombus

  3. parallelogram

  4. rectangle
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Keeping the above points as vertices and using the distance formula, 
$D=\sqrt{(x _{2}-x _{1})^{2}+(y _{2}-y _{1})^{2}+(z _{2}-z _{1})^{2}}$
We get that the sides of the parallelogram formed by the above lines are equal and all the sides are of $7$ units.
Hence the parallelogram will be either a rhombus or a square. For the parallelogram to be square, all the adjacent sides of the parallelogram should make an angle of $\dfrac{\pi}{2}$.
Consider the vector equation of $AB$ as $(5-(-1))i'+(-1-(-3))j'+(1-4)k'$
$6i'+2j'-3k'$
Consider the vector equation of $BC$ as $(7-5)i'+(-4-(-1))j'+(7-1)k'$
$2i'-3j'+6k'$
Taking dot product, we get $6(2)+2(-3)-3(6)$
$=12-6-18$
$=-12$
Thus the parallelogram is a rhombus.

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

A hall has dimensions $24 m \times 8 m \times 6 m$. The length of the longest pole which can be accommodated in the hall is

  1. 26 m

  2. 28 m

  3. 30 m

  4. 36 m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given that,

Dimensions of the hall x $=24cm\times 8cm\times 6cm$

Now Leght of the longest pole which can be accommodated in the hall x $=\sqrt{{{24}^{2}}+{{8}^{2}}+{{6}^{2}}}=26cm$

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