Tag: distance between two points in space

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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

Perimeter of triangle whose vertices are $(0,4,0), (3,4,0)$ and $(0,4,4)$, is

  1. $10$

  2. $12$

  3. $25$

  4. $15$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

 Vertices of triangle are $(0,4,0),(3,4,0)$ and $(0,4,4)$

then perimeter=?
here AB=$\sqrt{(3-0)^2+(4-4)^2+(0-0)^2}$
$=3$
BC$=\sqrt{(3-0)^2+(4-4)^2+(0-4)^2}$
$=\sqrt{9+16}$
$=5$
CA$=\sqrt{(0-0)^2+(4-4)^2+(0-4)^2}=4$
used distance formula b/w two points
$(x _1,y _1,z _1) and (x _2,y _2,z _2)$
$=\sqrt{(x _2-x _2)^2+(y _2-y _1)^2+(z _2-z _1)^2}$
$ perimeter =ABC+BC+CA$
$=3+5+4$
$=12\ units$

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

Let the distance between vectors are given as follows :
$(i)4i +3j-6k, -2i+j-k$  be $\displaystyle \sqrt{k}$
$(ii) -2i+3j+5k, 7i-k $  be  $\displaystyle m\sqrt{n}$
Find $k-(m*n)$ ?

  1. 20

  2. 21

  3. 22

  4. 23
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

(i) Distance between $4i + 3j - 6k$ and $-2i + j - k$ is given by $\sqrt{(4 + 2)^2 + (3 - 1)^2 + (-6 + 1)^2} = \sqrt{36 + 4 + 25} = \sqrt{65}$

$\Rightarrow k = 65$

(ii) Distance between $-2i + 3j + 5k$ and $7i - k$ would be $\sqrt{(-2 - 7)^2 + (3)^2 + (5 + 1)^2} = \sqrt{81 + 9 + 36} = \sqrt{126} = 3\sqrt{14}$
$\Rightarrow m = 3, n = 14$

$\therefore k - (m*n) = 65 - (3 \times 14) = 65 - 42 = 23$

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

Find the distance between the pairs of points whose cartesian coordinates are $(2,3,-1), (2,6,2).$

  1. $\displaystyle 3\sqrt{2}.$

  2. $\displaystyle 2\sqrt{3}.$

  3. $\displaystyle 5\sqrt{2}.$

  4. $\displaystyle 2\sqrt{5}.$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Distance Between two points $(a,b,c)$ and $(x,y,z)$ is given by
$\sqrt { \left( a-x \right) ^{ 2 }+\left( b-y \right) ^{ 2 }+\left( c-z \right) ^{ 2 } } $
Given $(2,3,-1)$, $(2,6,2)$
Distance $=\sqrt { \left( 2-2 \right) ^{ 2 }+\left( 3-6 \right) ^{ 2 }+\left( -1-2 \right) ^{ 2 } } $
                $=\sqrt { 18 } $
                $=3\sqrt { 2 }$

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

Find the distance between the points whose position vectors are given as follows

$(-1,1,3), (0,5,6)$

  1. $\displaystyle \sqrt{118}$

  2. $8$

  3. $\displaystyle \sqrt{26}$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Distance Between two points $(a,b,c)$ and $(x,y,z)$ is given by
$\sqrt { \left( a-x \right) ^{ 2 }+\left( b-y \right) ^{ 2 }+\left( c-z \right) ^{ 2 } } $
Given $(-1,1,3)$, $(0,5,6)$
Distance $=\sqrt { \left( -1-0 \right) ^{ 2 }+\left( 1-6 \right) ^{ 2 }+\left( 3-6 \right) ^{ 2 } } $
                $=\sqrt { 26 } $

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

Find the distance between the points whose position vectors are given as follows

$-2\hat i+3\hat j+5\hat k, 7\hat i-\hat k$

  1. $\displaystyle 3\sqrt{14}$

  2. $\displaystyle \sqrt{54}$

  3. $\displaystyle 3\sqrt{19}$

  4. $\displaystyle \sqrt{57}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Distance Between two position vectors $a\hat i+b\hat j+c\hat k$ and $x\hat i+y\hat j+z\hat k$ is given by
$\sqrt { \left( a-x \right) ^{ 2 }+\left( b-y \right) ^{ 2 }+\left( c-z \right) ^{ 2 } } $
Given $-2\hat i+3\hat j+5\hat k$, $7\hat i-\hat k$
Distance $=\sqrt { \left( -2-7 \right) ^{ 2 }+\left( 3-0 \right) ^{ 2 }+\left( 5+1 \right) ^{ 2 } } $
                $=\sqrt { 126 } $
                $=3\sqrt { 14 } $

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

Find the distance between the points whose position vectors are given as follows

$4\hat i+3\hat j-6\hat k, -2\hat i+\hat j-\hat k$

  1. $\displaystyle \sqrt{65}$

  2. $\displaystyle \sqrt{69}$

  3. $13$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Distance Between two position vectors $a\hat i+b\hat j+c\hat k$ and $x\hat i+y\hat j+z\hat k$ is given by
$\sqrt { \left( a-x \right) ^{ 2 }+\left( b-y \right) ^{ 2 }+\left( c-z \right) ^{ 2 } } $
Given $4\hat i+3\hat j-6\hat k$, $-2\hat i+\hat j-\hat k$
Distance $=\sqrt { \left( 4+2 \right) ^{ 2 }+\left( 3-1 \right) ^{ 2 }+\left( -6+1 \right) ^{ 2 } } $
                $=\sqrt { 65 } $

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 $(3, -5, 1), (-1, 0, 8)$ and $(7, -10, -6)$ is

  1. a triangle

  2. a straight line

  3. an isosceles triangle

  4. an equilateral triangle
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $A=(3,-5,1), B = (-1,0,8)$ and $C=(7,-10,-6)$

Now, 
$AB =\sqrt{(3+1)^2+(-5-0)^2+(1-8)^2}=\sqrt {90}=3\sqrt{10}$,
$BC =\sqrt{(-1-7)^2+(0+10)^2+(8+6)^2}=\sqrt{360}=6\sqrt{10}$ 
and $CA =\sqrt{(7-3)^2+(-10+5)^2+(-6-1)^2}=\sqrt{90}=3\sqrt{10}$
Clearly $BC = AB+CA$
$\therefore  $ given points lies on straight line. 

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

If $(1, 1, a)$ is the centroid of the triangle formed by the points $(1, 2, -3)$ , $(\mathrm{b}, 0, 1)$ and $(-1, 1, -4)$ then $a-b$ $=$

  1. $-5$

  2. $-7$

  3. $5$

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

The coordinates of the vertices of the triangle are given by $(1,2,-3) , (b,0,1), (-1,1,-4)$

Accordingly the coordinates of the centroid of this triangle will be given by 

($ \dfrac{b}{3}, 1 , -2 $)

Hence, $ \dfrac{b}{3} $ $= 1$ Or, $b= 3$

and $a = -2$

So $a- b = -5$

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

The circum centre of the triangle formed by the points $(2, 5, 1), (1, 4, -3)$ and $(-2, 7, -3)$ is

  1. $(6,0,1)$

  2. $(0,6,-1)$

  3. $(-1,6,2)$

  4. $(6,1,-2)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the points $A(2,5,1) B(1,4,-3)$ and $C(-2,7,-3)$
Now using distance formula in $3D$, we have
$AB {=}$$\sqrt{18}$, $BC {=}$$\sqrt{18}$, and $AC {=}$$\sqrt{36}$
Since, ${AB}^{2}+{BC}^{2}$${=}$${AC}^{2}$
Hence, it is right angle triangle and as we know that the circumcentre of right angled triangle is at the midpoint of hypotenuse i.e $AC$.
Therefore by using section formula (1:1), circumcentre ${=}(0,6,-1)$

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

Assertion (A): The points $A(2,9,12) ,B(1,8,8) ,C(2,11,8) D(1,12,12)$ are the vertices of a rhombus
Reason (R): $AB = BC = CD = DA$ and $AC = BD$

  1. Both A and R are individually true and R is the correct explanation of A

  2. Both A and R individually true but R is not the correct explanation of A

  3. A is true but R is false

  4. Both A and R false
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given: The points $A(2,9,12) ,B(1,8,8) ,C(2,11,8) D(1,12,12)$ are the vertices of a rhombus. 
So using distance formula Reason is not true. 
Thus both A and R false.