Tag: introduction to three dimensional geometry

Quizzes (6) Questions (119)

Questions Related to introduction to three dimensional geometry

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. 
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 $=\sqrt{{{24}^{2}}+{{8}^{2}}+{{6}^{2}}}=26cm$