Tag: introduction to three dimensional geometry

Quizzes (6) Questions (119)

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

The equation of motion of a rocket are: $x=2t,y=-4t,z=4t,$ where the time $t$ is given in seconds and the coordinate of a moving point in kilometers. At what distance will the rocket be from the starting point $O(0,0,0)$ in $10$ seconds ?

  1. $60$ km

  2. $30$ km

  3. $45$ km

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

Eliminating t from the given equation, we get the equation of the path $\dfrac{x}{2}=\dfrac{y}{-4}=\dfrac{z}{4}=t $

Thus the path of the Rocket represents a straight line passing through the origin for $t=10sec.$
we have $x=20,y=-40,z=40$
Let $\vec r=x\vec i+y\vec j+z\vec k$
$\Longrightarrow |\vec r|=\sqrt{{x^2}+{y^2}+{z^2}}=\sqrt{400+1600+1600}=60km$

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

If $A= \left ( 5,-1,1 \right ),B= \left ( 7,-4,7 \right ),C= \left ( 1,-6,10 \right ),D= \left ( -1,-3,4 \right )$. Then $ABCD$ is a

  1. square

  2. rectangle

  3. rhombus

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

AB${=}$ $\sqrt{{(7-5)}^{2}+{(-4+1)}^{2}+{(7-1)}^{2}}$
AB${=}$ $\sqrt{{(2)}^{2}+{(-3)}^{2}+{(6)}^{2}}$
AB${=}$ $\sqrt{49}$
AB${=}$ 7
Similarly you find that BC${=}$ $\sqrt{49}$  CD${=}$ 7  and DA${=}$7
Hence all sides of quadrilateral are equal, Now we check the diagonals
AC${=}$ $\sqrt{{(1-5)}^{2}+{(-6+1)}^{2}+{(10-1)}^{2}}$
AC${=}$ $\sqrt{122}$
similarly BD${=}$ $\sqrt{74}$ 
Diagonals are not equal
direction ratio of line passing through AC is (-4,-5,9)
direction ratio of line passing through  BD is (-8,1,-3), As the dot product dr of AC and BD are equal to 0 which means AC is perpendicular to BD,
All sides are equal and diagonal are not equal but bisect each other at right angle
hence it is rhombus

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

Let $A= \left ( 1,2,3 \right )B= \left ( -1,-2,-1 \right )C= \left ( 2,3,2 \right )$ and $ D= \left ( 4,7,6 \right )$. Then $ABCD$ is a

  1. rectangle

  2. square

  3. parallelogram

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

AB${=}$ $\sqrt{{(-1-1)}^{2}+{(-2-2)}^{2}+{(-1-3)}^{2}}$
AB${=}$ $\sqrt{{(-2)}^{2}+{(-4)}^{2}+{(-4)}^{2}}$
AB${=}$ $\sqrt{36}$
AB${=}$ 6
Similarly you find that BC${=}$ $\sqrt{43}$  CD${=}$ 6  and DA${=}$ $\sqrt{43}$
Hence opposite sides of quadrilateral are equal, Now we check the diagonals
AC${=}$ $\sqrt{{(2-1)}^{2}+{(3-2)}^{2}+{(2-3)}^{2}}$
AC${=}$ $\sqrt{3}$
similarly BD${=}$ $\sqrt{155}$ 
Diagonals are not equal
direction ratio of line passing through AB is (-2,-4,-4)
direction ratio of line passing through  CD is (2,4,4), As the dr of AB and CD are proportional which means AB is parallel to CD,
Similarly check for BC and DA then you will find that they are also parallel
hence it is parallelogram

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

If $A= \left ( 0,0,2 \right ),B= \left (\sqrt{2},\sqrt{2},2 \right ),C= \left ( \sqrt{2},\sqrt{2},0 \right )$ and $D= \left ( \displaystyle \frac{8\sqrt{2}-20}{17},\frac{12\sqrt{2}+4}{17},\frac{20-8\sqrt{2}}{17} \right )$, then $ABCD$ is a

  1. rhombus

  2. square

  3. parallelogram

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

Given points are, $A= \left ( 0,0,2 \right ),B= \left (\sqrt{2},\sqrt{2},2 \right ),C=

\left ( \sqrt{2},\sqrt{2},0 \right )$ and $D= \left ( \displaystyle

\frac{8\sqrt{2}-20}{17},\frac{12\sqrt{2}+4}{17},\frac{20-8\sqrt{2}}{17}

\right )$

Length $AB = \sqrt{\sqrt{2}^2 + \sqrt{2}^2 + (2-2)^2} = 2$
Length $BC = \sqrt{(\sqrt{2}-\sqrt{2})^2 + (\sqrt{2}-\sqrt{2})^2 + 2^2} = 2$
Length $CD = \sqrt{\left(

\dfrac{8\sqrt{2}-20}{17}-\sqrt{2}\right)^2 + \left(\dfrac{12\sqrt{2}+4}{17}-\sqrt{2}\right)^2 + \left(\dfrac{20-8\sqrt{2}}{17}\right)^2} = 2$
Length $AD = \sqrt{\left(

\dfrac{8\sqrt{2}-20}{17}\right)^2 + \left(\dfrac{12\sqrt{2}+4}{17}\right)^2 + \left(\dfrac{20-8\sqrt{2}}{17} - 2\right)^2} = 2$

Angle between two vectors $\overline{AB} = p _{1}\hat{i}+q _{1}\hat{j}+r _{1}\hat{k} = \sqrt{2}\hat{i} + \sqrt{2}\hat{j} + 0\hat{k}$ and $\overline{BC} = p _{2}\hat{i}+q _{2}\hat{j}+r _{2}\hat{k} = 0\hat{i} + 0\hat{j} + 2 \hat{k}$ is $cos\theta = 0 \Rightarrow \theta = 90^o$

All sides are equal and angle between $AB$ and $BC$ is $90^o$, Hence, its a square.

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

The points $A(1,2,-1),B(2,5,-2),C(4,4,-3)$ and $D(3,1,-2)$ are

  1. collinear

  2. vertices of a rectangle

  3. vertices of a square

  4. vertices of a rhombus
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

AB${=}$ $\sqrt{{(2-1)}^{2}+{(5-2)}^{2}+{(-2+1)}^{2}}$
AB${=}$ $\sqrt{{(1)}^{2}+{(3)}^{2}+{(-1)}^{2}}$
AB${=}$ $\sqrt{11}$
Similarly you find that BC${=}$ $\sqrt{6}$  CD${=}$ $\sqrt{11}$  and DA${=}$$\sqrt{6}$
Hence opposite sides of quadrilateral are equal, Now we check the diagonals
AC${=}$ $\sqrt{{(4-1)}^{2}+{(4-2)}^{2}+{(-3+1)}^{2}}$
AC${=}$ $\sqrt{17}$
similarly BD${=}$ $\sqrt{17}$ 
Diagonals are not equal
direction ratio of line passing through AB is (1,3,-1)
direction ratio of line passing through  BC is (2,-1,-1), As the dot product dr of AB and BC are equal to 0 which means AB is perpendicular to BC,similarly check for others sides too
opposite sides are equal and diagonal are equal
hence it is rectangle

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

A rectangular parallelopiped is formed by drawing planes through the points $(-1,2,5)$ and $(1,-1,-1)$ and parallel to the coordinate planes. the length of the diagonal of the parallelopiped is

  1. $2$

  2. $3$

  3. $6$

  4. $7$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The plane forming the parallelipiped are

$x=-1,x=1;y=2,y=-1$ and $z=5,z=-1$
Hence, the lengths of the edges of the parallelopiped are
$1-\left( -1 \right) =2,\left| -1-2 \right| =3$ and $\left| -1-5 \right| =6$
$($ length of an edge of the parallelopiped is the distance between the parallel plane sperpendicular to the edge$)$
$\therefore$ Length of diagonal of the parallelopiped
$=\sqrt { { 2 }^{ 2 }+{ 3 }^{ 2 }+{ 6 }^{ 2 } } =\sqrt { 49 } =7$

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

The coordinates of a point which is equidistant from the point $(0,0,0),(a,0,0),(0,b,0)$ and $(0,0,c)$ are given by

  1. $\displaystyle \left( \frac { a }{ 2 } ,\frac { b }{ 2 } ,\frac { c }{ 2 }  \right) $

  2. $\displaystyle \left( \frac { -a }{ 2 } ,\frac { -b }{ 2 } ,\frac { c }{ 2 }  \right) $

  3. <span>$\displaystyle&nbsp;\left( \frac { a }{ 2 } ,\frac { -b }{ 2 } ,\frac { -c }{ 2 } &nbsp;\right) $</span>

  4. <span>$\displaystyle&nbsp;\left( \frac { -a }{ 2 } ,\frac { b }{ 2 } ,\frac { -c }{ 2 } &nbsp;\right) $</span>
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $P(x,y,z)$ be the required point.

Then $OP=PA=PB=PC$
Now $OP=PA \Longrightarrow OP^2+PA^2 \Longrightarrow x^2+y^2+z^2$

= $(x-a)^2+(y-0)^2+(z-0)^2 \Longrightarrow x=\dfrac{a}{2}$
Similarily, $OP=PB=y=\dfrac{b}{2}$ and $OP=PC=z=\dfrac{c}{2}$

hence the coordinate of the required point are $\left(\dfrac{a}{2},\dfrac{b}{2},\dfrac{c}{2}\right)$

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

What is the distance in space between $(1,0,5)$ and $(-3,6,3)$?

  1. $4$

  2. $6$

  3. $2\sqrt { 11 } $

  4. $2\sqrt { 14 } $

  5. $12$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The distance between the points $(1,0,5)$ and $(-3,6,3)$ is given below:

$D=\sqrt { \left( 3-5 \right) ^{ 2 }+\left( 6-0 \right) ^{ 2 }+\left( -3-1 \right) ^{ 2 } } $
$=\sqrt { \left( -2 \right) ^{ 2 }+\left( 6 \right) ^{ 2 }+\left( -4 \right) ^{ 2 } } $
$=\sqrt { 4+36+16 } =\sqrt { 56 } $
$=2\sqrt { 14 }$ 

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

$L _1:\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}$
$L _2:\dfrac{x-2}{3}=\dfrac{y-4}{2}=\dfrac{z-5}{5}$ be two given lines, point P lies on $L _1$ and Q lies on $L _2$ then distance between P and Q can be

  1. $\dfrac{1}{3}$

  2. $\dfrac{1}{9}$

  3. $15$

  4. $30$
Reveal answer Fill a bubble to check yourself
A,C,D Correct answer
Explanation

$\dfrac{|\begin{vmatrix}2-1&4-2  &5-3 \2 & 3 &4 \ 3& 2 & 5\end{vmatrix}|}{|\begin{vmatrix}i&j  &k \2 & 3 & 4\3 & 2 & 5\end{vmatrix}|}=\dfrac{1}{\sqrt{78}}$