Tag: introduction to three dimensional geometry

Questions Related to introduction to three dimensional geometry

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 ?

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

  1. $60$ km

  2. $30$ km

  3. $45$ km

  4. None of these

Show answer
Correct Option: A
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 $\vec r=x\vec i+y\vec j+z\vec k$
$\Longrightarrow |\vec r|=\sqrt{{x^2}+{y^2}+{z^2}}=\sqrt{400+1600+1600}=60km$

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

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

  1. square

  2. rectangle

  3. rhombus

  4. none of these

Show answer
Correct Option: C
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

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

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

  1. rectangle

  2. square

  3. parallelogram

  4. none of these

Show answer
Correct Option: C
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

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

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

  1. rhombus

  2. square

  3. parallelogram

  4. none of these

Show answer
Correct Option: B
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.

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

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

  1. collinear

  2. vertices of a rectangle

  3. vertices of a square

  4. vertices of a rhombus

Show answer
Correct Option: B
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

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

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

  1. $2$

  2. $3$

  3. $6$

  4. $7$

Show answer
Correct Option: D
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
$=\sqrt { { 2 }^{ 2 }+{ 3 }^{ 2 }+{ 6 }^{ 2 } } =\sqrt { 49 } =7$

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

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

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

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

Show answer
Correct Option: A
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 $\left(\dfrac{a}{2},\dfrac{b}{2},\dfrac{c}{2}\right)$

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

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

  1. $4$

  2. $6$

  3. $2\sqrt { 11 } $

  4. $2\sqrt { 14 } $

  5. $12$

Show answer
Correct Option: D
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 } $
$=2\sqrt { 14 }$ 

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

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

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

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

  3. $15$

  4. $30$

Show answer
Correct Option: A,C,D
Explanation:

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