Tag: introduction to three dimensional geometry

Questions Related to introduction to three dimensional geometry

The distance of the point $P(3,8,2)$ from the line $\dfrac{x-1}{2}=\dfrac{y-3}{4}=\dfrac{z-2}{3}$ measured parallel to the plane $3x+2y-2z+15=0$ is 

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

  1. $5\sqrt[2}$

  2. $18$

  3. $9\sqrt{3}$

  4. $7$

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Correct Option: A

If the shortest distance between the line 
$\dfrac {x-1}{\alpha}=\dfrac {y+1}{-1}=\dfrac {z}{1}(\alpha \neq 1)$ and $x+y+z+1=0=2x-y+z+3$ is $\dfrac {1}{\sqrt {3}}$, then a value $\alpha$ is:

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

  1. $-\dfrac {16}{19}$

  2. $-\dfrac {19}{16}$

  3. $\dfrac {32}{19}$

  4. $\dfrac {19}{23}$

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Correct Option: A

The shortest distance between line $y-x=1$ and curve $x=y^{2}$ is :-

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

  1. $\dfrac{8}{3\sqrt{2}}$

  2. $\dfrac{4}{\sqrt{3}}$

  3. $\dfrac{\sqrt{3}}{4}$

  4. $\dfrac{3\sqrt{2}}{8}$

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Correct Option: A

A point $Q$ at a distance $3$ from the point $P(1,1,1)$ lying on the line joining the points $A(0,-1,3)$ and $P$, has the coordinates

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

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

  2. $(4,7,-5)$

  3. $(0,-1,3)$

  4. $(-2,-5,7)$

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Correct Option: A,C
Explanation:

Let the coordinates of point $Q$ be $(a,b,c)$


The distance of $Q$ from $P(1,1,1)$ is $3$


The equation of line $AP$ is $\displaystyle \frac{x-1}{1}=\frac{y-1}{2}=\frac{z-1}{-2}$

Therefore the point on the line $AP$ will look like $Q(t+1,2t+1,1-2t)$

$|QP|=\sqrt{(t+1-1)^2+(2t+1-1)^2+(1-2t-1)^2}=\sqrt{9t^2}=3$

So $|QP| = \pm3t=3$

$\Rightarrow t=\pm1$

So the possible coordinates of $Q$ are $(2,3,-1)$ and $(0,-1,3)$

Therefore the correct options are $A$ and $C$

The distance of the point $\left( 1,-2,3 \right) $ from the plane $x-y+z=5$ measured parallel to the line $\displaystyle \frac { x }{ 2 } =\frac { y }{ 3 } =\frac { z-1 }{ -6 } $ is

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

  1. $1$

  2. $2$

  3. $4$

  4. None of these

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Correct Option: A
Explanation:

Equation of the line through $\left( 1,-2,3 \right) $ parallel to the line $\displaystyle \dfrac { x }{ 2 } =\dfrac { y }{ 3 } =\dfrac { z-1 }{ -6 } $ is


$\displaystyle \dfrac { x-1 }{ 2 } =\dfrac { y+2 }{ 3 } =\dfrac { z-1 }{ -6 } =r$ (say)   ...$(1)$


Then any point on $(1)$ is $\left( 2r+1,3r-2,-6r+3 \right) $.


If this point lies on the plane $x-y+z=5$, then 


$\displaystyle \left( 2r+1 \right) -\left( 3r-2 \right) +\left( -6r+3 \right) =5\Rightarrow -7r+6=5\Rightarrow r=\dfrac { 1 }{ 7 } $


Hence, the point is $\displaystyle \left( \dfrac { 9 }{ 7 } ,-\dfrac { 11 }{ 7 } ,\dfrac { 15 }{ 7 }  \right) $


Distance between $\left( 1,-2,3 \right) $ and $\displaystyle \left( \dfrac { 9 }{ 7 } ,-\dfrac { 11 }{ 7 } ,\dfrac { 15 }{ 7 }  \right) $


$\displaystyle =\sqrt { \left( \dfrac { 4 }{ 49 } +\dfrac { 9 }{ 49 } +\dfrac { 36 }{ 49 }  \right)  } =\sqrt { \dfrac { 49 }{ 49 }  } =1$

The points $(4, -5, 1)$, $(3, -4, 0)$, $(6, -7, 3)$, $(7, -8, 4)$ are vertices of 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. parallelogram

  3. rectangle

  4. rhombus

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Correct Option: B
Explanation:

Let $A=(4,-5,1)$
$B=(3,-4,0)$
$C=(6,-7,3)$
$D=(7,-8,4)$
Now let the quadrilateral be $ABCD$.
Then
$AB=-i+j-k$ $|AB|=\sqrt{3}$
$BC=3i-3j+3k$ $|BC|=3\sqrt{3}$
$CD=i-j+k$  $|CD|=\sqrt{3}$
$AD=3i-3j+3k$ $|AD|=3\sqrt{3}$.
Hence opposite sides are equal and parallel.
Therefore the above points form a parallelogram.
Now all the sides are not equal.
Hence it cannot qualify as a rhombus or square.
Now dot product of $AB$ and $BC$ is not zero.
Hence adjacent sides are not perpendicular to each other.
Therefore it is not a rectangle also.

$A, B, C$ are three points on the axes of $x, y$ and $z$ respectively at distance $a, b, c$ from the origin $O$; then the co - ordinates of the point which is equidistant from $A, B, C$ and $O$ is

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

  1. $\displaystyle \left ( a,b,c \right )$

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

  3. $\displaystyle \left ( \frac{a}{3},\frac{b}{3},\frac{c}{3} \right )$

  4. None of these

Show answer
Correct Option: B
Explanation:

Let $P$ be the required point $\displaystyle \left ( x,y,z \right )$ and the point
$A, B, C$ and $O$ are $\displaystyle \left ( a,0,0 \right ),\left ( 0,b,0 \right ),\left ( 0,0,c \right )$ and $\displaystyle \left ( 0,0,0 \right )$ ;

We are given that $\displaystyle PO=PA=PB=PC.$
Taking $\displaystyle PO=PA$ or $\displaystyle PO^{2}=PA^{2},$ we get
$\displaystyle x^{2}+y^{2}+z^{2}=\left ( x-a \right )^{2}+y^{2}+z^{2}$
$\displaystyle 0=a^{2}-2ax$  i.e.  $\displaystyle x=\dfrac {a}{2}$
Similarly taking $\displaystyle PO^{2}=PB^{2}$ and $\displaystyle PO^{2}=PC^{2},$ we get 
$\displaystyle y=\dfrac {b}{2}$ and $\displaystyle z=\dfrac {c}{2}$

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

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

  1. $10$

  2. $12$

  3. $25$

  4. $15$

Show answer
Correct Option: B
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$
$=3+5+4$
$=12\ units$

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 $k-(m*n)$ ?

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

  1. 20

  2. 21

  3. 22

  4. 23

Show answer
Correct Option: D
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 $\sqrt{(-2 - 7)^2 + (3)^2 + (5 + 1)^2} = \sqrt{81 + 9 + 36} = \sqrt{126} = 3\sqrt{14}$
$\Rightarrow m = 3, n = 14$

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

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

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

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

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

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

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

Show answer
Correct Option: A
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 } $
                $=3\sqrt { 2 }$