Tag: distance between two points in 3d

Quizzes (0) Questions (82)

Questions Related to distance between two points in 3d

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

If $L _1$ is the line of intersection of the plane $2x-2y+3z-2=0, x-y+z+1=0$ and $L _2$ is the line of intersection of the plane   $x+2y-z-3=0, 3x-y+2z-1=0$, then the distance of origin from from the plane containing the lines $L _1$ + $L _2$ is :

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

  2. $\dfrac{1}{4\sqrt{2}}$

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

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

The lines L1 and L2 are found by solving the systems of plane equations. The plane containing these lines is determined, and the distance from the origin is calculated using the standard point-to-plane formula.

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 plane which is passing through the point $(1,2,3)$ and which is at maximum distance from the point $(-1,0,2)$ is

  1. $2x+2y+z=9$

  2. $2x+z=5$

  3. $3x+y-z=2$

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

The plane passes through (1,2,3). The distance from (-1,0,2) is maximized when the normal vector is parallel to the vector from the plane point to the external point. Normal = (1-(-1), 2-0, 3-2) = (2,2,1). Equation: 2(x-1)+2(y-2)+1(z-3)=0 => 2x+2y+z=9.

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

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:

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

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

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

  4. $\dfrac {19}{23}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The shortest distance between two skew lines is given by the formula involving the cross product of direction vectors and the vector connecting points on the lines.

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

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

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

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

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

  4. $(-2,-5,7)$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
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 $C$

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

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

  1. $1$

  2. $2$

  3. $4$

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


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

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 $(4, -5, 1)$, $(3, -4, 0)$, $(6, -7, 3)$, $(7, -8, 4)$ are vertices of a

  1. square

  2. parallelogram

  3. rectangle

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

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

$A, B, C$ 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

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $\displaystyle z=\dfrac {c}{2}$