Tag: distance between two points in space

Questions Related to distance between two points in space

The circum radius of the triangle formed by the points $(0, 0, 0)$, $(0, 0, 12)$ and $(3, 4, 0)$ is

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

  1. $\sqrt{156}$

  2. $13$

  3. $\displaystyle \frac { 13 }{ 2 } $

  4. $8$

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Correct Option: C
Explanation:

Center of the circum circle of a right angle triangle is on the mid of the hypotenuse. 

Hence the radius is the half of the length of the hypotenuse.
$\dfrac{\sqrt{3^2+4^2+12^2}}{2}=\dfrac{13}{2}$ 

The distance between the points $P(x,\,-1)$ and $Q(3,\,2)$ is $5$ units. Find the value of $x$.

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

  1. $2,8$

  2. $-2,9$

  3. $1,8$

  4. $-1,7$

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Correct Option: D
Explanation:

Distance=$\sqrt{(x _2-x _1)^2+(y _2-y _1)^2)}$


P$=(x _1,y _1)=(x,-1)$

Q$=(x _2,y _2)=(3,2)$

$25=(3-x)^2+(2+1)^2$

$25=(9+x^2-6x+9)$

$x^2-6x-7=0$

$x^2-7x+x-7=0$

$x(x-7)+1(x-7)=0$

$x=-1,7$

Find the coordinates of the point on the $x$-axis that is equidistant from $P(4,3,1)$ and $Q(-2,-6,-2)$.

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 { 3 }{ 2 } ,0,0 \right) $

  2. $\displaystyle \left( -\frac { 3 }{ 2 } ,0,0 \right) $

  3. $\displaystyle \left( 0,-\frac { 3 }{ 2 } ,0 \right) $

  4. $\displaystyle \left( 0,\frac { 3 }{ 2 } ,0 \right) $

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Correct Option: B
Explanation:

Let $R(x,0,0)$ be the point on $x$-axis which is equidistant from $P(4,3,1)$ and $Q(-2,-6,-2)$

$\Rightarrow { \left( x-4 \right)  }^{ 2 }+{ \left( -3 \right)  }^{ 2 }+{ \left( -1 \right)  }^{ 2 }={ \left( x+2 \right)  }^{ 2 }+{ 6 }^{ 2 }+{ 2 }^{ 2 }$ gives $-12x=18$ 
So $x=-1.5$ 
Hence, $\displaystyle  R\equiv \left( -\frac { 3 }{ 2 } ,0,0 \right) $

$P$ and $Q$ are points on the line joining $A(-2,5)$ and $B(3,1)$ such that $AP=PQ=QB$. Then, the distance of the midpoint of $PQ$ from the origin is

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

  1. $3$

  2. $\frac {\sqrt 37}{4}$

  3. $4$

  4. $3.5$

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Correct Option: B

A line passes through two point $A (2, -3, -1)$ and $B (8, -1, 2)$. The coordinates of a point on this line at a distance of $14$ units from $A$ are

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

  1. $(14, 1, 5)$

  2. $(-10, -7, 7)$

  3. $(86, 25, 41)$

  4. None of these

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Correct Option: A
Explanation:

Given points are $A(2,-3,-1)$ and $B(8,-1,2)$
Therefore direction  ratio of $AB$ are $l = \dfrac{6}{7}, m=\dfrac{2}{7},n= \dfrac{3}{7}$ or $l=\dfrac{-6}{7},m= \dfrac{-2}{7},n= \dfrac{-3}{7}$
Hence, coordinates of a point $14$ unit from point $A$ is given as,$(2+14l,-3+14m,-1+14n)$
$\Rightarrow (14,1,5)$ or $(-10,-7,-7)$
Hence, option 'A' is correct.

If  $C _1:{x^2+y^2}-20x+64=0$ and $C _2:{x^2+y^2}+30x+144=0$. Then the length of the shortest line segment $PQ$  which touches $C _1$ at $P$ and  to  $C _2$ at $Q$ 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. $15$

  3. $22$

  4. $27$

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Correct Option: A
Explanation:

Given $C _1 : x^2+y^2-20x+64=0 \Rightarrow (x-10)^2+y^2=36$
and $C _2 : x^2+y^2+30x+144=0 \Rightarrow (x+15)^2+y^2=81$
So centre and radius of $C _1$ and $C _2$ are $(10,0)$, $(-15,0)$ and $6,9$ respectively.
Then, distance between $C _1$ and $C _2$ is $\sqrt{(15+10)^{2}+(0-0)^{2}}=25$. 

$PQ$ touches $C _{1}$ at $P$ and $C _{2}$ at $Q$.
Then, shortest length of $PQ$ is $=25-(9+6)=25-15=10$

The distance of the point $(4,7)$ from the $x-$ axis is 

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

  1. $4$

  2. $7$

  3. $11$

  4. $\sqrt{65}$

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Correct Option: A

Minimum distance between the curves
$y^{2}=4x$ & $x^{2}+y^{2} -12x+31=0$ is -

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

  1. $\sqrt {21}$

  2. $\sqrt {26}-\sqrt {5}$

  3. $\sqrt {20}-\sqrt {5}$

  4. $\sqrt {21}-\sqrt {5}$

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Correct Option: A

The distance of the point $(2,3)$ form the line $x-2y+5=0$ measured in a direction parallel to the line $x-3y=0$ is

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

  1. $2\sqrt{10}$

  2. $\sqrt{10}$

  3. $2\sqrt{5}$

  4. $None\ of\ these$

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Correct Option: A

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

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

  1. $\sqrt{10}$

  2. $\sqrt{20}$

  3. $\sqrt{5}$

  4. $\sqrt{30}$

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Correct Option: A