Tag: distance between two points in space

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

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

  1. $\sqrt{156}$

  2. $13$

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

  4. $8$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.
$\dfrac{\sqrt{3^2+4^2+12^2}}{2}=\dfrac{13}{2}$ 

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

  1. $2,8$

  2. $-2,9$

  3. $1,8$

  4. $-1,7$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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


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

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

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

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

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

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

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

$x=-1,7$

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

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

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

  2. <span>$\displaystyle&nbsp;\left( - \frac { 3 }{ 2 } ,0,0 \right) $</span>

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

  4. <span>$\displaystyle&nbsp;\left( 0,\frac { 3 }{ 2 } ,0 \right) $</span>
Reveal answer Fill a bubble to check yourself
B Correct answer
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) $

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

$P$ 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

  1. $3$

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

  3. $4$

  4. $3.5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Points P and Q divide AB into three equal parts. P = (-2+1/3(3-(-2)), 5+1/3(1-5)) = (-1/3, 11/3). Q = (-2+2/3(5), 5+2/3(-4)) = (4/3, 7/3). Midpoint of PQ = (1/2, 3). Distance from origin = sqrt(1/4 + 9) = sqrt(37)/2 = sqrt(37/4).

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

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

  1. $(14, 1, 5)$

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

  3. $(86, 25, 41)$

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

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

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

  1. $10$

  2. $15$

  3. $22$

  4. $27$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $=25-(9+6)=25-15=10$

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 $(2,1,-1)$ 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

  1. $\sqrt{10}$

  2. $\sqrt{20}$

  3. $\sqrt{5}$

  4. $\sqrt{30}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The distance is found by projecting the vector from the point to a point on the line onto the normal of the plane containing the line and the direction parallel to the given plane.