Tag: distance between two points in space

Questions Related to distance between two points in space

The distance between two points $(1,1)$ and $\left( {\dfrac{{2{t^2}}}{{1 + {t^2}}},\dfrac{{{{\left( {1 - t} \right)}^2}}}{{1 + {t^2}}}} \right)$ is 

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

  1. 4t

  2. 3t

  3. 1

  4. none of these

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Correct Option: C
Explanation:
Two point are $A(1,1)$ and $B\left(\dfrac{2t^{2}}{1+t^{2}}, \dfrac{(1+t)^{2}}{1+t^{1}}\right)$

$AB=\sqrt{\left(1-\dfrac{2t^{2}}{1+t^{2}}\right)^{2}+\left(1-\dfrac{(1-t^{2})}{1+t^{2}}\right)^{2}}$

$=\sqrt{\left(\dfrac{1+t^{2}-2t^{2}}{1+t^{2}}\right)^{2}+\left(\dfrac{1+t^{2}-1-t^{2}+2t}{1+t^{2}}\right)^{2}}$

$=\sqrt{\left(\dfrac{1-t^{2}}{1+t^{2}}\right)^{2}+\left(\dfrac{RT}{1+t^{2}}\right)^{2}}$

$=\sqrt{\dfrac{1+t^{4}-2t^{2}+4t^{2}}{(1+t^{2})^{2}}}$

$=\sqrt{\dfrac{(1+t^{2})^{2}}{(1+t^{2})^{2}}}$

$=1$

The plane passing through the point $\left(-2,-2,2\right)$ and containing the line joining the points $\left(1,1,1\right)$ and $\left(1,-1,2\right)$ makes intercepts on the coordinates axes, the sum whose lengths 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. $4$

  3. $6$

  4. $12$

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Correct Option: A

The points $A(1,2,3); B-(-1,-2,-1); C(2,3,2)$ and $D(4,7,6)$ form 

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. Parallelogram

  4. Rhombus

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Correct Option: A

30 consider at three dimensional figure represented by $xy{z^2} = 2$, then its minimum distance from origin 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. 4

  3. 3

  4. 1

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Correct Option: B

Perpendicular distance from the origin to the line joining the points $(a\cos{\theta},a\sin{\theta})(a\cos{\theta},a\sin{\theta})$ is

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

  1. $2a\cos{(\theta-\phi)}$

  2. $a\cos { \left( \cfrac { \theta -\phi }{ 2 } \right) } $

  3. $4a\cos { \left( \cfrac { \theta -\phi }{ 2 } \right) } $

  4. $a\cos { \left( \cfrac { \theta +\phi }{ 2 } \right) } $

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Correct Option: A

From which of the following the distance of the point $(1, 2, 3)$ is $\sqrt{10}$?

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

  1. Origin

  2. $x-$axis

  3. $y-$axis

  4. $z-$axis

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Correct Option: C
Explanation:

The point is $P(1,2,3)$ so distance of point from $y$ axis is
${=}$ $\sqrt{{(1)}^{2} + {(3)}^{2} } $  

$=\sqrt{10}$

Hence, option C is correct.

If the sum of the squares of the distance of a point from the three coordinate axes be $36$, then its distance 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. $6$ units

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

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

  4. none of these

Show answer
Correct Option: B
Explanation:

Let $(x, y, z)$ be the point.


Given sum of the squares of distance from point to the axes is $36$. 

$ \Rightarrow (x^2+y^2) +(y^2+z^2) + (z^2+x^2) = 36 $

$ \Rightarrow 2(x^2+y^2+z^2) = 36 \Rightarrow x^2 + y^2 + z^2 = 18  $

So the distance of the point from the origin is $ = \sqrt{x^2 + y^2 + z^2} = 3\sqrt{2}$

Hence, option B; is correct.

The perimeter of the triangle formed by the points $(1,0,0),(0,1,0),(0,0,1)$ is 

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

  1. $\sqrt 2 $

  2. $2\sqrt 2 $

  3. $3\sqrt 2 $

  4. $4\sqrt 2 $

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Correct Option: C
Explanation:
Given points $A(1, 0, 0), B(0, 1, 0), C(0, 0, 1)$ is
$AB=\sqrt{1+1}=\sqrt{2}$
$BC=\sqrt{1+1}=\sqrt{2}$
$CA=\sqrt{1+1}=\sqrt{2}$
Perimeter of the triangle is $AB+BC+CA=\sqrt{2}+\sqrt{2}+\sqrt{2}=3\sqrt{2}$.

If the distance of a point $(a,a,a)$ from the origin is $ \sqrt { 108 } $, then the value of $a$ is

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

  1. $9$

  2. $6$

  3. $-9$

  4. $-6$

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Correct Option: B,D
Explanation:
Distance between $(x _1,y _1)$ and $(x _2,y _2)$ is $\sqrt { { ({ x } _{ 1 }-{ x } _{ 2 }) }^{ 2 }+{ ({ y } _{ 1 }-{ y } _{ 2 }) }^{ 2 } } $
Distance between $(0,0,0)$ and $(a,a,a )$ is $\sqrt { {(a-0) }^{ 2 }+{ (a-0) }^{ 2 } +{ (a-0) }^{ 2 } } $

$\sqrt { 3\times { (a) }^{ 2 } } =\sqrt { 108 } =6\sqrt { 3 } \\ \Rightarrow a=\pm 6\\$

If the extremities of a diagonal of a square are $(1, -2, 3)$ and $(4, 2, 3)$ then the area of the square is

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

  1. $25$

  2. $50$

  3. $\displaystyle \frac{25}{2}$

  4. $\sqrt{50}$

Show answer
Correct Option: C
Explanation:

If $a$ is the length of a side of square then the length of diagonal is given by $\sqrt{2}a$. Distance between two given  points is $\sqrt{(1-4)^2+(-2-2)^2+(3-3)^2}=5=\sqrt{2}a$. Hence the area is given by $a^2=\dfrac{25}{2}$.