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 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 

  1. 4t

  2. 3t

  3. 1

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Two point are $A(1,1)$ and $B\left(\dfrac{2t^{2}}{1+t^{2}}, \dfrac{(1+t)^{2}}{1+t^{1}}\right)$

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

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

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

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

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

$=1$

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

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

  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) } $
Reveal answer Fill a bubble to check yourself
A Correct answer
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 sum of the squares of the distance of a point from the three coordinate axes be $36$, then its distance from the origin is

  1. $6$ units

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

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

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

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

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

  1. $\sqrt 2 $

  2. $2\sqrt 2 $

  3. $3\sqrt 2 $

  4. $4\sqrt 2 $
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $AB+BC+CA=\sqrt{2}+\sqrt{2}+\sqrt{2}=3\sqrt{2}$.
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 distance of a point $(a,a,a)$ from the origin is $ \sqrt { 108 } $, then the value of $a$ is

  1. $9$

  2. $6$

  3. $-9$

  4. $-6$
Reveal answer Fill a bubble to check yourself
B,D Correct answer
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 } } $

$\sqrt { 3\times { (a) }^{ 2 } } =\sqrt { 108 } =6\sqrt { 3 } \\ \Rightarrow a=\pm 6\\$
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 extremities of a diagonal of a square are $(1, -2, 3)$ and $(4, 2, 3)$ then the area of the square is

  1. $25$

  2. $50$

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

  4. $\sqrt{50}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $a^2=\dfrac{25}{2}$.