Tag: introduction to three dimensional geometry

Questions Related to introduction to three dimensional geometry

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

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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 $a^2=\dfrac{25}{2}$.

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. $-1,7$

Show answer
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) $

Show answer
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) $