Tag: introduction to three dimensional geometry

Questions Related to introduction to three dimensional geometry

The equation of plane passing through $(-1,0,-1)$ parallel to $xz$ plane is

maths introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

  1. $y=-2$

  2. $y=0$

  3. $-x-z=0$

  4. None of the above

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Correct Option: B
Explanation:

Given that the plane is parallel to $xz$ plane and the plane passes through $(-1,0,-1)$

Since the plane is parallel to $xz$ plane , the $y$ coordinate should be constant
Given that it passes through point $(-1,0,-1)$ , therefore the plane lies on $xz$ plane
Therefore the equation of plane is $y=0$
The correct options are $B$

The planes $2x-y+4z=5$ and $5x-2.5y+10z=6$ are

maths introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

  1. Parallel

  2. Perpendicular

  3. Intersect

  4. intersect $x$ axis

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Correct Option: A
Explanation:

Planes are $2x-y+4z=5$ 

and $5x-2.5y+10z=6$
Multiply both sides by 2 to the second equation
$\Rightarrow 10x-5y+20=12$
Now divide both sides by $2$
$\Rightarrow 2x-y+4z=\dfrac{12}{5}$

Clearly both planes are parallel 

In a three-dimensional space, the equation $3x - 4y = 0$ represents.

maths introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

  1. A plane containing $Z-axis$

  2. A plane containing $X-axis$

  3. A plane containing $Y-axis$

  4. Passing through $(0, 0)$

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Correct Option: A
Explanation:

If we consider the z-part also, then the equation is $3x-4y+0z=0$

Which means on putting any value of z equation will have no effect as $0\times z=0$
$\therefore$ it will be a plane containing $Z-axis$.($\because$ it will pass through all points z if it satisfy condition for$ (x,y) $)
(D) is not right because its plane and not a line so it will pass through $(0,0,0)$ not $(0,0)$.
Hence, $(A)$


The point $(3, 0, -4)$ lies on the

maths introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

  1. Y-axis

  2. Z-axis

  3. XY-plane

  4. XZ-plane

  5. YZ-plane

Show answer
Correct Option: D
Explanation:
$(3, 0, -4)$   $\rightarrow$   Given point
Clearly, $y = 0$ and $ x$ and $z$ have non-zero value.
If the point lies on $x-z$ plane, this condition is possible.
Hence, the answer is $XZ$- plane.

Which of the following is true for a plane?

maths introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

  1. A locus is called a plane if the line joining any two arbitrary points on the locus is also a part of the locus.

  2. Value of $y$ in a $zx$ plane is non-zero.

  3. Value of $z$ in a $xy$ plane is zero.

  4. None of the above

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Correct Option: A,C
Explanation:

Option A and C are correct 

A locus is called a plane if the line joining any two arbitrary points on the locus is also a part of the locus. and also Value of z in a xy plane is zero.

There are three points with position vectors $ -2a+3b+5c, a+2b+3c $ and$ 7a-c$. What is the relation between the three points?

maths introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

  1. Collinear

  2. Forms a triangle

  3. In different plane

  4. None of the above

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Correct Option: A
Explanation:

The relation between the three points are collinear

Thus option A is correct answer 

The coordinates of the point where the line through $(3, -4, -5)$ and $(2, -3, 1)$ crosses the plane passing through three points $(2, 2, 1),(3, 0, 1)$ and $(4, -1, 0)$ is

maths introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

  1. $(1, 2, 7)$

  2. $(-1, 2, -7)$

  3. $(1, -2, 7)$

  4. None of these

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Correct Option: A

The distance of origin from the image of (1, 2, 3) in plane x - y + z = 5 is 

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

  1. $\sqrt{17}$

  2. $\sqrt{29}$

  3. $\sqrt{34}$

  4. $\sqrt{41}$

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Correct Option: C
Explanation:

$P(1,2,3),$ Plane :$x-y+z=5$

$F$ is foot of perpendicular form $P$ to plane and $I$ is image,then $PF=FI$
$\therefore$ If $(x,y,z)=(r+1,-r+2,r+3)$ are foot of perpendicular.
$ \Rightarrow (r+1)-(-r+2)+r+3=5\quad \quad \Rightarrow r=1\ \therefore F=(2,1,4)\ \therefore I=(3,0,5)$
$ \therefore$ distance of $I$ from origin  $=\sqrt { { 3 }^{ 2 }+{ 0 }^{ 2 }+{ 5 }^{ 2 } } =\sqrt { 34 } $

The equation of the set of points which are equidistant from the points $(1, 2, 3)$ and $(3, 2, -1)$.

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

  1. $x-2z=0$

  2. $2x-z=0$

  3. $2x+y=0$

  4. $x-2y=0$

Show answer
Correct Option: A
Explanation:

Let the given points be A($1,2,3$) and B($3,2,-1$) and let the point equidistant from A and B be P($x,y,z$) 

then  $PA=PB$
$\sqrt {{{(x - 1)}^2} + {{(y - 2)}^2} + {{(z - 3)}^2}}  = \sqrt {{{(x - 3)}^2} + {{(y - 2)}^2} + {{(z + 1)}^2}} $
Squaring both sides
${(x - 1)^2} + {(y - 2)^2} + {(z - 3)^2} = {(x - 3)^2} + {(y - 2)^2} + {(z + 1)^2}$
${x^2} + 1 - 2x + {z^2} + 9 - 6z = {x^2} + 9 - 6x + {z^2} + 1 + 2z$
$-2x-6z+10=-6x+2z+10$
$4x-8z=0$
$x-2z=0$

If the distance between a point $P$ and the point $(1, 1, 1)$ on the line $\dfrac {x - 1}{3} = \dfrac {y - 1}{4} = \dfrac {z - 1}{12}$ is $13$, then the coordinates of $P$ are

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

  1. $(3, 4, 12)$

  2. $\left (\dfrac {3}{13}, \dfrac {4}{13}, \dfrac {12}{13}\right )$

  3. $(4, 5, 13)$

  4. $(40, 53, 157)$

Show answer
Correct Option: C
Explanation:

Let the given points be $A(1,1,1)$

Consider,
$\dfrac{{x - 1}}{3} = \dfrac{{y - 1}}{4} = \dfrac{{z - 1}}{{12}} = \lambda $

$\begin{array}{l} x=3\lambda +1 \  \ y=4\lambda + 1\  \ z=12\lambda +1 \end{array}$

General point on the line is 
$3\lambda  + 1,\,4\lambda  + 1,\,12\lambda  + 1$

Given that,
$AP=13$ 

$\sqrt {{{\left( {3\lambda  + 1 - 1} \right)}^2} + {{\left( {\,4\lambda  + 1 - 1} \right)}^2} + {{\left( {12\lambda  + 1 - 1} \right)}^2}}  = 13$

$13\lambda =13$

$\lambda =1$

$\begin{array}{l} 3\lambda +1=4 \  \ 4\lambda +1=5 \  \ 12\lambda +1=13 \end{array}$

Therefore, required point $P$ is $(4,5,13)$.
Hence the correct option is $C$.