Tag: introduction to three dimensional geometry

Quizzes (6) Questions (119)

Questions Related to introduction to three dimensional geometry

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

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

  1. $y=-2$

  2. $y=0$

  3. $-x-z=0$

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $B$

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

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

  1. Parallel

  2. Perpendicular

  3. Intersect

  4. intersect $x$ axis
Reveal answer Fill a bubble to check yourself
A Correct answer
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 

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

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

  1. A plane containing $Z-axis$

  2. A plane containing $X-axis$

  3. A plane containing $Y-axis$

  4. Passing through $(0, 0)$
Reveal answer Fill a bubble to check yourself
A Correct answer
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, $(A)$


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

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

  1. Y-axis

  2. Z-axis

  3. XY-plane

  4. XZ-plane

  5. YZ-plane
Reveal answer Fill a bubble to check yourself
D Correct answer
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.
Multiple choice maths introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

Which of the following is true for a plane?

  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
Reveal answer Fill a bubble to check yourself
A,C Correct answer
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.

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

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

  1. Collinear

  2. Forms a triangle

  3. In different plane

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The relation between the three points are collinear

Thus option A is 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

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

  1. $\sqrt{17}$

  2. $\sqrt{29}$

  3. $\sqrt{34}$

  4. $\sqrt{41}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 } $

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

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

  1. $x-2z=0$

  2. $2x-z=0$

  3. $2x+y=0$

  4. $x-2y=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
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
${(x - 1)^2} + {(y - 2)^2} + {(z - 3)^2} = {(x - 3)^2} + {(y - 2)^2} + {(z + 1)^2}$
${x^2} + 1 - 2x + {z^2} + 9 - 6z = {x^2} + 9 - 6x + {z^2} + 1 + 2z$
$-2x-6z+10=-6x+2z+10$
$4x-8z=0$
$x-2z=0$

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 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

  1. $(3, 4, 12)$

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

  3. $(4, 5, 13)$

  4. $(40, 53, 157)$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $C$.