Mathematics

3D Geometry and Coordinate Distance

140 Questions

Three dimensional geometry and coordinate distance problems involve calculating spatial measurements between points and lines. Questions feature vector distances, perpendicular distances, and coordinate geometry theorems. This advanced topic is typically found in mathematics examinations.

Point distance calculationsPerpendicular distancesVector coordinatesLine ratiosAxis distances

3D Geometry and Coordinate Distance Questions

Multiple choice

The perpendicular distance of (- 2, - 4) from the x-axis is ______.
    • 4 units
    • 2 units

  3. 2 units

  4. 4 units
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The perpendicular distance of a point (- 2, - 4) from x-axis will be y co-ordinate but with a positive sign, as the distance is never negative. So, the perpendicular distance of (- 2, - 4) from x-axis is 4 units.

Multiple choice

The perpendicular distance of (11, –9) from the x–axis is

  1. 11 units

  2. –11 units

  3. 9 units

  4. –9 units
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The perpendicular distance of the point (11, –9) from the x-axis will be y coordinate of the point, but with a positive sign as the distance is never negative. So, the perpendicular distance of (11, –9) from the x-axis is 9 units.

Multiple choice maths introduction to three-dimensional geometry distance between two parallel planes three dimensional geometry - ii distance formula

The distance between the planes given by $\vec{r}.\left ( i:+:2j:-:2k \right ):+:5= 0$ and $\vec{r}.\left ( i:+:2j:-:2k \right ):-:8= 0$ is

  1. $1$ unit
  2. $13/3$ units
  3. $13$ units

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given planes
$\vec{r}\cdot(\hat{i}+2\hat{j}-2\hat{k})+5=0$---------(1)
$\vec{r}\cdot(\hat{i}+2\hat{j}-2\hat{k})-8=0$---------(2)
from Here $a=1,b=2,c=-2$ and $d _{1}=-5\ and\ d _{2}=8$
$distance=\left | \dfrac{d _{1}-d _{2}}{\sqrt{a^2+b^2+c^2}} \right |$
$distance=\left | \dfrac{-5-8}{\sqrt{1^2+2^2+(-2)^2}} \right |$
$distance=\left | \dfrac{-13}{\sqrt{1+4+4}} \right |$
$distance=\left | \dfrac{-13}{\sqrt{9}} \right |$
$distance=\left | \dfrac{-13}{3} \right |$
$distance=\dfrac{13}{3}$
Multiple choice maths compound measures and motion kinetic graphs time - calcuation of distance travel graphs

Choose the correct answers from the alternative given :
In a row. 25 trees are planted at equal distance from each other. The distance between 1st and 25th tree is 30 m.
What is the distance between 3rd and 15th tree?

  1. $8$ m
  2. $15$ m
  3. $16$ m
  4. $18$ m
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let distance between two consecutive trees be x m.
So, distance between 1st and 25th tree = 24x m = $30$ m
Distance between 3rd and 15th = 12x m = $15$

Multiple choice maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

In the complex plane, what is the distance of $4-2i$ from the origin?

  1. $2$
  2. $3.46$
  3. $4.47$
  4. $6$
  5. $12$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $z=4-2i$

In a complex plane, distance of $z$ from any point origin is given by $|z|=\sqrt{(x _1)^2+(y _1)^2}$
$\therefore$ Distance of $z=4-2i$ from origin is given by $|z|=|4-2i|=\sqrt{4^2+(-2)^2}=\sqrt{20}\approx 4.47$
Hence, the answer is $4.47$.

Multiple choice maths trigonometry trigonometric ratios of acute angles compound angles, multiple angles, sub multiple angles and transformation formulae trigonometric identities


lf the distance between the points $(a \cos\theta, a \sin\theta), (a \cos\phi, a \mathrm{sin} \phi)$ is $2\mathrm{a}$ then $\theta=$.

  1. $2n\pi\pm\pi+\phi, n\in z$
  2. $n\displaystyle \pi\pm\frac{\pi}{2}+\phi, n\in z$
  3. $n\pi-\phi, n\in z$
  4. $2n\pi+\phi, n\in z$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given points $(a=cos\theta, a\sin\theta), (a\cos\phi, a\sin \phi)$

distance $2a$

distance between $(x _{1}, y _{1}), (x _{2}, y _{2})$

$\sqrt{(x _{2}-x _{1})^{2}+(y _{2}-y _{2})^{2}}$

So, 
$\sqrt{(a\cos\theta-a\cos\theta)^{2}+(a\sin\phi-a\sin\theta)^{2}}=2a$

Squaring on both sides

$\Rightarrow a^{2}(\cos^{2}\theta+\cos^{2}\phi-2\cos\theta\cos\phi)+a(\sin^{2}\phi+\sin^{2}\theta-2\sin\theta-\sin\phi)=4a^{2}$

$\Rightarrow a^{2}(\sin^{2}\theta+\cos^{2}\theta+\sin^{2}\theta+\cos^{2}\theta-2(\sin\theta-\sin\phi+\cos\theta\cos\phi))=4a^{2}$

$\Rightarrow a^{2}(1+1-2(\cos(\theta-\phi))=4a^{2}$

$\Rightarrow 2-2\cos(\theta-\phi)=4$

$\Rightarrow 2\cos(\theta-\phi)=-2$

$\Rightarrow \cos(\theta-\phi)=-1=\cos(\pi)$

$\Rightarrow \theta-\phi=2n\pi\pm \pi, n\in z$

$\Rightarrow \boxed{\theta=2n\pi\pm\pi+\phi, n\in z}$
Multiple choice physics motion and measurement measuring length measurement of small and large distances measurement of distance

A star's distance ($d$) and its parallax angle ($p$) are related to each other as:

  1. $d=\dfrac{1}{p}$
  2. $d=\dfrac{1}{p^2}$
  3. $p=\dfrac{1}{d^2}$

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

Astronomers use an effect called parallax to measure distances to nearby stars. Parallax is the apparent displacement of an object because of a change in the observer's point of view.

The relationship between a star's distance and its parallax angle:

$d=\dfrac{1}{p}$

The distance $d$ is measured in parsecs and the parallax angle $p$ is measured in arc seconds.

Multiple choice maths how much does it weigh? basic operations with same units operations involving units of length calculations define weight and units of weight

Subtract $778\ m\ 78\ cm$ from $2\ km\ 768\ cm$

  1. $1228.9\ m$
  2. $571.1\ m$
  3. $1221.988\ m$
  4. $189.22\ m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

First we have to convert both the quantities in $m$

$1\ \ m=100\ \ cm$
$\Rightarrow 1\ \ cm=.01 \ m$
$778 \ \ m\ \ 78\ \ cm =778+.01\times 78=778.78\ \ m$
$1\ \ km=1000\ \ m$
$2\ \ km\ \ 768\ \ cm =2\times1000+.01\times768=2007.68\ \ m$
Now subtracting both 
$2007.86-778.8=1128.9\ \ m$
Option $A$ is correct.

Multiple choice maths how much does it weigh? basic operations with same units operations involving units of length calculations define weight and units of weight

Subtract :
$23.67\ m$ from $1.4\ km$

  1. $116.33$
  2. $1376.33\ m$
  3. $13976.33\ m$
  4. $22.27\ m$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

First we have to convert $km$ into $m$

$1\ \ km=1000\ \ m$
$\Rightarrow 1.4\ \ km=1.4\times1000=1400\ \ m$
Now subtracting both
$1400 \ \ m-23.67\ \ m=1376.33\ \ m$
So option $C$ is correct.

Multiple choice maths how much does it weigh? basic operations with same units operations involving units of length calculations define weight and units of weight

If Harry runs $23\ m\ 5\ cm$ and Joy runs $14.37\ m$ from the same starting point, then how far is Joy from Harry.

  1. $8.68\ m$
  2. $9.13\ m$
  3. $86.8\ m$
  4. $91.3\ m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We low that $1 m=100 cm$

Hence distance run by Harry $=23 m + 5 cm\=23 m+ \dfrac{5}{100} m\=23.05 m$
Distance run by Joy $=14.37 m$
Distance between them $=23.05 m -14.37 m\=8.68 m$
                                                 

Multiple choice physics constellations and galaxies light year evolution and end stages of stars in the world of stars

If the light from star $A$ takes $15$ min to reach star $B$ and the speed of light is $3\times {10}^{8}m/s$, then what is the distance between the stars?

  1. $2.7\times {10}^{8}km$
  2. $2,9\times {10}^{11}km$
  3. $2.7\times {10}^{8}m$
  4. $36\times {10}^{9}km$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given $t=15$min
Speed of light $=3\times {10}^{8}m/s$
$\therefore$ $t=15min=15\times 60=900 sec$
By using the formula
$Speed=\cfrac{Distance}{Time}$
$Distance=Speed \times time$ $=3\times {10}^{8}m/s\times 900 sec$ $=2700\times {10}^{8}m$ $=2.7\times {10}^{8}km$

Multiple choice maths linear graphs quadrants the cartesian plane the cartesian system

 If the distance between the points $\left( {a\,\cos {{48}^ \circ },0} \right)$ and $\left( {\,0,a\,\cos {{12}^ \circ }} \right)$ is d,then ${d^2} - {a^2} = $

  1. ${a^2}\left( {\sqrt 5 - 1} \right),/4$
  2. ${a^2}\left( {\sqrt 5 + 1} \right),/4$
  3. ${a^2}\left( {\sqrt 5 - 1} \right),/8$
  4. $\dfrac{a^2( {\sqrt 5 + 1} )}{8}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Using distance formula,${d}^{2}={\left(a\cos{{48}^{\circ}}-0\right)}^{2}+{\left(0-a\cos{{12}^{\circ}}\right)}^{2}$

${d}^{2}-{a}^{2}={\left(a\cos{{48}^{\circ}}-0\right)}^{2}+{\left(0-a\cos{{12}^{\circ}}\right)}^{2}-{a}^{2}$

$={a}^{2}{\cos}^{2}{{48}^{\circ}}+{a}^{2}{\cos}^{2}{{12}^{\circ}}-{a}^{2}$

$={a}^{2}{\cos}^{2}{{48}^{\circ}}-{a}^{2}+{a}^{2}{\cos}^{2}{{12}^{\circ}}$

$={a}^{2}{\cos}^{2}{{48}^{\circ}}-{a}^{2}\left(1-{\cos}^{2}{{12}^{\circ}}\right)$

$={a}^{2}{\cos}^{2}{{48}^{\circ}}-{a}^{2}{\sin}^{2}{{12}^{\circ}}$

We know that ${\cos}^{2}{B}-{\sin}^{2}{A}=\cos{\left(A+B\right)}\cos{\left(A-B\right)}$

$={a}^{2}\cos{\left({12}^{\circ}+{48}^{\circ}\right)}\cos{\left({12}^{\circ}-{48}^{\circ}\right)}$

$={a}^{2}\cos{{60}^{\circ}}\cos{\left(-{36}^{\circ}\right)}$

$={a}^{2}\cos{{60}^{\circ}}\cos{{36}^{\circ}}$ since $\cos{\left(-\theta\right)}=\cos{\theta}$

$={a}^{2}\times\dfrac{1}{2}\times\dfrac{\sqrt{5}+1}{4}$

$=\dfrac{{a}^{2}\left(\sqrt{5}+1\right)}{8}$