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 maths vectors: lines in two and three dimensions distance from a point to line perpendicular distance of a point from a plane the distance from a point to a line

$A = (0, 1, 2), B=(3, 0, 1), C=(4, 3, 6), D=(2, 3, 2)$ are the rectangular cartesian co-ordinates. Find the perpendicular distance from $A$ to the line $BC$.

  1. $\displaystyle \left ( \dfrac {6}{7} \right )\sqrt{14}$
  2. $\displaystyle \left ( \dfrac {6}{7} \right )\sqrt{18}$
  3. $\displaystyle \left ( \dfrac {4}{7} \right )\sqrt{14}$
  4. $\displaystyle \left ( \dfrac {4}{7} \right )\sqrt{18}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Vector BC = (4-3, 3-0, 6-1) = (1, 3, 5). Vector AB = (3-0, 0-1, 1-2) = (3, -1, -1). The distance is |AB x BC| / |BC|. AB x BC = i(3*-1 - 5*-1) - j(1*-1 - 5*3) + k(1*-1 - 3*3) = i(2) - j(-16) + k(-10) = (2, 16, -10). Magnitude = sqrt(4 + 256 + 100) = sqrt(360) = 6*sqrt(10). |BC| = sqrt(1+9+25) = sqrt(35). Distance = 6*sqrt(10)/sqrt(35) = 6*sqrt(2)/sqrt(7) = 6*sqrt(14)/7.

Multiple choice maths vectors: lines in two and three dimensions distance from a point to line perpendicular distance of a point from a plane the distance from a point to a line

The distance between a point $P$ whose position vector is $5\hat{i}+\hat{j}+3\hat{k}$ and the line $\vec{r}=(3\hat{i}+7\hat{j}+\hat{k})+\lambda(\hat{j}+\hat{k})$ is

  1. $3$
  2. $4$
  3. $5$
  4. $6$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\vec{r}= 3\vec{i}+(7+\lambda )\hat{j}+(1+\lambda )\hat{k}$
$(\hat{r}-\hat{a})= -2\hat{i}+(6+\lambda )\hat{j}+(\lambda -2)\hat{k}$
$(\vec{r}-\vec{a})\perp (\hat{j}+\hat{k})$
$(\vec{r}-\vec{a}).(\hat{j}+\hat{k})= 0$
$6+\lambda +\lambda -2= 0$
$\lambda =-2$
$(\vec{r}-\vec{a})= -2\hat{i}+4\hat{j}-4\hat{k}$
$\left | \vec{r}-\vec{a} \right |= 6$

Multiple choice maths vectors: lines in two and three dimensions distance from a point to line perpendicular distance of a point from a plane the distance from a point to a line

The perpendicular distance of a corner of unit cube from a diagonal not passing through it is

  1. $\sqrt{\dfrac{2}{3}}$
  2. $\dfrac{2}{3}$
  3. $\dfrac{1}{3}$
  4. $

    1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let P.V. of
$\overline{A}=\overline{0}$
$\overline{B}=\overline{b}$
$\overline{C}=\overline{c}+\overline{b}$
$\overline{D}=\overline{c}$
$\overline{G}=\overline{d}$
$\overline{E}=\overline{c}+\overline{b}+\overline{a}$
$\overline{AE}$ is the main diagonal$=\overline{c}+\overline{b}+\overline{d}$
& $\overline{c}$ be a Corner$=\overline{c}+\overline{b}$
So line equation of $AC=\overline{o}+t(\overline{c}+\overline{b}+\overline{d})$
So, $\bot$ distance $=\dfrac{|(\overline{c}+\overline{b})\times(\overline{c}+\overline{b})+\overline{d}|}{|\overline{c}+\overline{b}+\overline{d}|}$
$=\dfrac{|(\overline{c}+\overline{b})\times\overline{d}|}{\sqrt{3}}$
$=\dfrac{|\overline{c}+\overline{b}||\overline{d}|\sin \theta}{\sqrt{3}}$
$=\dfrac{\sqrt{2}}{3}$

Multiple choice maths vectors: lines in two and three dimensions distance from a point to line perpendicular distance of a point from a plane the distance from a point to a line

The perpendicular distance from $(4, -3, 2)$ to the line $\displaystyle \dfrac{x-2}{3}=\dfrac{y-3}{-2}=\dfrac{z-5}{6}$ is

  1. $7\sqrt{2}$
  2. $14$
  3. $7$
  4. $49$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

D.R of line are $3,-2,6$
Now in general the coordinates of any point on line can be represented in the form 
$(3\lambda+2 , -2\lambda + 3 , 6\lambda + 5)$
DR of line perpendicular to the given line $(3\lambda -2 , -2 \lambda + 6 , 6\lambda + 3)$
Now,
 $3(3\lambda -2) -2 (-2\lambda + 6) +6 (6\lambda + 3) = 0$
$\lambda = 0$
Hence, the coordinates are $(2,3,5)$
Distance between points $(2,3,5)$ and $(4,-3,2) = \sqrt {2^2+6^2+3^2} = 7$
Hence, option C is correct.

Multiple choice maths vectors: lines in two and three dimensions distance from a point to line perpendicular distance of a point from a plane the distance from a point to a line

State the following statement is True or False

Distance of the point $P(x, y, z)$ from the plane $X  Y$ is $\sqrt {x^{2} + y^{2} + z^{2}}$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Co-ordinates of plane $XY$ is $(0,0,z)$, since $Z$ is not included. 
$\therefore $  distance of pt $P(x,y,z)$ from xy plane $(0,0,z)$ is given by
$\sqrt { { (x-0) }^{ 2 }+{ (y-0) }^{ 2 }+{ (z-z) }^{ 2 } } $

$= \sqrt { { x }^{ 2 }+{ y }^{ 2 } } $

$\therefore $ The given statement is FALSE.
Multiple choice maths vectors: lines in two and three dimensions distance from a point to line perpendicular distance of a point from a plane the distance from a point to a line

The perpendicular distance of$\overrightarrow A $ (1,4,-2) from the segment BC where$\overrightarrow B $  (2,1,-2) and $\overrightarrow C $ (o,-5,1) is 

  1. $\frac{3}{7}\sqrt {26} $
  2. $\frac{6}{7}\sqrt {26} $
  3. $\frac{4}{7}\sqrt {26} $
  4. $\frac{2}{7}\sqrt {26} $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
We have to find a perpendicular distance $AP$
Now $\vec{AB}=(2, 1, -2)-(1, 4, -2)=(1, -3, 0)$
$\vec{AC}=(0, -5, 1)-(1, 4, -2)=(-1, -9, 3)$
$\vec{AB}\times \vec{AC}\begin{vmatrix} i & j & k \\ 1 & -3 & 0 \\ -1 & -9 & 3 \end{vmatrix}=-9i-3j-12k$
$|\vec{AB}\times \vec{AC}|=\sqrt{(-9)^{2}+(-3)^{2}+(-12)^{2}}=3\sqrt{26}$ units
Area of $\Delta =\dfrac{1}{2}|\vec{AB}\times \vec{AC}|=\dfrac{3}{2}\sqrt{26}$ sq.units
Now $\vec{BC}=(0, -5, 1)-(2, 1, -2)$
$=(-2, -6, 3), |\vec{BC}|=\sqrt{(2)^{2}+(6)^{2}+3^{2}}=7$ units
$=\dfrac{1}{2}\times 7\times AP=\dfrac{3}{2}\sqrt{26}$
$\therefore Ap=\dfrac{3}{7}\sqrt{26}$ units
Multiple choice maths vectors: lines in two and three dimensions distance from a point to line perpendicular distance of a point from a plane the distance from a point to a line

The perpendicular distance of the point $P(1,2,3)$ from the straight line passing through the point $A(-1,4,7)$ and $B(2,8,7)$

  1. $\displaystyle \frac{\sqrt{149}}{25}$
  2. $\displaystyle \frac{149}{25}$
  3. $\displaystyle \frac{2\sqrt{149}}{25}$
  4. $\displaystyle \frac{2\sqrt{149}}{5}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The drs of line joining points $A(-1,4,7)$ and $B(2,8,7)$ are $3,4,0$

The equation of line $AB$ is $\frac{x+1}{3}=\frac{y-4}{4}=\frac{z-7}{0}$
So the point on line $AB$ will look like $C(3t-1,4t+4,7)$
Assume that $CP$ is perpendicular to $AB$
So we get $(3t-2)3+(4t+2)4=0$
$\Rightarrow t=-\frac{2}{25}$
So the point $C$ is $(-\frac{31}{25}\frac{92}{25},7)$
The length of $CP$ is $\sqrt { { (\frac { 56 }{ 25 } ) }^{ 2 }+{ (\frac { 42 }{ 25 } ) }^{ 2 }+{ 7 }^{ 2 } } =\frac{2\sqrt{149}}{5}$
Therefore the correct option is $D$

Multiple choice maths vectors: lines in two and three dimensions distance from a point to line perpendicular distance of a point from a plane the distance from a point to a line

The distance from the point $(1,6,3)$ to the line $\bar{r}=(\hat{j}+2\hat{k})+\lambda(\hat{i}+2\hat{j}+3\hat{k})$ is

  1. $\sqrt{13}$
  2. $13$
  3. $2\sqrt{13}$

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

Using fact the $\perp$ distance from $(\alpha,\beta,\gamma)$ in the line

$\displaystyle \dfrac{x-x _{1}}{1}=\dfrac{y-y _{1}}{m}=\dfrac{z-z _{1}}{n}$ is given by 

$\sqrt{(\alpha-x _{1})^{2}+(\beta-y _{1})^{2}+(\gamma-z _{1})^{2}}$

$\sqrt{-[l(\alpha-x _{1})+m(\beta-y _{1})+n(\gamma-z _{1})]^{2}}$

Now $(\alpha,\beta,\gamma)=(1,6,3),(x _{1},y _{1}.z _{1})=0,1,2,l,m,n=\dfrac{1}{\sqrt{14}},\dfrac{1}{\sqrt{14}},\dfrac{1}{\sqrt{14}}$

$\therefore$ Required distance

$\displaystyle =\sqrt{(1-0)^{2}+(6-1)^{2}+(3-2)^{2}-\left[\dfrac{1}{\sqrt{14}}(1+2(5)+3(1)\right]^{2}}=\sqrt{27-14}=\sqrt{13}$

Multiple choice maths mapping your way mapping mapping space around us bearing and drawings

In a ground, the distance between two consecutive trees is $4$ m and distance between next $2$ trees is $5$ m. Then calculate the distance between first and third tree.

  1. $9$ m
  2. $5$ m
  3. $4$ m
  4. $3$ m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the first Tree be at point A, second at point B and  third one a C.

$AB=4$ m and $BC=5$ m.
$AC= AB+BC$
$\therefore AC = 9$ m.

Multiple choice maths mapping your way mapping mapping space around us bearing and drawings

Distance between two houses is $40$ m. If a new house with area $4\times 4$ is constructed in the middle of $2$ houses, then find the distance between middle house and one of the corner house.

  1. $20$ m
  2. $19$ m
  3. $18$ m
  4. $17$ m
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
The distance of middle point of middle house from the corner house is $\dfrac { 40 }{ 2 } =20$ and house to house distance will be $\therefore 20-2=18$
The distance between he Middle house and Corner house will be $18$ m.

Multiple choice maths bearing and drawings mapping mapping space around us changing scale

On a scale map $0.7\ cm$ represents $8.4\ km$. If the distance between two points on the map is $46.5\ cm$, what is the actual distance between the points?

  1. $56\ km$
  2. $55.80\ km$
  3. $62.80\ km$
  4. $72\ km$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given $0.7$cm represents $8.4$ km 

$46.5$ cm represents $x$
By proportionality 

$\dfrac{0.7}{46.5} = \dfrac{8.4}{x}$

$x = \dfrac{8.4\times46.5}{0.7}$
 
$x = 55.80km$

Multiple choice maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

The distance of the point $(-1,-5,-10)$ from the point of intersection of the line $\dfrac{x-2}{2}=\dfrac{y+1}{4}=\dfrac{z-2}{12}$ and the plane $x-y+z=5$ is

  1. $2\sqrt{11}$
  2. $\sqrt{126}$
  3. $13$
  4. $14$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given line is, $\dfrac{x-2}{2} = \dfrac{y+1}{4} = \dfrac{z-12}{2} = k$ (say)


So any point on this line is given by, $(2k+2, 4k-1, 12k+2)$

Now line intersects the plane $x-y+z=5$

$\Rightarrow 2k+2-(4k-1)+12k+2=5 \Rightarrow k = 0$

Thus point of intersection is, $(2, -1, 2)$  

Therefore required distance is $= \sqrt{(2+1)^2+ (-1+5)^2 + (2+10)^2} = 13$

Hence, option 'C' is correct.

Multiple choice maths binomial theorem, sequence and series series introduction to series introduction to sequences and series

A, B, C, D are four points in a straight line. Distance from A to  B is 10, B to C is 5, C to D  is 4 and A to D is 1. Which  one of the following is the correct sequence of the  points ?

  1. A- B - C - D

  2. A - C - B - D

  3. A - D - C - B

  4. A - C - D - B
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given:$A,B,C,D$ are four points in a staraight line
Distance from $A$ to $B$ is $AB=10$ units
Distance from $B$ to $C$ is $BC=5$ units
Distance from $C$ to $D$ is $CD=4$ units
Distance from $A$ to $D$ is $AD=1$ unit
Here $10>5>4>1\Rightarrow\,AB>BC>CD>DA$
$AB$ is the longest line,the sequence is $A-B$
$\Rightarrow\,BC<CD\Rightarrow\,C$ lies between $A$ and $B$
$\therefore\,$the sequence is $A-C-B$
$AC=5$ units then $CB=5$ units
Now,$D$ may lie between $AC$ or $AB$
$\because \,CD<AC,\,CD<BC$
Since $AD$ starts from $A$ and lies between $A$ and $C$
and $AD=1$ unit
$\therefore\,D$ cannot lie between $B$ and $C$
Hence $D$ lies between $A$ and $C$
The correct sequence is $A-D-C-B$

Multiple choice maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

The distance between the top of two trees $20$m and $28$m high is $17$m. The horizontal distance between the trees is:

  1. $11$m
  2. $31$m
  3. $15$m
  4. $9$m
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Consider the two trees as perpendiculars on the ground forming a right triangle, where the distance between their bases is the base $(B)$, the distance between their tops is the Hypotenuse $(H)$ and difference between their lengths is the perpendicular $(P)$.
Hypotenuse, $H = 17$ m
Perpendicular, $P=$ distance between their lengths $= 28 - 20 = 8$ m
Thus, By Pythagoras Theorem,
$H^2 = P^2 + B^2$
$17^2 = 8^2 + B^2$
$B^2 = 289 - 64$
$B = 15m$.