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 direct proportion and inverse proportion rule of three types of proportions direct proportion

On a scale of map, $0.6$ cm represents $ 6.6$km. If the distance between the points on the map is  $80.5$  cm, the actualdistance between these points is

  1. $9$ km
  2. $72.5$ km
  3. $190.75$ km
  4. $885.5$ km
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the actural distance be x km. Then, more distance on the map, more is the actual distance. 
$\therefore$ 0.6 : 80.5 :: 6.6 : x
$\Rightarrow\, 0.6x\, =\, 80.5\, \times\, 6.6$
$\Rightarrow\, x\, =\, \displaystyle \frac{80.5\, \times\, 6.6}{0.6}\, \Rightarrow\, x\, =\, 885.5$

Multiple choice chemistry coordination chemistry bonding in metal carbonyls metal carbonyls coordination compounds

The V-C distance in ${\text{V}}{\left( {{\text{CO}}} \right) _{\text{6}}}\;{\text{and}}\;\left[ {{\text{V}}{{\left( {{\text{CO}}} \right)} _{\text{6}}}} \right]$ are respectively (in pm) -

  1. 200, 200

  2. 193, 200

  3. 200, 193

  4. 193, 193
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In V(CO)6 (neutral), the V-C bond is longer due to less back-bonding compared to the [V(CO)6]- anion, where increased back-bonding strengthens the M-C bond and shortens the distance.

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 $(2,4,-1)$ from the line $\dfrac{x+5}{1}=\dfrac{y+3}{4}=\dfrac{z-6}{-9}$ is

  1. $3$
  2. $5$
  3. $7$
  4. $9$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let P(2,4,−1)P(2,4,−1) be the given point and the Line L→x−51=y−34=z−6−9=λL→x−51=y−34=z−6−9=λ
⇒d.r.⇒d.r. of the line L=(1,4,−9)L=(1,4,−9)
Any point Q(x,y,z)Q(x,y,z) on the line L is given by Q(λ+5,4λ+3,−9λ+6)Q(λ+5,4λ+3,−9λ+6)
Let this point QQ be the foot of ⊥⊥ of the point P on the line LL.
d.r of $\overrightarrow {PQ} $ the ⊥⊥
=(λ+5−2,4λ+3−4,−9λ+6+1)=(λ+5−2,4λ+3−4,−9λ+6+1)
=(λ+3,4λ−1,−9λ+7)
∴$\overrightarrow {PQ} $(1,4,−9)=0∴PQ¯.(1,4,−9)=0
(λ+3,4λ−1,−9λ+7).(1,4,−9)=0(λ+3,4λ−1,−9λ+7).(1,4,−9)=0
⇒(λ+3)1+(4λ−1)4−9(−9λ+7)=0⇒(λ+3)1+(4λ−1)4−9(−9λ+7)=0
⇒λ+3+16λ−4+81λ−63=0⇒λ+3+16λ−4+81λ−63=0
⇒98λ=64⇒98λ=64
⇒λ=3249
$\overrightarrow {PQ} $ is ⊥⊥⊥ to the line L
Substituting λλ in Q we get the foot of ⊥Q⊥Q as
Q(3249Q(3249+5,12849+5,12849+3,−28849+3,−28849+6)+6)

⊥⊥ distance≡$\overrightarrow {PQ} $ =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 perpendicular distance from a point $P$ with position vector $5\vec {i}+\vec {j}+3\vec {k} $ to the line $\vec {r}=(3\vec {i}+7\vec {j}+\vec {k})+t(\vec {j}+\vec {k})$ is

  1. $3$
  2. $6$
  3. $9$
  4. $12$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Point Pis $(5, 1, 3)$  and line is given

$\vec r = \left( {3\hat i + 7\hat j + \hat k} \right) + t\left( {\hat j + \hat k} \right)$
 let the foot of the perpendicular from point +p to the given line be $\theta \left( {x{,^1}y{,^1},{z^1}} \right)$ then direction ratios of $pq$ are
$\left( {{x^1} - 5,{y^1} - 1,{z^1} - 3} \right)$
since $pq$is $ \bot $ to the given line so 
$b \times \left( {{x^1} - 5} \right) + 1 \times \left( {{y^1} - 11 + 1 \times \left( {{z^1} - 31} \right)} \right) = 0$
${y^1} + 1 + {z^1} - 3 = 0$
${y^1} + {z^1} = 4 = 0$ 
artesian equation of the given line is 
$\frac{{x - 3}}{0} = \frac{{y - 7}}{1} = \frac{{z - 1}}{1}$ 
let $\frac{{x - 3}}{0} = \frac{{y - 7}}{1} = \frac{{z - 1}}{1} = r$
$x=3, y= r+7, z=r+1$
so general point on this line is (3,r+7,r+1)
since $\left( {{x^1},{y^1},{z^1}} \right)$
lies on this line so ${x^1} = 3\,\,{y^1} = r + 7\,\,{z^1} = r + 1$
putting values of ${y^1}& $ in
${y^1} + {z^1} = 4$
$r + 7 + r + 1 = 4$
$2r=-4$
$=r=-2$
thus ${x^1} = 3,{y^1} =  - 2 + 7 = 5$
${z^1} =  - 2 + 1 =  - 1$
so point Q is $(3,5,-1)$
therefore perpendicular distance $\sqrt {{{\left( {5 - 3} \right)}^2} + {{\left( {1 - 5} \right)}^2} + {{\left( {3 + 1} \right)}^2}} $
$\sqrt {4 + 16 + 16} $
$\sqrt {36} $
$=6units$

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 $(6, -4, 4)$ on to the line joining the points $A(2, 1, 2), B(3, -1, 4)$ is?

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

Let the point be $P(6,-4,4)$ 


Equation of the line joining $A$ and $B$ is 

$\dfrac{x-2}{3-2}=\dfrac{y-1}{-1-1}=\dfrac{z-2}{4-2}$

$\dfrac{{x - 2}}{1} = \dfrac{{y - 1}}{{ - 2}} = \dfrac{{z - 2}}{2}$

Let
$\dfrac{{x - 2}}{1} = \dfrac{{y - 1}}{{ - 2}} = \dfrac{{z - 2}}{2}=r$

$x=r+2,\,y=-2r+1,\,z=2r+2,\,z=2r+2$

General point on the given line is 
$(r+2,-2r+1,2r+2)$

Let the foot of perpendicular from $P$ to the given line be $Q(r _1+2,-2r _1+1,2r _1+2)$

Direction ratios the line $PQ$ are 
$(r _1+2-6,\,-2r _1+1+4,\,2r _1+2-4)$ which is 
$r _1-4,\,\,-2r _1+5,\,\,2r _1-2$

Since the line $PQ$ is perpendicular to the given line 

$1(r _1-2)-2(-2r _1+5)+(2r _1-2)=0$

$r _1=2$

$\begin{array}{l} { r _{ 1 } }+2=4 \  \ -2{ r _{ 1 } }+1=-3 \  \ 2{ r _{ 1 } }+2=6 \end{array}$
Therefore, 
Foot of perpendicular is $Q(4,-3,6)$

Perpendicular distance $=PQ$

$=\sqrt {(4-6)^2+(-3+4)^2+(6-4)^2}=3$

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 of $p _1, p _2, p _3$ of points $({a^2}, 2a), \, (ab, a + b), \, ({b^2}, 2b)$ respectively from straight line $x + y\tan \theta + {{tan}^2} \theta = 0$ are in :

  1. A.P

  2. G.P

  3. H.P

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

The distance of a point (x,y) from x + y*tan(theta) + tan^2(theta) = 0 is |x + y*tan(theta) + tan^2(theta)| / sqrt(1 + tan^2(theta)) = |x + y*tan(theta) + tan^2(theta)| / sec(theta). Substituting the points, one finds the distances form a geometric progression.

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

Distance of the point $P(\vec p)$ from the line $\vec r=\vec a+\lambda \vec b$ is-

  1. $\mid (\vec a-\vec p)+\dfrac {((\vec p-\vec a)\cdot \vec b)\vec b}{\mid \vec b\mid^2}\mid$
  2. $\mid (\vec b-\vec p)+\dfrac {((\vec p-\vec a)\cdot \vec b)\vec b}{\mid \vec b\mid^2}\mid$
  3. $\mid (\vec a-\vec p)+\dfrac {((\vec p-\vec b)\cdot \vec b)\vec b}{\mid \vec b\mid^2}\mid$

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

Let $Q(\vec{q})$ be the foot of perpendicular drawn from $P(\vec{p})$ to the line $\vec{r} = \vec{a} + \lambda\vec{b}$
$\Rightarrow (\vec{q}-\vec{p}) \cdot \vec{b} = 0$ and $ \vec{q} = \vec{a} + \lambda\vec{b}$
$\Rightarrow ( \vec{a} + \lambda\vec{b}-\vec{p}) \cdot \vec{b} = 0$
$\Rightarrow ( \vec{a} -\vec{p}) \cdot \vec{b} +  \lambda \mid \vec{b} \mid^2 = 0$
$\Rightarrow   \lambda  = \dfrac{( \vec{p} -\vec{a}) \cdot \vec{b} }{\mid \vec{b} \mid^2}$
$\Rightarrow   \vec{q} -\vec{p} = \vec{a} + \dfrac{( \vec{p} -\vec{a}) \cdot \vec{b} }{\mid \vec{b} \mid^2} - \vec{p}$
$\Rightarrow   \mid \vec{q} -\vec{p}\mid =\mid (\vec{a} - \vec{p})+ \dfrac{( \vec{p} -\vec{a}) \cdot \vec{b} }{\mid \vec{b} \mid^2} \mid$

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 of the point $P(3,8,2)$ from the line $\cfrac{1}{2}(x-1)=\cfrac{1}{4}(y-3)=\cfrac{1}{3}(z-2)$ measured parallel to the plane $3x+2y-2z+15=0$ is

  1. $7$
  2. $9$
  3. $\sqrt{7}$
  4. $49$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Lines equation : $\dfrac{x-1}{2}= \dfrac{y-3}{4} = \dfrac{z-2}{3}= \lambda $
General point $(B)= (2\lambda+1, 4\lambda+3, 3 \lambda+2)$
$\overrightarrow {BP} = (2 \lambda-2, 4 \lambda -5, 3 \lambda)$
As $\overrightarrow {BP}$ is parallel to the plane, it is perpendicular to its normal, $\Rightarrow (2 \lambda - 2, 4\lambda-5, 3\lambda).(3,2,-2)=0$
$\Rightarrow 6 \lambda-6+8 \lambda-10-6 \lambda=0 \Rightarrow 8\lambda=16 \Rightarrow \lambda=2$
$\Rightarrow B= (5,11,8)= |\overrightarrow {BP}|= \sqrt{4+9+36}= \sqrt{49} = 7 \Rightarrow (A)$
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 shortest distance of the points $(a, b, c)$ from the x-axis is

  1. $\sqrt{(a^2 + b^2)}$
  2. $\sqrt{(b^2 + c^2)}$
  3. $\sqrt{(c^2 + a^2)}$
  4. $\sqrt{(a^2 + b^2 + c^2)}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The point is $(a,b,c)$
the perpendicular distance from $x$ axis will be the shortest distance

$\therefore$ distance is $\sqrt{{b}^{2}+{c}^{2}}$

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

Perpendicular distance of the point $(3,4,5)$ from the $y$-axis, is

  1. $\sqrt { 34 } $
  2. $\sqrt { 41 } $
  3. $4$
  4. $5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Distance of $\left( \alpha ,\beta ,\gamma  \right) $ from $y$-axis is given by,

$d=\sqrt { { \alpha  }^{ 2 }+{ \gamma  }^{ 2 } } $
$\therefore$ distance $(d)$ of $(3,4,5)$ from $y$-axis is
$d=\sqrt { { 3 }^{ 2 }+{ 5 }^{ 2 } } =\sqrt { 9+25 } =\sqrt { 34 } $

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 $\displaystyle -\hat i + 2\hat j + 6\hat k$ to the straight line passing through the point with position vector $\displaystyle 2\hat i + 3\hat j - 4\hat k$ and parallel to the vectors $\displaystyle 6\hat i + 3\hat j - 4\hat k$ is

  1. $10$
  2. $7$
  3. $5$
  4. $3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let $A=(2+6t,3+3t,-4-4t)$ is the point on the line which is least distance from the point $B=(2,3,-4)$. 
Then the vector $BA=(3+6t)i+(2+3t)j+(-10-4t)k$ is perpendicular to $6i+3j-4k$.
$\Rightarrow (3+6t)\times 6+(2+3t)3-(10-4t)4=0$
$\Rightarrow t=-1$
$\Rightarrow A=(-4,0,0)$.
The distance from A to $(-1,2,6)$ is $\sqrt{(-3)^2+2^2+6^2}=7$.

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 perpendicular distance of point $(2, -1, 4)$ from the line $\dfrac{x + 3}{10} = \dfrac{y - 2}{-7} = \dfrac{z}{1}$ lies between 

  1. $(2, 3)$
  2. $(3, 4)$
  3. $(4, 5)$
  4. $(1, 2)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the foot of perpendicular from $P (2, -1, 4)$ to the given line be $A(10 \lambda - 3, -7 \lambda + 2, \lambda) \overrightarrow{PA} . (10 \hat{i} - 7 \hat{j} + k) = 0$
$\Rightarrow 10 (10 \lambda - 5) - 7 (-7 \lambda + 3) + 1 (\lambda - 4) = 0$
$\Rightarrow 150 \lambda = 75 \Rightarrow  \lambda = \dfrac{1}{2}$
$|\overrightarrow{PA}| = \sqrt{(10 \lambda - 5)^2 + (-7 \lambda + 3)^2 + (\lambda - 4)^2}$
$= \sqrt{0 + \left(\dfrac{1}{2}\right)^2 + \left(\dfrac{7}{2} \right)^2} = \sqrt{\dfrac{50}{4}}$
Which lies in $(3, 4)$

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 $\left ( x,\, y,\, z \right )$ from the x-axis is 

  1. $\sqrt{x^{2}\, +\, y^{2}}$
  2. $\sqrt{y^{2}\, +\, z^{2}}$
  3. $\sqrt{z^{2}\, +\, x^{2}}$
  4. $\sqrt{x^{2}\, +\, y^{2}\, +\, z^{2}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The given point is $(x,y,z)$

$\therefore$ by distance formula
Distance = $ \sqrt{(x-x)^2+(y-0)^2+(z-0)^2}=\sqrt{y^2+z^2}$

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 of the point $B$ with position vector $i +2j +3k$ from the line passing through the point $A$ with position vector $4i + 2j + 2k$ and parallel to the vector $2i + 3j + 6k$ is

  1. $\sqrt{10}$
  2. $\sqrt{5}$
  3. $\sqrt{6}$

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

Equation of the line passing through the point $A\left(4i+2j+2k\right)$ and parallel to $\left(2i+3j+6k\right)$ is
$\overrightarrow { r } = \left(4i+2j+2k\right)+t\left(2i+3j+6k\right)$
any point on the line is of the form $\left(4+2t,2+3t,2+6t\right)$
let $BC$ is perpendicular to the given line, $B\left(i+2j+3k\right)$ and $C\left(4+2t,2+3t,2+6t\right)$
applying perpendicularity condition
$(2t+3)2+(3t)3+(6t-1)6=0$
$\Rightarrow t=0$
$\therefore$ $BA$ is perpendicular to the line
Hence,required distance is $\sqrt{10}$

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

Perpendicular distance of the point $(3,4,5)$ from the $y$-axis is

  1. $\sqrt { 34 } $
  2. $\sqrt { 41 } $
  3. $4$
  4. $5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given point is $(3,4,5)$

Distance of $\left( \alpha ,\beta ,\gamma  \right) $, from $y$-axis is given by
$d=\sqrt { { \alpha  }^{ 2 }+{ \gamma  }^{ 2 } } $

$\therefore$ distance $(d)$ of $(3,4,,5)$ from $y$-axis is
$d=\sqrt { { 3 }^{ 2 }+{ 5 }^{ 2 } } =\sqrt { 9+25 } =\sqrt { 34 } $