Tag: distance from a point to line

Questions Related to distance from a point to 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

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

  1. $7\sqrt{2}$

  2. $14$

  3. $7$

  4. $49$

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Correct Option: C
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.

The distance of the point $A(-2,3,1)$ from the line $BC$ passing through $B(-3,5,2)$ which makes equal angles with the axes is

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

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

  2. $\sqrt{\dfrac{14}{3}}$

  3. $\displaystyle \dfrac{16}{\sqrt{3}}$

  4. $\displaystyle \dfrac{5}{\sqrt{3}}$

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Correct Option: B
Explanation:

Since $ \alpha = \beta = \gamma $
$\Rightarrow l = m = n = \dfrac{1}{\sqrt{3}}$, where $l,m,n$ are direction cosines of line $PQ$
Let $M$ be a point on the line $PQ$ such that $AM \perp PQ$
So, $PM$  $ =$   Projection of $AP$ on $ PQ$
          

 $ = \mid (-2 + 3)\dfrac{1}{\sqrt{3}} + (3 - 5)\dfrac{1}{\sqrt{3}} +(1 - 2)\dfrac{1}{\sqrt{3}} \mid = \dfrac{2}{\sqrt{3}}$

and $ AP= \sqrt{(-2+3)^2 + (3-5)^2 + (1-2)^2} = \sqrt{6}$
Hence required distance is,
$AM = \sqrt{PQ^2 - QM^2} = \sqrt{\dfrac{14}{3}} $

State the following statement is True or False

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

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

  1. True

  2. False

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Correct Option: B
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.

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 

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

  1. $\frac{3}{7}\sqrt {26} $

  2. $\frac{6}{7}\sqrt {26} $

  3. $\frac{4}{7}\sqrt {26} $

  4. $\frac{2}{7}\sqrt {26} $

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Correct Option: A
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

Find the length of perpendicular from $ P(2, -3, 1)$ to the line $\displaystyle \frac{x- 1}{2} = \frac{y - 3}{3} = \frac{z + 2}{-1}$

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

  1. $5$

  2. $\displaystyle \sqrt{\dfrac{531}{14}}$

  3. $\sqrt{50}$

  4. $\sqrt{\dfrac{221}{3}}$

Show answer
Correct Option: B
Explanation:

Let take a point on line $(2\lambda +1,3\lambda +3,-\lambda -2)$


Direction ratio's of line which is  perpendicular to given line, 

$(2\lambda +1-2,3\lambda +3+3,-\lambda -2-1)$

$(2\lambda -1,3\lambda +6,-\lambda -3)$

And the direction ratio's of given line are $2,3,-1$.

These two lines are perpendicular, so

$(2\lambda -1)\cdot 2+(3\lambda +6)\cdot 3+(-\lambda -3)+(-1)=0$


$\lambda=\dfrac{-19}{14}$

So point is $\left (\dfrac{-24}{14},\dfrac{-15}{14},\dfrac{-9}{14}\right)$

So distance between $\left (\dfrac{-24}{14},\dfrac{-15}{14},\dfrac{-9}{14}\right) $ and $(1,3,-2)$ is $\sqrt{\dfrac{531}{14}}$...................(by distance formula) 

A line is drawn from $P(x _1 , y _1)$ in the direction $\theta$ with the X - axis, to meet $ax + by + c = 0$ at $Q$. Then length $PQ$ is equal to :

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

  1. $\dfrac{|ax _1 + by _1 + c|}{\sqrt{(a^2 + b^2)}}$

  2. $\left|\dfrac{ax _1 + by _1 + c}{a \, cos \theta + b \, sin \theta} \right|$

  3. $\dfrac{bx _1 + ay _1 + c}{a cos \theta + b sin \theta}$

  4. $ - \dfrac{ax _1 + by _1 + c}{a sin \theta + b cos \theta}$

Show answer
Correct Option: B
Explanation:
Equation of a line drawn through a point $P\left( { x } _{ 1 }{ y } _{ 1 } \right) $ at an angle $\theta $ with the $x-axis$
$\dfrac { x-{ x } _{ 1 } }{ \cos\theta  } =\dfrac { y-{ y } _{ 1 } }{ \sin\theta  } \quad \longrightarrow \left( 1 \right) $
$Q$ is a point the above line and also lies on the line $ax+by+cz=0$
Say $\left| PQ \right| =r$ and the coordinates of $Q$ are $\left( h,k \right) $
Then,
$h={ x } _{ 1 }+r\cos\theta $
$k={ y } _{ 1 }+r\sin\theta $
$\because$   $Q$ lies on $ax+by+cz=0$
$\therefore$   $ah+bk+c=0$
$\Rightarrow a\left( { x } _{ 1 }+r\cos\theta  \right) +b\left( { y } _{ 1 }+r\sin\theta  \right) +c=0$
$\Rightarrow { ax } _{ 1 }+{ by } _{ 1 }+c+r\left( a\cos\theta +b\sin\theta  \right) =0$
$\Rightarrow \quad r=\dfrac { -\left( { ax } _{ 1 }+{ by } _{ 1 }+c \right)  }{ a\cos\theta +b\sin\theta  } $
$\because$   $'r'$ is the magnitude of length of line segment $PQ$, it cannot be negative.
So, $r=\left| \dfrac { { ax } _{ 1 }+{ by } _{ 1 }+c }{ a\cos\theta +b\sin\theta  }  \right| $

The $\perp $ distance of a corner of a unit cube on a diagonal not passing through is 

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

  1. $\displaystyle \frac{\sqrt{6}}{3}$

  2. $3\sqrt{3}$

  3. $2\sqrt{3}$

  4. None of these

Show answer
Correct Option: A
Explanation:
Let the edges Of unit cube are OA along z axis,OB along x axis and OC along y axis.
Let CM be the perpendicular from corner C on diagonal OP where point P is (1,1,1)
OP = $ \vec i+ \vec j+ \vec k$

Unit vector in the direction of OP is  = $\dfrac {\vec i+ \vec j+ \vec k}{\sqrt 3}$
OM= projection of OC on OP
OM=  OC . (unit  vector of OP)

OM=$\vec j \dfrac {((\vec i+\vec j+\vec k)}{\sqrt3}$=$1 \sqrt 3$

Now $CM^2=OC^2−OM^2$

$CM^2$=$1-\dfrac {1} {3}$ = $\dfrac {2} {3}$

So CM = $\dfrac {\sqrt 2}{\sqrt 3}$ = $\dfrac {\sqrt 6}{ 3}$


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

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

  1. $\displaystyle \frac{\sqrt{149}}{25}$

  2. $\displaystyle \frac{149}{25}$

  3. $\displaystyle \frac{2\sqrt{149}}{25}$

  4. $\displaystyle \frac{2\sqrt{149}}{5}$

Show answer
Correct Option: D
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 $D$

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

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

  1. $\sqrt{13}$

  2. $13$

  3. $2\sqrt{13}$

  4. None of these

Show answer
Correct Option: A
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

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

If $\vec {a},\vec {b},\vec {c}$ are position vectors of the non-collinear points $A, B, C$ respectively, then the shortest distance of $A$ from $BC$ is

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

  1. $\vec {a}.(\vec {b}-\vec {c})$

  2. $|\displaystyle \vec {b}-\vec {a}|-\left(\dfrac{(\vec {a}-\vec {b}).(\vec {c}-\vec {b})}{|\vec {c}-\vec {b}|}\right)^{2}$

  3. $|\vec {b}-\vec {a}|$

  4. $\displaystyle \sqrt {(|\vec {b}-\vec {a}|)^2-\left(\dfrac{(\vec {b}-\vec {a}).(\vec {c}-\vec {b})}{|\vec {c}-\vec {b}|}\right)^{2}}$

Show answer
Correct Option: D
Explanation:

We know that shortest distance is $\bot$ distance
$=\dfrac{|(\overline{b}-\overline{a})\times(\overline{c}-\overline{b})|}{|(\overline{c}-\overline{b})|}$


$=\dfrac{1}{|\overline{c}-\overline{b}|}\sqrt{|\overline{b}-\overline{a}|^2|\overline{c}-\overline{b}|^2-((\overline{b}-\overline{a})\cdot(\overline{c}-\overline{b}))^2}$

$=\sqrt{|\overline{b}-\overline{a}|^2-\left ( \dfrac{(\overline{b}-\overline{a})\cdot(\overline{c}-\overline{b})}{|\overline{c}-\overline{b}|} \right )^2}$