Tag: distance from a point to line

Questions Related to distance from a point to line

Perpendicular distance of the point $(3,4,5)$ from the $y$-axis, 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 { 34 } $

  2. $\sqrt { 41 } $

  3. $4$

  4. $5$

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Correct Option: A
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 } $

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

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. $10$

  2. $7$

  3. $5$

  4. $3$

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Correct Option: B
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 $\sqrt{(-3)^2+2^2+6^2}=7$.

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

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. $(2, 3)$

  2. $(3, 4)$

  3. $(4, 5)$

  4. $(1, 2)$

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Correct Option: B
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 $(3, 4)$

The perpendicular distance of the point $\left ( x,\, y,\, z \right )$ from the x-axis 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. $\sqrt{x^{2}\, +\, y^{2}}$

  2. $\sqrt{y^{2}\, +\, z^{2}}$

  3. $\sqrt{z^{2}\, +\, x^{2}}$

  4. $\sqrt{x^{2}\, +\, y^{2}\, +\, z^{2}}$

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Correct Option: B
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}$

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

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. $\sqrt{10}$

  2. $\sqrt{5}$

  3. $\sqrt{6}$

  4. none of these

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Correct Option: A
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 $\sqrt{10}$

Perpendicular distance of the point $(3,4,5)$ from the $y$-axis 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 { 34 } $

  2. $\sqrt { 41 } $

  3. $4$

  4. $5$

Show answer
Correct Option: A
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 } $

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

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. $\displaystyle \left ( \dfrac {6}{7} \right )\sqrt{14}$

  2. $\displaystyle \left ( \dfrac {6}{7} \right )\sqrt{18}$

  3. $\displaystyle \left ( \dfrac {4}{7} \right )\sqrt{14}$

  4. $\displaystyle \left ( \dfrac {4}{7} \right )\sqrt{18}$

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Correct Option: A

The length of the perpendicular drawn from $(1,2,3)$ to the line $\displaystyle \frac { x-6 }{ 3 } =\frac { y-7 }{ 2 } =\frac { z-7 }{ -2 } $ 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. $4$

  2. $5$

  3. $6$

  4. $7$

Show answer
Correct Option: D
Explanation:

Direction cosines of the given line are  

$\displaystyle \frac { 3 }{ \sqrt { 17 }  } ,\frac { 2 }{ \sqrt { 17 }  } ,\frac { -2 }{ \sqrt { 17 }  } $
$\displaystyle \therefore AM=\left| \left( 6.1 \right) .\frac { 3 }{ \sqrt { 17 }  } +\left( 7-2 \right) .\frac { 2 }{ \sqrt { 17 }  } +\left( 7-3 \right) .\frac { -2 }{ \sqrt { 17 }  }  \right| =17$
$AP=\sqrt { { \left( 16-1 \right)  }^{ 2 }+{ \left( 7-2 \right)  }^{ 2 }+{ \left( 7-3 \right)  }^{ 2 } } $
$=\sqrt { 25+25+16 } =\sqrt { 66 } $
$\therefore$ length of the perpendicular is
$PM=\sqrt { { AP }^{ 2 }-{ AM }^{ 2 } } $
$=\sqrt { 66-17 } =\sqrt { 49 } =7$

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

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. $3$

  2. $4$

  3. $5$

  4. $6$

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Correct Option: D
Explanation:

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

The perpendicular distance of a corner of unit cube from a diagonal not passing through it 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. $\sqrt{\dfrac{2}{3}}$

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

  3. $\dfrac{1}{3}$

  4. $

    1$

Show answer
Correct Option: A
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 $=\dfrac{|(\overline{c}+\overline{b})\times(\overline{c}+\overline{b})+\overline{d}|}{|\overline{c}+\overline{b}+\overline{d}|}$
$=\dfrac{|(\overline{c}+\overline{b})\times\overline{d}|}{\sqrt{3}}$
$=\dfrac{|\overline{c}+\overline{b}||\overline{d}|\sin \theta}{\sqrt{3}}$
$=\dfrac{\sqrt{2}}{3}$