Tag: perpendicular distance of a point from a plane

Questions Related to perpendicular distance of a point from a plane

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

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Correct Option: D
Explanation:

Let us take a point on line $(3\lambda +6,2\lambda +7,-2\lambda +7)$.
Direction ratio's of line which is  perpendicular to given line 
$(3\lambda +6-1,2\lambda +7-2,-2\lambda +7-3)$
$(3\lambda +5,2\lambda +5,-2\lambda +4)$
and the direction ratio's of given line are 3,2,-2
These two lines are perpendicular, so$(3\lambda +5)*3+(2\lambda +5)*2+(-2\lambda +4)+(-2)=0$
$\lambda=-1$
So, point is $(2,4,6)$
So distance between $(3,5,9)$ and $(1,2,3)$ is $7$. (by distance formula)

Distance of the point $P(\vec p)$ from the line $\vec r=\vec a+\lambda \vec b$ 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. $\mid (\vec a-\vec p)+\dfrac {((\vec p-\vec a)\cdot \vec b)\vec b}{\mid \vec b\mid^2}\mid$

  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.

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Correct Option: A
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}$
$\Rightarrow ( \vec{a} + \lambda\vec{b}-\vec{p}) \cdot \vec{b} = 0$
$\Rightarrow ( \vec{a} -\vec{p}) \cdot \vec{b} +  \lambda \mid \vec{b} \mid^2 = 0$
$\Rightarrow   \lambda  = \dfrac{( \vec{p} -\vec{a}) \cdot \vec{b} }{\mid \vec{b} \mid^2}$
$\Rightarrow   \vec{q} -\vec{p} = \vec{a} + \dfrac{( \vec{p} -\vec{a}) \cdot \vec{b} }{\mid \vec{b} \mid^2} - \vec{p}$
$\Rightarrow   \mid \vec{q} -\vec{p}\mid =\mid (\vec{a} - \vec{p})+ \dfrac{( \vec{p} -\vec{a}) \cdot \vec{b} }{\mid \vec{b} \mid^2} \mid$

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

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. $7$

  2. $9$

  3. $\sqrt{7}$

  4. $49$

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Correct Option: A
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, $\Rightarrow (2 \lambda - 2, 4\lambda-5, 3\lambda).(3,2,-2)=0$
$\Rightarrow 6 \lambda-6+8 \lambda-10-6 \lambda=0 \Rightarrow 8\lambda=16 \Rightarrow \lambda=2$
$\Rightarrow B= (5,11,8)= |\overrightarrow {BP}|= \sqrt{4+9+36}= \sqrt{49} = 7 \Rightarrow (A)$

The length of the perpendicular from (1,6,3) to the line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ 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. $\sqrt{11}$

  3. $\sqrt{13}$

  4. 5

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Correct Option: C

The shortest distance of the points $(a, b, c)$ 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. $\sqrt{(a^2 + b^2)}$

  2. $\sqrt{(b^2 + c^2)}$

  3. $\sqrt{(c^2 + a^2)}$

  4. $\sqrt{(a^2 + b^2 + c^2)}$

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Correct Option: B
Explanation:

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

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

A line is perpendicular to the plane $x+2y+2z=0$ and passes through $(0, 1, 0)$. The perpendicular distance of this line from the origin 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. $\displaystyle \frac {\sqrt 5}{3}$

  2. $\displaystyle \frac {\sqrt 7}{3}$

  3. $\displaystyle \frac {2}{3}$

  4. $3$

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Correct Option: A
Explanation:

Since the line is perpendicular to the plane $x+2y+2z=0$, so directon of line will be $(1,2,2)$ and it passes through $(0,1,0)$,
Therefore equation of line is $\dfrac{x}{1}=\dfrac{y-1}{2}=\dfrac{z}{2}$
So general point on the line is $(r,2r+1,2r)$
Now direction ratio of line passing through this point and origin which perpendicular to given line is $(r,2r+1,2r)$
Since they are perpendicular, so dot product will be $0$ .
$r+2\times (2r+1)+2\times 2r{=}0$
$r+4r+2+4r{=}0$
$\therefore r{=}$$\dfrac{-2}{9}$
So points on line which perpendicular from origin is $(-2/9,5/9,-4/9)$
Now applying distance formula,
${=}$ $\sqrt{{(-2/9)}^{2}+{(5/9)}^{2}+{(-4/9)}^{2}}$
${=}$ $\sqrt{4/81+25/81+16/81}$
${=}$ $\dfrac{\sqrt{5}}{3}$

If $(a-a\prime )^2+(b-b\prime )^2+(c-c\prime )^2=p$ and $(ab\prime -a\prime b)^2+(bc\prime -b\prime c)^2+(ca\prime -c\prime a)^2=q,$ then the perpendicular distance of the line $ax+by+cz=1,$ $a\prime x+b\prime y+c\prime z=1$ from origin, 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 { p }{ q } } $

  2. $\sqrt { \dfrac { q }{ p } } $

  3. $\dfrac { p }{ \sqrt { q } } $

  4. $\dfrac { q }{ \sqrt { p } } $

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Correct Option: A

The length of the perpendicular drawn from the point $(3,\ -1,\ 11)$ to the line $\dfrac {x}{2}=\dfrac {y-2}{3}=\dfrac {z-3}{4}$

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. $\sqrt {53}$

  2. $\sqrt {66}$

  3. $\sqrt {29}$

  4. $\sqrt {33}$

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Correct Option: A

If $\vec{AB}=\vec{b}$ and $\vec{AC}=\vec{c}$, then the length of perpendicular from $A$ to the line $BC$ 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. $\displaystyle \dfrac{\left | \vec{b}\times \vec{c} \right |}{\left | \vec{b}+\vec{c} \right |}$

  2. $\displaystyle \dfrac{\left | \vec{b}\times \vec{c} \right |}{\left | \vec{b}-\vec{c} \right |}$

  3. $\displaystyle \dfrac{1}{2}\frac{\left | \vec{b}\times \vec{c} \right |}{\left | \vec{b}-\vec{c} \right |}$

  4. None of these

Show answer
Correct Option: B
Explanation:

Let length of perpendicular be $h$,
Then area of triangle $ABC$ is, 

$=\dfrac{1}{2}\times h\times BC $
$= \dfrac{1}{2}.|\vec{AB}\times \vec{AC}|$
$\Rightarrow h = \cfrac{|\vec{b}\times \vec{c}|}{|\vec{b}-\vec{c}|}$

Perpendiculars AP, AQ and AR are drawn to the $x-,y-$ and $z-$axes, respectively, from the point $A\left ( 1,-1,2 \right )$. The A.M. of $AP^2,$ $AQ^2$ and $AR^2$ 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. $3$

  4. $2$

Show answer
Correct Option: A
Explanation:
$A=(1,-1,2), P=(1,0,0), ,Q=(0,-1,0),R=(0,0,2)$
$AP^2 = [(-1)^2+2^2] = 5$,
$ AQ^2  = [1^2+2^2] = 5$,
$ AR^2 = [(-1)^2+1^2] = 2$
Hence required A.M is $=\cfrac{5+5+2}{3}=4$