Perpendicular distance of a point from a plane - class-XII

According to the question:

A point on the line is given by (2+t, 3+t, 4+t). For option A, 2+t=2014 => t=2012. Then 3+2012=2015 and 4+2012=2016. This matches (2014, 2015, 2016).

The perpendicular distance of a plane and origin is give as 

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

Let the point be $P(6,-4,4)$ 

The line is x/2 = (y-2)/3 = (z-3)/4 = k. Point P = (2k, 3k+2, 4k+3). Vector from P to (3, -1, 11) is (3-2k, -3-3k, 8-4k). This must be perpendicular to the line direction (2, 3, 4). 2(3-2k) + 3(-3-3k) + 4(8-4k) = 0 => 6-4k-9-9k+32-16k = 0 => 29-29k=0 => k=1. Point P = (2, 5, 7). Distance = sqrt((3-2)^2 + (-1-5)^2 + (11-7)^2) = sqrt(1 + 36 + 16) = sqrt(53).

The distance of a point (x,y) from x + ytan(theta) + tan^2(theta) = 0 is |x + ytan(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.

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)

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

Line: x=k, y=2k+1, z=3k+2. Vector from (1,6,3) to (k, 2k+1, 3k+2) is (k-1, 2k-5, 3k-1). Dot product with (1, 2, 3) = 0 => k-1 + 4k-10 + 9k-3 = 0 => 14k = 14 => k=1. Point is (1, 3, 5). Distance = sqrt((1-1)^2 + (6-3)^2 + (3-5)^2) = sqrt(0 + 9 + 4) = sqrt(13).

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

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,
${=}$ $\sqrt{{(-2/9)}^{2}+{(5/9)}^{2}+{(-4/9)}^{2}}$
${=}$ $\sqrt{4/81+25/81+16/81}$
${=}$ $\dfrac{\sqrt{5}}{3}$

The distance of the intersection line of two planes from the origin involves the coefficients of the planes. The given expressions p and q relate to the distance formula for the line of intersection.

This is a duplicate of 430686. The calculation yields sqrt(53).

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

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

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

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 $\sqrt{10}$

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

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 - 53) + k(1-1 - 33) = i(2) - j(-16) + k(-10) = (2, 16, -10). Magnitude = sqrt(4 + 256 + 100) = sqrt(360) = 6sqrt(10). |BC| = sqrt(1+9+25) = sqrt(35). Distance = 6sqrt(10)/sqrt(35) = 6sqrt(2)/sqrt(7) = 6*sqrt(14)/7.

Direction cosines of the given line are  

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

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 $=\dfrac{|(\overline{c}+\overline{b})\times(\overline{c}+\overline{b})+\overline{d}|}{|\overline{c}+\overline{b}+\overline{d}|}$
$=\dfrac{|(\overline{c}+\overline{b})\times\overline{d}|}{\sqrt{3}}$
$=\dfrac{|\overline{c}+\overline{b}||\overline{d}|\sin \theta}{\sqrt{3}}$
$=\dfrac{\sqrt{2}}{3}$

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.

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$
          

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

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

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

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

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