3D Geometry - Lines and Planes

The equation of a plane at distance d from origin with normal vector n is r dot (n/|n|) = d. Here n = i + 2j - 2k, so |n| = sqrt(1+4+4) = 3. The equation is x + 2y - 2z = 10 * 3 = 30.

The dr's of the normal to the plane are $(2,-2,1)$.
The dc's will be $\left ( \dfrac{2}{3} , \dfrac{-2}{3} , \dfrac{1}{3} \right)$
Hence, the equation of the plane in the normal form will be,
$ \dfrac{2x}{3} - \dfrac{2y}{3} + \dfrac{z}{3} $ = $ \dfrac{5}{3} $

$2x-z+1=0$
$\therefore$ $2x+0y-1z=-1$
$\therefore$ $(x,y,z).(2,0,-1)=-1$
$\therefore$ $F.(2,0,-1)+1=0$

$L:\cfrac { x-1 }{ 1 } =\cfrac { y-1 }{ -2 } =\cfrac { z-2 }{ 3 } \ P(1,0,-1),Q(3,2,2)$

Vector normal to plane $P _1$ is $\overrightarrow{n _1}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\3&4&2\4&2&3\end{vmatrix}$

Clearly, the normal to the plane has DR's $\equiv(1, 2, -3)$

Given, $\vec{r} = (1+\lambda-\mu)\hat{i}+(2-\lambda)\hat{j}+(3-2\lambda+2\mu)\hat{k}$
$\Rightarrow x\hat{i}+y\hat{j}+z\hat{k} = (1+\lambda-\mu)\hat{i}+(2-\lambda)\hat{j}+(3-2\lambda+2\mu)\hat{k}$
Comparing coefficient, we get
$ 1+\lambda-\mu = x, 2-\lambda=y, 3-2\lambda+2\mu=z$
$\Rightarrow\lambda = 2-y, \mu=1+\lambda - x = 3-y-x$
Eliminating $\mu$ and $\lambda$, we get
$2x+z=5$ which is required equation of plane in cartesian form.

Any point on the first line is $P\left( 3\lambda +1,\lambda +2,2\lambda +3 \right) $
and on the second line is $Q\left( u+3,2u+1,3u+2 \right) $
$P$ and $Q$ represent the same point if $\lambda =u=1$
And the point of intersection of the given line is $P\left( 4,3,5 \right) $
The plane given in (a),(b),(c) and (d) all pass through $P$.
The plane at greatest distance is one which is at a distance equalt to $OP$ from the origin.
So, the distance of the plane from origin is $\sqrt { { 4 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 } } =\sqrt { 50 } $
The equation of plane is $4x+3y+5z=50$

$\vec{AB}=i+2j+4k$
$\vec{BC}=3i+3j+3k$

The plane passes through (3, 2, 0) and contains the line with direction (1, 5, 4). The normal vector is the cross product of the vector from (3, 2, 0) to a point on the line (3, 6, 4) and the line's direction vector.

Equation of plane passes through $(2,2,1)$ is given by,
$a(x-2)+b(y-2)+c(z-1) = 0.......(A)$
Given it also passes through $(9,3,6)$
$\Rightarrow 7a+b+5c=0 ......(1)$
and this plane is perpendicular to plane $2x+6y+6z-1=0$
$\Rightarrow 2a+6a+6c=0 .....(2)$
Solving (1) and (2), we get $ a= \dfrac{-3a}{5}, b = \dfrac{-4a}{5}$
Putting these values in (A) our required plane is,
$3x+4y-5z=9$

We have $\overrightarrow { r } =\left( 1+\lambda -\mu  \right) i+\left( 2-\lambda  \right) j+\left( 3-2\lambda +2\mu  \right) k$

In the normal form lx + my + nz = p, l, m, and n are the direction cosines of the normal to the plane, which satisfy the condition l^2 + m^2 + n^2 = 1.

Let $ax+by+cz=1$ be the desired plane.
Since, it is perpendicular to $x+2y+3z=7$ & $2x-3y+4z=0$
Therefore, $a+2b+3c=0$        .... (1)
and $2a-3b+4c=0$       ...(2)
and it passes through $(1,1,1)$
Therefore, $a+b+c=1$      ...(3)
Solving $(1),(2)$ and $(3)$ simultaneously, we get

Since $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$ is the proof any vector $(x,y,z)$on the plane. The given equation can be written as.

$x\hat{i}+y\hat{j}++z\hat{k}=(s—2t)\hat{i}+(3-t)\hat{j}+(2s+t)\hat{k}\\x=(s-2t)\quad y=(3-t)\quad z=2s+t$

Similarly We get $x-2y=s-6$ and $y+z=3+2s$

Now eliminating $s$ we get $2(x-2y)-(y+z)=-15$

$2x-5y-z+15=0$ is the required form of the equation.

Since, the direction ratios of the normal are $ \alpha , \beta , \gamma $
The equation of the plane will be of the form, $ \alpha x + \beta y + \gamma z  = d$
And the plane passes through the point $(2,1,4)$.
Hence, equation of plane is $ \alpha x + \beta y + \gamma z $ = $ 2 \alpha + \beta  + 4 \gamma  $.

Since the plane is perpendicular to the zx-plane (y=0), its normal vector must be parallel to the y-axis (0, 1, 0). Thus, the equation is of the form by + d = 0. Given the points, we solve for the coefficients.

Equation of plane passing through origin is given by,
$ax+by+cz=0$
Also this line containing line whose d.cs are $(1,-2,2)$ and $(2,3,-1)$
$\Rightarrow a-2b+2c=0$ and $2a+3b-c=0$
Solving these, $ b= \dfrac{5c}{7}, a = -\dfrac{4c}{7}$
Hence, plane equation is
$4x-5y-7z=0$

The vector equation of plane passing through the intersection of planes $\vec r.\vec {n _1}=d _1$ and $\vec r.\vec {n _2}$ and also through the point $x _1,y _1,z _1$ is 

$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$

The planes equally inclined to both the lines will have their normals along the angle bisectors of the two lines.  
The unit vector along the first line is $ \dfrac{2 \hat{i} -2 \hat{j} -\hat{k} } { 3 }$.
The unit vector along the second line is  $ \dfrac{ 8\hat{i}+ \hat{j} -4\hat{k} }{ 9} $.

The vectors along the angle bisectors can be written as : 
$ (\dfrac{2}{3} +\dfrac{8}{9}) \hat{i}  + ( \dfrac{-2}{3} + \dfrac{1}{9} )\hat{j} + (\dfrac{-1}{3} + \dfrac{-4}{9} )\hat{k}  = \dfrac{1}{9} (14\hat{i} -5\hat{j} -7\hat{k}) =0 $ 
and 
$ (\dfrac{2}{3} -\dfrac{8}{9}) \hat{i}  + ( \dfrac{-2}{3} - \dfrac{1}{9} )\hat{j} + (\dfrac{-1}{3} - \dfrac{-4}{9} )\hat{k} = \dfrac{1}{9} (-2\hat{i} -7\hat{j} +\hat{k} ) $.

Hence, options A and B represent equations of planes which have normals along the angle bisector and pass through the origin. 

The plane passing through the points $A(2,2,1),B(3,4,2) and C(7,0,6)$.