Wize University Linear Algebra Textbook > Equations of Lines and Planes

Planes in $\reals^3$

\reals^n

Previous Section

HyperplanesNext Section

0:00 / 0:00

Planes in R3\colorOne{\reals^3}R3
A plane is an infinite flat surface in 3D space.

Planes can be described using various pieces of information:
A point on the plane and a normal vector (orthogonal to the plane)
Two lines on the plane
Three points on the plane
PAGE BREAK

Point-Normal Form
A plane in R3\mathbb{R}^3R3 can be described in point-normal form (or vector form) using one point on the plane, and one normal vector.

Given a known point on the plane P(x1,y1,z1)P(x_1,y_1,z_1)P(x1​,y1​,z1​)  and a normal vector n⃗=(a,b,c)\vec{n}=(a,b,c)n=(a,b,c): 
n⃗⋅(x⃗−P⃗)=0\boxed{\quad
\vec{n}\cdot (\vec x - \vec P)=0 
\quad}
n⋅(x−P)=0​
  ⟹  a(x−x1)+b(y−y1)+c(z−z1)=0\quad\implies\quad a(x-x_1)+b(y-y_1)+c(z-z_1)=0⟹a(x−x1​)+b(y−y1​)+c(z−z1​)=0
Standard Form
Simplifying the point-normal form gives the standard form (or scalar form) of a plane in R3\reals^3R3, using  d=n⃗⋅P⃗d=\vec n \cdot \vec Pd=n⋅P: 
ax+by+cz=d\boxed{\quad
ax+by+cz=d
\quad}ax+by+cz=d​

Geometrically

PAGE BREAK
Parametric Form
Given a point on the plane PPP, and two non-collinear vectors in the plane d⃗1, d⃗2\vec d_1 , \ \vec d_2d1​, d2​, we may introduce two parameters, s ,t∈Rs\ ,t \in \realss ,t∈R:
X⃗=P⃗+s d⃗1+t d⃗2\boxed{\quad
\vec X = \vec P + s\ \vec d_1 + t\ \vec d_2
\quad}X=P+s d1​+t d2​​
Geometrically


PAGE BREAK
Given Two Lines on the Plane
Steps
Find the normal vector n⃗=d1⃗×d2⃗\vec n = \vec{d_1} \times \vec{d_2}n=d1​​×d2​​  where d1⃗, d2⃗\vec{d_1}, \ \vec{d_2}d1​​, d2​​ are the direction vectors of the lines.
Find any point PPP on one of the lines (this is a known point on the plane).
Use n⃗\vec nn and PPP to write the equation of the plane in point-normal form or standard form,
or use PPP  and d⃗1, d⃗2\vec d_1,\  \vec d_2d1​, d2​ to write the plane in parametric form.
Given Three Points on the Plane
Knowing three points P,Q,RP, Q, RP,Q,R  that lie in the plane allows us to find two lines on the plane, so we can proceed as above.
Steps
Find direction vectors in the plane: for example, let d1⃗=PQ→\vec{d_1} = \overrightarrow{PQ}d1​​=PQ​ and d2⃗=PR→\vec{d_2} = \overrightarrow{PR}d2​​=PR.
Find the normal vector n⃗=d1⃗×d2⃗\vec n = \vec{d_1} \times \vec{d_2}n=d1​​×d2​​  where d1⃗, d2⃗\vec{d_1}, \ \vec{d_2}d1​​, d2​​ are the direction vectors of the lines.
Use n⃗\vec nn and PPP to write the equation of the plane in point-normal form or standard form,
or use PPP  and d⃗1, d⃗2\vec d_1,\  \vec d_2d1​, d2​ to write the plane in parametric form.

0:00 / 0:00

Example: Planes in R3\colorOne{\reals^3}R3
Find the equation of the plane with the normal vector ⟨3,1,−1⟩\lang 3,1,−1 \rang⟨3,1,−1⟩ that passes through point P(0,2,1)P(0,2,1)P(0,2,1).
Point-Normal Form
n⃗⋅(x⃗−P⃗)=0\vec n \cdot (\vec x - \vec P) = 0n⋅(x−P)=0

n⃗=[31−1]\vec n=
\begin{bmatrix}
  3\\ 1\\ -1\\
\end{bmatrix}n=​31−1​​
Substituting:
[31−1]⋅([xyz]−[021])=0\boxed{
\begin{bmatrix}
  3\\ 1\\ -1\\
\end{bmatrix}
\cdot 
\left(
\begin{bmatrix}
  x\\ y\\ z\\
\end{bmatrix}
−
\begin{bmatrix}
  0\\ 2\\ 1\\
\end{bmatrix}
\right)

= 0
}​31−1​​⋅​​xyz​​−​021​​​=0​
We could simplify to obtain a scalar equation:
3(x−0)+(y−2)−(z−1)=0
3(x-0) + (y-2) - (z-1) = 03(x−0)+(y−2)−(z−1)=0

Standard form
ax+by+cz=dax+by+cz=dax+by+cz=d  where n⃗=⟨a,b,c⟩\vec n = \lang a,b,c \rangn=⟨a,b,c⟩ and d=n⃗⋅P⃗d=\vec{n}\cdot\vec{P}d=n⋅P

n⃗=[abc]=[31−1]\vec n
=
\begin{bmatrix}
  a\\ b\\ c\\
\end{bmatrix}
=
\begin{bmatrix}
  3\\ 1\\ -1\\
\end{bmatrix}n=​abc​​=​31−1​​
d=n⃗⋅P⃗=[31−1]⋅[021]=(3)(0)+(1)(2)+(−1)(1)=1\begin{aligned}
d 
&= \vec n \cdot \vec P\\[0.3em]
&= 
\begin{bmatrix}
  3\\ 1\\ -1\\
\end{bmatrix}
\cdot
\begin{bmatrix}
  0\\ 2\\ 1\\
\end{bmatrix}\\[1.7em]
&=(3)(0)+(1)(2)+(−1)(1)\\[0.3em]
&=1
\end{aligned}d​=n⋅P=​31−1​​⋅​021​​=(3)(0)+(1)(2)+(−1)(1)=1​

Now we substitute into the standard form formula using the normal vector where a=3a=3a=3, b=1b=1b=1, c=−1c=-1c=−1:
3x+y−z=1\boxed{3x+y−z=1}3x+y−z=1​

0:00 / 0:00

Example: Planes in R3\colorOne{\reals^3}R3
Find the parametric, point-normal, and standard form equations of the plane containing the points A(−6,−3,−1)A(-6,-3,-1)A(−6,−3,−1) , B(0,0,−5)B(0,0,-5)B(0,0,−5), and C(−9,−3,0)C(-9,-3,0)C(−9,−3,0).

We start by finding two direction vectors:
d1⃗=AB→=[00−5]−[−6−3−1]=[63−4]\begin{aligned}
\vec{d_1} 
&= 
\overrightarrow{AB}\\

&= 
\begin{bmatrix}
  0\\ 0\\ -5\\
\end{bmatrix}
-\begin{bmatrix}
  -6\\ -3\\ -1\\
\end{bmatrix}\\[1.5em]

&=
\begin{bmatrix}
  6\\ 3\\ -4\\
\end{bmatrix}
\end{aligned}d1​​​=AB=​00−5​​−​−6−3−1​​=​63−4​​​           d2⃗=BC→=[−9−30]−[00−5]=[−9−35]\begin{aligned}
\vec{d_2} 
&= 
\overrightarrow{BC}\\

&= 
\begin{bmatrix}
  -9\\ -3\\ 0\\
\end{bmatrix}
-\begin{bmatrix}
  0\\ 0\\ -5\\
\end{bmatrix}\\[1.5em]

&=
\begin{bmatrix}
  -9\\ -3\\ 5\\
\end{bmatrix}
\end{aligned}d2​​​=BC=​−9−30​​−​00−5​​=​−9−35​​​

We can now write the plane in parametric form with point  B(0,0,−5)\colorOne{B(0,0,-5)}B(0,0,−5) and vectors  d⃗1=⟨6,3,−4⟩,   d⃗2=⟨−9,−3,5⟩\colorThree{\vec d_1 = \lang 6,3,-4 \rang},\ \ \ 
\colorFive{\vec d_2 = \lang -9,-3,5 \rang}d1​=⟨6,3,−4⟩,   d2​=⟨−9,−3,5⟩:
⟨x,y,z⟩=⟨0,0,−5⟩+s⟨6,3,−4⟩+t⟨−9,−3,5⟩\boxed{
\lang x,y,z \rang 
= \colorOne{\lang 0,0,-5 \rang} 
+ s \colorThree{\lang 6,3,-4 \rang}
+ t \colorFive{\lang -9,-3,5 \rang}
}⟨x,y,z⟩=⟨0,0,−5⟩+s⟨6,3,−4⟩+t⟨−9,−3,5⟩​

To write the plane in point-normal or standard form, we need to find a normal vector so we find the cross product of the direction vectors:

n⃗=d1⃗×d2⃗=[63−4]×[−9−35]=[(3)(5)−(−4)(−3)(−4)(−9)−(6)(5)(6)(−3)−(3)(−9)]=[369]\begin{aligned}
\vec{n}
&= 
\vec{d_1} \times \vec{d_2}\\

&= \begin{bmatrix}
  6\\ 3\\ -4\\
\end{bmatrix}
\times
\begin{bmatrix}
  -9\\ -3\\ 5\\
\end{bmatrix}\\[1.5em]

&=
\begin{bmatrix}
  (3)(5)-(-4)(-3)\\[0.5em] 
  (-4)(-9)-(6)(5)\\[0.5em]
 (6)(-3)-(3)(-9)\\[0.5em]
\end{bmatrix}\\[1.5em]

&=
\begin{bmatrix}
  3\\ 6\\ 9\\
\end{bmatrix}


\end{aligned}n​=d1​​×d2​​=​63−4​​×​−9−35​​=​(3)(5)−(−4)(−3)(−4)(−9)−(6)(5)(6)(−3)−(3)(−9)​​=​369​​​

Using n⃗=⟨3,6,9⟩\colorFour{\vec n = \lang 3,6,9 \rang}n=⟨3,6,9⟩ and point B(0,0,−5)\colorOne{B(0,0,-5)}B(0,0,−5) we can write:

⟨3,6,9⟩⋅(⟨x,y,z⟩−⟨0,0,−5⟩)=0[point-normal form]⟨3,6,9⟩⋅⟨x−0, y−0, z−(−5)⟩=03(x−0)+6(y−0)+9(z−(−5))=03x+6y+9(z+5)=03[ x+2y+3(z+5) ]=0x+2y+3(z+5)=0x+2y+3z=−15[standard form]\begin{array}{rlr}
\colorFour{\lang 3,6,9 \rang} \cdot (\lang x,y,z \rang - \colorOne{\lang 0,0,-5 \rang}) &= 0 \quad \text{[point-normal form]}\\[0.5em]

\colorFour{\lang 3,6,9 \rang} \cdot 
\lang
x-\colorOne{0},\ 
y-\colorOne{0},\ 
z-\colorOne{(-5)} 
\rang  &= 0\\[0.5em]

\colorFour{3}(x-\colorOne{0})
+\colorFour{6}(y-\colorOne{0})
+\colorFour{9}(z-\colorOne{(-5)}) &= 0\\[0.5em]

3x + 6y + 9(z+5) &= 0\\[0.5em]
3\big[\ x + 2y + 3(z+5)\ \big] &= 0\\[0.5em]
x + 2y + 3(z+5)&= 0\\[0.5em]
x+2y+3z &= -15 \quad \text{[standard form]}\\[0.5em]

\end{array}⟨3,6,9⟩⋅(⟨x,y,z⟩−⟨0,0,−5⟩)⟨3,6,9⟩⋅⟨x−0, y−0, z−(−5)⟩3(x−0)+6(y−0)+9(z−(−5))3x+6y+9(z+5)3[ x+2y+3(z+5) ]x+2y+3(z+5)x+2y+3z​=0[point-normal form]=0=0=0=0=0=−15[standard form]​

Alternatively
We could have jumped straight to standard form using the formula:
n⃗⋅x⃗=n⃗⋅B⃗3x+6y+9z=[369]⋅[00−5]3x+6y+9z=−45(multiply by 13)x+2y+3z=−15\begin{aligned}
\vec n \cdot \vec x &= \vec n \cdot \vec B\\[0.5em]

\colorFour{3}x
+\colorFour{6}y
+\colorFour{9}z 
&=
\colorFour{
\begin{bmatrix}
  3\\ 6\\ 9\\
\end{bmatrix}}
\cdot
\colorOne{
\begin{bmatrix}
  0\\ 0\\ -5\\
\end{bmatrix}}
\\[1.5em]

3x
+6y
+9z 
&= -45 \qquad \text{\small (multiply by $\dfrac{1}{3}$)}\\[0.5em]

\colorTwo{x
+2y
+3z} 
&\colorTwo{= -15}
\end{aligned}n⋅x3x+6y+9z3x+6y+9zx+2y+3z​=n⋅B=​369​​⋅​00−5​​=−45(multiply by 31​)=−15​

Practice: Parallel Planes in R3\colorOne{\reals^3}R3
Find the values of aaa and bbb such that the following planes are parallel:
Π1:  3x+2y+z=8Π_1:\ \ 3x+2y+z=8Π1​:  3x+2y+z=8 
Π2:  ax+6y−bz=9Π_2:\ \ ax+6y−bz=9Π2​:  ax+6y−bz=9 

a=

b=

I don't know

Practice: Planes in R3\colorOne{\reals^3}R3
Find the standard form equation of the plane that contains the points A(1,−1,0),  B(1,1,6), C(2,0,3)A\left(1,-1,0\right),\ \ B\left(1,1,6\right),\ C\left(2,0,3\right)A(1,−1,0),  B(1,1,6), C(2,0,3).

Answer

I don't know

Practice: Planes in R3\colorOne{\reals^3}R3 (Tricky)
Consider the points A(−1,3,2)A(-1,3,2)A(−1,3,2), B(−3,5,0)B(-3,5,0)B(−3,5,0), and C(xC,yC,zC)C(x_C,y_C,z_C)C(xC​,yC​,zC​), and suppose we know the vector AC→=⟨3,−4,−5⟩\overrightarrow{AC}=\lang 3,-4,-5\rangAC=⟨3,−4,−5⟩.

Find the equation of the plane containing point CCC, and perpendicular to the line passing through points AAA and BBB.

3x−4y−5z=−13x-4y-5z=-13x−4y−5z=−1

−2x+2y−2z=1-2x+2y-2z=1−2x+2y−2z=1

3x−4y−5z=33x-4y-5z=33x−4y−5z=3

x+y+z=1x+y+z=1x+y+z=1

−x+y−z=0-x+y-z=0−x+y−z=0

I don't know

Extra Practice

Write down the equation of a plane defined by this parametric equation: P: (-2,0,5)+s(3,1,2)+t(-2,2,0)

Consider the two lines, L_1:(1,0,0)+s(1,1,1) and L_2:(-1,2,3)+t(4,0,c).
a) For what value of c two lines intersects.
b) Find the equation of the plane containing both lines.

Write down equation of Line L: (1,3,5)+t(-4,1,3) as the intersection of two planes.

Find a point whose distance to the plane 3x+4y−10z=303x+4y-10z=303x+4y−10z=30 is equal to 10.

Consider two planes with equations 3x+2y−z=8 and ax−6y+bz=8.3x+2y-z=8\ and\ ax-6y+bz=8.3x+2y−z=8 and ax−6y+bz=8.
a) For what values of a and b are two planes parallel?
b) For what values of a and b is the angle between two planes 60 degrees?

If two planes 2x+ay−2z=2y+62x+ay-2z=2y+62x+ay−2z=2y+6 and x=y+z+3x=y+z+3x=y+z+3 are parallel, at which point does the first plane intersect with x-axis.

Consider two lines: L1:(1,0,0)+s(1,1,1) and L2:(-1,2,3)+t(4,0,c).
a) For what value of c two lines intersects.
b) Find the equation of the plane containing both lines.

Find gen form plane thru pt parallel to 2 vecs

Given that the plane PPP passes through the point (1,3,−1)\left(1,3,-1\right)(1,3,−1) and is parallel to both vectors
𝑢⃗=(0,1,1)𝑢⃗ = (0,1,1)u⃗=(0,1,1) and 𝑣⃗=(3,2,1),\vec𝑣 = (3,2,1),v=(3,2,1),find the standard form equation of the plane.
(type your answer in the form ax+by+cz=d, with no spaces)

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 59.1

 Given the points P1=(0,0,1)\bcb{P_1 = (0,0,1)}P1​=(0,0,1) and  P2=(1,−2,3)\bcb{P_2 = (1,-2,3)}P2​=(1,−2,3) and the vector v⃗=<2,−3,1>\bcb{\vec{v} = \left< 2, -3, 1 \right>}v=⟨2,−3,1⟩
Find the equation of the plane Π\bcb{\Pi}Π perpendicular to v⃗\bcb{\vec{v}}v, containing the point P1\bcb{P_1}P1​. 

Planes in $\reals^3$

Practice: Normal Vector
Which of the following vectors is normal to the plane 3x+z=73x+z=73x+z=7?

A plane is parallel to the vector v=⟨1,1,0⟩v = \langle1, 1, 0\ranglev=⟨1,1,0⟩. If it passes through points (1,2,−1)(1,2,-1)(1,2,−1) and (2,1,0)(2,1,0)(2,1,0) then find the equation of the plane.

Which one of the following is a vector that is orthogonal to the line that passes through the points P(1,−1,0)P\left(1,-1,0\right)P(1,−1,0) and Q(5,0,−3)Q\left(5,0,-3\right)Q(5,0,−3)?

Which of the following is a vector orthogonal/ perpendicular to the plane 2x−z=32x-z=32x−z=3?

133 - FML 3 - 18.1W e.g. 30

Find the equation of a plane passing through the point (2,0,1)\bcb{(2, 0, 1)}(2,0,1), perpendicular to the straight line passing through the points (1,1,0)\bcb{(1, 1, 0)}(1,1,0) and (4,−1,−2)\bcb{(4, -1, -2)}(4,−1,−2).

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 60.7

 Given the planes
π1 : 4x−3y+z=2\bcb{\pi_1 \, : \, 4x - 3y + z = 2}π1​:4x−3y+z=2 and π2 : x+2y+2z=0\bcb{\pi_2 \, : \, x + 2y + 2z = 0}π2​:x+2y+2z=0, and the lines
L1 : {x=3t+2y=−2t+1z=t, L2 : {x=−2t+1y=t+1z=t+3, L3 : {x=t+3y=−tz=2t+1\bcb{L_1 \, : \, \left\{ \begin{matrix} x & = & 3t + 2 \\ y & = & -2t + 1 \\ z & = & t \end{matrix} \right., ~ 
L_2 \, : \, \left\{ \begin{matrix} x & = & -2t + 1 \\ y & = & t + 1 \\ z & = & t + 3 \end{matrix} \right., ~ 
L_3 \, : \, \left\{ \begin{matrix} x & = & t + 3 \\ y & = & -t \\ z & = & 2t +1 \end{matrix} \right.}L1​:⎩⎨⎧​xyz​===​3t+2−2t+1t​, L2​:⎩⎨⎧​xyz​===​−2t+1t+1t+3​, L3​:⎩⎨⎧​xyz​===​t+3−t2t+1​,

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 60.4

 Given the planes
π1 : 4x−3y+z=2\bcb{\pi_1 \, : \, 4x - 3y + z = 2}π1​:4x−3y+z=2 and π2 : x+2y+2z=0\bcb{\pi_2 \, : \, x + 2y + 2z = 0}π2​:x+2y+2z=0, and the lines
L1 : {x=3t+2y=−2t+1z=t, L2 : {x=−2t+1y=t+1z=t+3, L3 : {x=t+3y=−tz=2t+1\bcb{L_1 \, : \, \left\{ \begin{matrix} x & = & 3t + 2 \\ y & = & -2t + 1 \\ z & = & t \end{matrix} \right., ~ 
L_2 \, : \, \left\{ \begin{matrix} x & = & -2t + 1 \\ y & = & t + 1 \\ z & = & t + 3 \end{matrix} \right., ~ 
L_3 \, : \, \left\{ \begin{matrix} x & = & t + 3 \\ y & = & -t \\ z & = & 2t +1 \end{matrix} \right.}L1​:⎩⎨⎧​xyz​===​3t+2−2t+1t​, L2​:⎩⎨⎧​xyz​===​−2t+1t+1t+3​, L3​:⎩⎨⎧​xyz​===​t+3−t2t+1​,

The following linear equations represent planes in R3R^3R3. Describe the intersection of these planes (if any).
2x+y−z=3x−y−z=33x+z=6\begin{array}{c}
2x+y-z=3\\
x-y-z=3\\
3x+z=6
\end{array}2x+y−z=3x−y−z=33x+z=6​

Planes in $\reals^3$

Practice: Planes in R3\colorOne{\reals^3}R3 (Tricky)
Consider the points A(−1,3,2)A(-1,3,2)A(−1,3,2), B(−3,5,0)B(-3,5,0)B(−3,5,0), and C(xC,yC,zC)C(x_C,y_C,z_C)C(xC​,yC​,zC​), and suppose we know the vector AC→=⟨3,−4,−5⟩\overrightarrow{AC}=\lang 3,-4,-5\rangAC=⟨3,−4,−5⟩.
Find the equation of the plane containing point CCC, and perpendicular to the line passing through points AAA and BBB.

133 - FML 3 - 18.1W e.g. 31

Find a point on, and a vector normal to, the plane 2x+y+3z=6\bcb{2x + y + 3z = 6}2x+y+3z=6.

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 57

Find the equation of a plane that contains the three points P1=(1,−2,3)\bcb{P_1 = (1,-2, 3)}P1​=(1,−2,3), P2=(0,−1,4)\bcb{P_2 = (0, -1, 4)}P2​=(0,−1,4) and P3=(4,3,−2)\bcb{P_3 = (4,3,-2)}P3​=(4,3,−2). 

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 50

 Find the equation of a plane perpendicular to w⃗=<1,−2,1>\bcb{\vec{w} = \left< 1, -2, 1\right>}w=⟨1,−2,1⟩ that contains the point P=(−1,1,2)\bcb{P = (-1, 1, 2)}P=(−1,1,2). 

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 59.3

 Given the points P1=(0,0,1)\bcb{P_1 = (0,0,1)}P1​=(0,0,1) and  P2=(1,−2,3)\bcb{P_2 = (1,-2,3)}P2​=(1,−2,3) and the vector v⃗=<2,−3,1>\bcb{\vec{v} = \left< 2, -3, 1 \right>}v=⟨2,−3,1⟩, find the point Po\bcb{P_o}Po​ on the plane Π: 2x−3y+z=1\bcb{\Pi: \ 2x - 3y + z = 1}Π: 2x−3y+z=1 that is closest to the point P2\bcb{P_2}P2​.

133 - FML 3 - 18.1W e.g. 31

Find a point on, and a vector normal to, the plane 2x+y+3z=6\bcb{2x + y + 3z = 6}2x+y+3z=6.

Planes in R^3

Which one of the following is the standard form equation of the plane in R3\reals^3R3 which contains the lines (x,y,z)=(2,0,−1)+t(2,−5,−1)(x,y,z)=(2,0,-1)+t(2,-5,-1)(x,y,z)=(2,0,−1)+t(2,−5,−1) and (x,y,z)=(3,1,2)+t(3,−5,1)(x,y,z)=(3,1,2)+t(3,-5,1)(x,y,z)=(3,1,2)+t(3,−5,1)

Wize University Linear Algebra Textbook > Equations of Lines and Planes

Planes in $\reals^3$

Popular Courses

Planes in $\colorOne{\reals^3}$

Point-Normal Form

Standard Form

Parametric Form

Given Two Lines on the Plane

Given Three Points on the Plane

Example: Planes in $\colorOne{\reals^3}$

Point-Normal Form

Standard form

Example: Planes in $\colorOne{\reals^3}$

Practice: Parallel Planes in $\colorOne{\reals^3}$

Practice: Planes in $\colorOne{\reals^3}$

Practice: Planes in $\colorOne{\reals^3}$ (Tricky)

Extra Practice

Find gen form plane thru pt parallel to 2 vecs

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 59.1

Planes in $\reals^3$

133 - FML 3 - 18.1W e.g. 30

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 60.7

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 60.4

Planes in $\reals^3$

133 - FML 3 - 18.1W e.g. 31

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 57

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 50

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 59.3

133 - FML 3 - 18.1W e.g. 31

Planes in R^3

Wize University Linear Algebra Textbook > Equations of Lines and Planes

Planes in R3\reals^3R3

Popular Courses

Planes in R3\colorOne{\reals^3}R3

Point-Normal Form

Standard Form

Parametric Form

Given Two Lines on the Plane

Given Three Points on the Plane

Example: Planes in R3\colorOne{\reals^3}R3

Point-Normal Form

Standard form

Example: Planes in R3\colorOne{\reals^3}R3

Practice: Parallel Planes in R3\colorOne{\reals^3}R3

Practice: Planes in R3\colorOne{\reals^3}R3

Practice: Planes in R3\colorOne{\reals^3}R3 (Tricky)

Extra Practice

Find gen form plane thru pt parallel to 2 vecs

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 59.1

Planes in $\reals^3$

133 - FML 3 - 18.1W e.g. 30

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 60.7

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 60.4

Planes in $\reals^3$

133 - FML 3 - 18.1W e.g. 31

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 57

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 50

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 59.3

133 - FML 3 - 18.1W e.g. 31

Planes in R^3

Planes in $\reals^3$

Planes in $\colorOne{\reals^3}$

Example: Planes in $\colorOne{\reals^3}$

Example: Planes in $\colorOne{\reals^3}$

Practice: Parallel Planes in $\colorOne{\reals^3}$

Practice: Planes in $\colorOne{\reals^3}$

Practice: Planes in $\colorOne{\reals^3}$ (Tricky)