Derivation of the Equation of a plane into vector, scalar, parametric and standard (parametric ) forms

1. EQUATION OF A PLANE
MUJUNGU HERBERT
National Teachers’ College
Kabale

2. Vector Equation of a plane
Let 𝑂𝑃 =< 𝑥 𝑜, 𝑦𝑜, 𝑧 𝑜 >= 𝒓 𝑜
and 𝑂𝑄 =< 𝑥, 𝑦, 𝑧 >= 𝒓
If 𝒏 and 𝑷𝑸 are orthonal, then
𝒏 ∙ 𝑷𝑸 = 𝟎
𝒏 ∙ 𝒓 − 𝒓 𝒐 = 𝟎
Or 𝒏 ∙ 𝒓 = 𝒏 ∙ 𝒓
Vector
Form
Vector Equation of a plane

3. Scalar Form of the Equation of a plane
From 𝒏 ∙ 𝒓 − 𝒓 𝒐 = 𝟎
a 𝒙 − 𝒙 𝒐 + b 𝒚 − 𝒚 𝒐 + c 𝒛 − 𝒛 𝒐 = 𝟎
Scalar
FormLinear
Form
Which is the standard form of the Equation
of a plane which is normal to a vector
𝑎
𝑏
𝑐
and d = −𝒂𝒙 𝒐 − 𝒃𝒚 𝒐 − 𝒄𝒛 𝒐.
𝒂𝒙 + 𝒃𝒚 + 𝒄𝒛 + 𝒅 = 𝟎
𝒂𝒙 + 𝒃𝒚 + 𝒄𝒛 − 𝒂𝒙 𝒐 − 𝒃𝒚 𝒐 − 𝒄𝒛 𝒐 = 𝟎
Scalar Form of the Equation of a plane
From 𝒏 ∙ 𝒓 − 𝒓 𝒐 = 𝟎
a 𝒙 − 𝒙 𝒐 + b 𝒚 − 𝒚 𝒐 + c 𝒛 − 𝒛 𝒐 = 𝟎

4. Parametric Form
Consider three points A, B and C with position vectors
𝒂 = 𝑎1, 𝑎2, 𝑎3 , 𝒃 = 𝑏1, 𝑏2, 𝑏3 and 𝐜 = 𝑐1, 𝑐2, 𝑐3 .
If a fourth point, P with position vector 𝐩 = 𝑥, 𝑦, 𝑧 , lies
on the same plane,
then scalars 𝜇 and 𝜆 can
be found such that,
𝑥 = 𝑎1 + 𝜇 𝑏1 − 𝑎1 + 𝜆(𝑐1 − 𝑎1)
𝑦 = 𝑎2 + 𝜇 𝑏2 − 𝑎2 + 𝜆(𝑐2 − 𝑎2)
𝑧 = 𝑎3 + 𝜇 𝑏3 − 𝑎3 + 𝜆(𝑐3 − 𝑎3)
𝐴𝑃 = 𝜇𝐴𝐵 + 𝜆𝐴𝐶
𝒑 − 𝒂 = 𝜇 𝒃 − 𝒂 + 𝜆(𝒄 − 𝒂)
𝒑 = 𝒂 + 𝜇 𝒃 − 𝒂 + 𝜆(𝒄 − 𝒂)

5. EXAMPLE 1
Find the parametric equation of the plane passing
through the points (1, 2, -3), (2, 3, 1), and (0, -2, -1).
Solution
Let points A = (1, 2, -3), B = (2, 3, 1), C = (0, -2, -1) and
P=(x,y,z) be the points on the plane
Recall Equation of the plane in the form 𝐴𝑃 = 𝜇𝐴𝐵 + 𝜆𝐴𝐶
And by using it,
We have,
𝑥 − 1
𝑦 − 2
𝑧 + 3
= 𝜇
2 − 1
3 − 2
1 + 3
+ 𝜆
0 − 1
−2 − 2
−1 + 3
𝑥 = 1 + 𝜇 − 𝜆
𝑦 = 2 + 𝜇 − 4𝜆
𝑧 = −3 + 4𝜇 + 2𝜆

6. Example 2
Find the equation of plane through point (1,-1,1) and
with normal vector i+j-k
Let A=(1,-1,1), P=(x,y,z) be another point on the plane
and normal vector n = 1,1, −1 .
𝑥 − 1
𝑦 + 1
𝑧 − 1
∙
1
1
−1
=0
Finally 𝑥 + 𝑦 − 𝑧 + 1 = 0
Solution:
Then 𝐴𝑃 ∙ 𝑛 = 0
or (𝑂𝑃 − 𝑂𝐴) ∙ 𝑛 = 0

7. Example 3
Find the equation of plane through the points (0,1,1),
(1,0,1) and(1,1,0)
First alternative solution
Let points A=(0,1,1), B=(1,0,1) , C=(1,1,0) and P=(x,y,z)
And using 𝐴𝑃 = 𝜇𝐴𝐵 + 𝜆𝐴𝐶
𝑥 − 0
𝑦 − 1
𝑧 − 1
= 𝜇
1 − 0
0 − 1
1 − 1
+ 𝜆
1 − 0
1 − 1
0 − 1
𝑥 = 𝜇 + 𝜆 … … … … . . (𝑖)
𝑦 = 1 − 𝜇 … … … … . (𝑖𝑖)
𝑧 = 1 − 𝜆 … … … … … (𝑖𝑖𝑖)
From Eqns (ii) and (iii) , 𝜇 = 1 − 𝑦 and 𝜆 = 1 − 𝑧 respectively and sub
in Eqn (iii),
That is 𝑥 = 1 − 𝑦 + 1 − 𝑧
𝑥 + 𝑦 + 𝑧 − 2 = 0Finally

8. Example 3
Find the equation of plane through the points (0,1,1),
(1,0,1) and(1,1,0)
Second alternative solution
𝒙 + 𝒚 + 𝒛 − 𝟐 = 0

9. TRIAL QUESTIONS
Find the equation of the plane passing through;
1. 𝑃𝑜 2,4, −1 having a normal vector 𝑛 = 2,3,4
2. P(1,3,2), Q(3,-1,6) and R(5,2,0).
3. (1,0,1), (0,1,1) and (1,1,0).
4. (5,1,0) and parallel to two lines x=2+t, y=-2t, z=1 and x=t,
y=2-t, z=2-3t.
5. (-1,2,1) and contains the line of intersection of two planes
x+y-z=2 and 2x-y+3z=1.
6. (-1,2,1) and is perpendicular to the line of intersection of
two planes x+y-z=2 and 2x-y+3z=1.
7. the line of intersection of the planes x-z=1 and y+2z=3, and
is perpendicular to the plane x+y-2z=1.
8.The point (-4,3,1) that is perpendicular to the vector
a=-4i+7j-2k
END

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