Normal to a Plane

The normal to the plane is obtained by the cross product: So, the plane , described by parameters and , has a normal: Similarly, for plane , described by parameters and the normal: Since, and are not parallel, the planes will intersect in a line. For a point in a plane with normal , equation of the plane is given by: Therefore, for plane with a point and normal , the equation of plane is: Similarly, for plane with a point

Q.
Answer:
a.
Planes Π1 and Π2 are in the form:⃗
x=⃗ r o+s⃗ d+t⃗ e
where, ⃗ ro is a position vector to a point o in the plane and ⃗ d are ⃗ e are line
vectors originating from o, in the plane. t and s are independent parameters.
The normal to the plane is obtained by the cross product: ⃗ d ×⃗ e
So, the plane Π1, described by parameters λ1 and λ2, has a normal:⃗
n1= ( 2
−2
1 ) × ( 3
0
3 )=
| i j k
2 −2 1
3 0 3|= ⟨ −6 −3 6 ⟩
Similarly, for plane Π2, described by parameters μ1 and μ2 the normal:⃗
n2 =
( 2
1
−2 ) × ( 3
3
0 )=
| i j k
2 1 −2
3 3 0 |= ⟨ 6 −6 3 ⟩
Since, ⃗ n1 and ⃗ n2 are not parallel, the planes will intersect in a line.
b.
For a point o= ( x0 , y0 , z0 ) in a plane with normal⃗ n= ⟨ a ,b ,c ⟩, equation of the plane
is given by:
a ( x1−x0 ) +b ( x2− y0 )+ c ( x3−z0 ) =0

Therefore, for plane Π1 with a point o= ( 1 ,−3 , 2 ) and normal⃗ n1= ⟨−6 ,−3 , 6 ⟩ , the
equation of plane is:
−6 ( x1−1 ) −3 ( x2+3 )+ 6 ( x3−2 )=0→−6 x1−3 x2 +6 x3=15 ( 1 )
Similarly, for plane Π2 with a point o= ( 1,0 ,−1 ) and normal⃗ n2 = ⟨ 6 ,−6 , 3 ⟩ , the
equation of plane is:
6 ( x1−1 ) −6 ( x2−0 ) +3 ( x3 +1 ) =0 →6 x1−6 x2+3 x3=3 ( 2 )
c.
Re-write (1) for x1:
x1=−x2
2 + x3−5
2
Re-write (2) for x1:
x1=x2− x3
2 +1
2
Since the planes intersect in a line, on the line of intersection we have:
x2− x3
2 +1
2 =−x2
2 + x3− 5
2
Multiplying throughout by 2:
2 x2 – x3 +1=−x2 +2 x3 – 5 → 3 x2=3 x3 – 6 → x2=x3 – 2
Re-substituting $x_2$ in $x_1$:
x1=−x3 −2
2 +x3− 5
2 → x1= x3
2 − 3
2
Let x3 be equal to the parameter ω. Therefore, the parametric form of line of
intersection is given by:
x1= ω
2 − 3
2 , x2=ω – 2 , x3=ω
In vector form the line is represented as:⃗
x= ( −3
2 ,−2, 0 ) +ω ⟨ 1
2 , 1 ,1 ⟩
d.
Plane Π1 in Cartesian form is:

−6 x1−3 x2+ 6 x3=15
For plane Π2:
x1=1+2 μ1+3 μ2 , x2 =μ1+ 3 μ2 , x3=−1−2 μ1
At the line of intersection, we can write:
−6 ( 1+2 μ1 +3 μ2 ) −3 ( μ1 +3 μ2 ) +6 ( −1−2 μ1 ) =15 →−27 μ1−27 μ2=27
Therefore, μ1=−1−μ2. Implies,
x1=−1+ μ2 , x2=−1+ 2 μ2 , x3 =1+ 2 μ2
By letting 1+2 μ2=ω , we get
x1= ω
2 − 3
2 , x2=ω−2 , x3 =ω
In vector form:⃗
x= ( −3
2 ,−2, 0 ) +ω ⟨ 1
2 , 1 ,1 ⟩
This is same as the result obtained in c.
e.
Even though the result in c and d is exactly same, there can be instances where
two parametric equations can represent the same curve.
For example, if instead we choose (refer d ¿ t=ω−2 as a parameter, then we can
write
x1= t
2 −5
2 , x2=t , x3=t+2
The vector form is:⃗
x= (−5
2 , 0 , 2 )+t ⟨ 1
2 , 1 ,1 ⟩
This is a different equation representing the same line.
f.⃗
m=⃗ n1 ×⃗ n2= | i j k
−6 −3 6
6 −6 3| = ⟨ 27 , 54 , 54 ⟩
The line of intersection:

⃗x= ( −3
2 ,−2, 0 ) +ω ⟨ 1
2 , 1 ,1 ⟩
has directional vector ⟨ 1
2 ,1 , 1 ⟩.⃗
m can be expressed as: ⃗ m=54 ⟨ 1
2 , 1 ,1 ⟩, which implies that it is parallel to the line
of intersection.
g.
Result from f can be interpreted in the following way:
Two planes are parallel if their normal are parallel to each other.
Two planes intersect in a line if their normal are not parallel. The line of
intersection is parallel to a direction, which is perpendicular to both the normal,
this direction is given by the cross product of the normal.