Then
¹ = 3 − 1 − ¸ = 3 − 1 −
5 − 3 + 1
= 23 − 5 − 2.
Put into (2):
4 = 1 + 2
5 − 3 + 1 − 2
23 − 5 − 2
,
or
4 = −63 + 45 + 7,
or:
63 + 4 − 45 = 7.

9/2/2015
6
HOW CAN WE GET THE PERPENDICULAR VECTOR OF A
PLANE IF WE ONLY HAVE 2 VECTORS ON THE PLANE??
•
³
= (1, 2, 2)
•
´
= (1. −2, 1)
•
Vectors
³
and
´
is ON the plane.
•
How can we find vector
0
perpendicular to the plane?
•
SOLUTION:
find the cross product of a and b:
³ 3 ´:
1
2
2
1
2
1
− 2
1
1
− 2
= (
6, 1, −4)
TEST IF (A X B) IS PERPENDICULAR
•
Perpendicular vectors gives
°. 0 = 0
Test
:
°. 0 =
1, 2, 2 .
6, 1, −4
= 6 + 2 − 8 = 0
YES
it is true for
°
Test
:
±. 0 =
1, −2, 1 .
6, 1, −4
= 6 − 2 − 4 = 0
YES
it is true for
±
SO:
° 3 ±
is perpendicular to the plane!!

9/2/2015
7
EXAMPLE
•
A point
L = (4, 5, −7)
lies on a plane.
Find the scalar
equation of the plane that is perpendicular to the directional
vector
0 = (1, 2, −3)
² − ²
:
. 0 = 0
SOLUTION:
² =
3, 4, 5
°08
²
:
=
4,5, −7
°08
0 = (1,2, −3)
Then:
(3, 4, 5
− (4,5, −7)). (1,2, −3) = 0
1
3 − 4 + 2
4 − 5 − 3
5 + 7
= 0
3 − 4 + 24 − 10 − 35 − 21 = 0
3 + 24 − 35 = 35
NOTE:
The values in front of x, y and z is the same as the
vector n.
ANGLE BETWEEN PLANES
.
Can we now find the angle between the planes
3 + 4 + 5 = 2
and
23 − 5 = 0
as well?

9/2/2015
8
ANGLE
BETWEEN
PLANES
The angle between the planes is the
same as the angle between the
perpendicular vectors.
The planes are
3 + 4 + 5 = 2
and
23 − 5 = 0
.
Vectors perpendicular to each are
3
=
1
1
1
and
4
=
2
0
−1
.
With
3 =
3,
4 =
5,
3
. 4
=
1
2 +
1
0 +
1
−1
= 1
the angle
]
between them is given by
cos ^ =
3
. 4
34
=
1
3
5
= 0,2582,
^ = 75,04°.

9/2/2015
9
NOW YOU:
Find the vector equation, the parametric
equations and the scalar equation of the
plane which contains vectors a and b and
a point P
0
°
=
1,2,2 ;
±
=
1, −2,1 ;
L
:
= (1,1,0)

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- Summer '20
- Vectors, Vector Space, Force, Euclidean vector