determine whether the planes are parallel, octagonal or neither
a. 4x + 5y + z = -1
3x + 5y - 37z = -3
b. 2x - y - 3z = -5
-2x - 6y - z = 3
I think you mean orthogonal not octagonal
a.
vector normal (perpendicular to) first plane 4 i + 5 j +1 k
second plane 3 i + 5 j -37 z
find the dot or cross product
dot is easier A dot B = |A||B| cos angle between
if cos is 0, they are othogonal because cos 90 degrees is 0
if cos is 1 or -1 they are parallel
in this case
dot product = 12 + 25 - 37
= ZERO
angle is 90 degrees, They are orthoganal
now you do the next one.
To determine whether the planes are parallel, orthogonal (perpendicular), or neither, we need to compare their normal vectors.
The normal vector of a plane is the coefficients of the variables (x, y, and z) in the equation of the plane. In other words, the normal vector is the vector (a, b, c) in the equation ax + by + cz + d = 0.
Let's start with the planes given:
a. Plane 1: 4x + 5y + z = -1
Normal vector of Plane 1: (4, 5, 1)
Plane 2: 3x + 5y - 37z = -3
Normal vector of Plane 2: (3, 5, -37)
To compare the normal vectors, we need to check whether they are scalar multiples of each other. Two vectors are scalar multiples if one vector can be multiplied by a scalar to obtain the other vector.
Let's compare the normal vectors:
(4, 5, 1)/(3, 5, -37) = (4/3, 5/5, 1/-37) = (4/3, 1, -1/37)
Since the ratio of the corresponding components is not a constant value, the planes are neither parallel nor orthogonal.
b. Plane 1: 2x - y - 3z = -5
Normal vector of Plane 1: (2, -1, -3)
Plane 2: -2x - 6y - z = 3
Normal vector of Plane 2: (-2, -6, -1)
Let's compare the normal vectors:
(2, -1, -3)/(-2, -6, -1) = (2/-2, -1/-6, -3/-1) = (-1, 1/6, 3)
Since the ratio of the corresponding components is not a constant value, the planes are neither parallel nor orthogonal.
In conclusion, for the given planes, neither of them are parallel nor octagonal.