A plane is defined by the equations x-4y+2z=16. Find two vectors parallel to the plane.
a) 2x+3y-4z+5=0 and b) -3x+2y+5z+6=0
All people come very close to being able to float in water. What therefore is the volume (in cubic meters) of a 50-kg woman?
Select one:
a. 0.007
b. 0.035
c. 0.050
d. 0.070
e. 0.085
To find two vectors parallel to the plane defined by the equation x - 4y + 2z = 16, we can rewrite the equation in vector form.
The equation can be written as:
(1) x - 4y + 2z - 16 = 0
We can rewrite equation (1) in vector form as:
(2) (x, y, z) • (1, -4, 2) = 16
where • denotes the dot product.
From equation (2), we can see that the vector (1, -4, 2) is a normal vector to the plane.
Now, to find two parallel vectors to the plane, we can use any two linearly independent vectors that are perpendicular to the normal vector (1, -4, 2).
One way to find two perpendicular vectors is to choose two variables and solve for the remaining variable such that the dot product of the resulting vector and the normal vector is zero.
Let's solve for x and y, and choose z = 0:
(3) x - 4y + 2(0) = 0
Solving equation (3) for x, we get:
(4) x = 4y
Now, let's solve for y and x, and choose z = 0:
(5) (1, -4, 2) • (x, y, 0) = 0
Expanding equation (5), we get:
(6) x - 4y + 0 = 0
Solving equation (6) for x, we get:
(7) x = 4y
From equations (4) and (7), we can see that both x and y can be any value as long as x = 4y.
Therefore, two vectors parallel to the plane are:
Vector 1: (1, 0, 0)
Vector 2: (0, 1, 0)
These vectors are parallel to the plane defined by the equation x - 4y + 2z = 16.