A plane is defined by the equations x-4y+2z=16. Find two vectors parallel to the plane.

You need 3 points.

Probably the easiest ones would be the intercepts
(0,0,8)
(0,-4,0) and
(16,0,0)
using the first two, a direction vector would be (0,4,8)
using the last two, a direction vector would be (16,4,0)
we could shorten these to (0,1,2) and (4,1,0)

check:
a normal to our plane is (1,-4,2)
taking the dot product of that normal with each of my two vectors yields zero, as expected.

If i want to find the x-, y- and z- intercepts do i write a new vector equation because we previously found the intercepts but the x-intercept =16 y-intercept = -4 and the z-intercept = 8

i tried writing a vector equation using the one vector as my point and the scalar equation as the direction vector but something seems to be wrong as i do not get the above values!

To find two vectors parallel to the plane defined by the equation x - 4y + 2z = 16, we can rewrite the equation in vector form.

Given the equation x - 4y + 2z = 16, we can rearrange it as follows:

x = 4y - 2z + 16

Now in vector form, this equation becomes:

[ x ] [ 4y - 2z + 16 ]
[ y ] = [ y ]
[ z ] [ z ]

Let's denote the vector form as R = [4y - 2z + 16, y, z].

To find two vectors parallel to the plane, we can choose values for y and z and calculate the resulting x, y, and z values for each choice.

First, let's choose y = 1 and z = 0:

R₁ = [4(1) - 2(0) + 16, 1, 0]
= [20, 1, 0]

Now, let's choose y = 0 and z = 1:

R₂ = [4(0) - 2(1) + 16, 0, 1]
= [14, 0, 1]

So, two vectors parallel to the plane are R₁ = [20, 1, 0] and R₂ = [14, 0, 1].

To find two vectors parallel to a plane, we need to find two linearly independent vectors that satisfy the equation of the plane.

The equation of the plane is x - 4y + 2z = 16. We can rewrite this equation in vector form:

P = (16, 0, 0) + a(1, -4, 0) + b(0, 2, 1)

Here, (1, -4, 0) and (0, 2, 1) are two vectors parallel to the plane, and a and b are scalar constants.

To find the two vectors parallel to the plane, we can choose values for a and b and then substitute them into the equation above. Let's choose a = 1 and b = 0:

P1 = (16, 0, 0) + 1(1, -4, 0) + 0(0, 2, 1)
= (16, 0, 0) + (1, -4, 0)
= (17, -4, 0)

So, vector P1 = (17, -4, 0) is parallel to the plane.

Now, let's choose a = 0 and b = 1:

P2 = (16, 0, 0) + 0(1, -4, 0) + 1(0, 2, 1)
= (16, 0, 0) + (0, 2, 1)
= (16, 2, 1)

Therefore, vector P2 = (16, 2, 1) is also parallel to the plane.

In summary, the two vectors parallel to the plane defined by the equation x - 4y + 2z = 16 are P1 = (17, -4, 0) and P2 = (16, 2, 1).