Thx for help tonight yall 3 more questions X_x

[PsychonautQQ]

Thx for help tonight yall :D 3 more questions X_x

Homework Statement

(1 pt)

First of all I'm a bit confused.. it wants the equation for a plane in all three questions.. the linear equation of a plane?

(A) If the positive z-axis points upward, an equation for a horizontal plane through the point (-5,4,-4) is

So it's a horizontal plane which means that the z direction number will be zero..
a(x-xi) + b(y-yi) + c(z-zi) = 0
ax - a5 + by - b4 = -
so i'll make a equal 4 and b = -5?
4x - 20 -5y +20 = 0
4x - 5y = 0

program says this is wrong..

.

(B) An equation for the plane perpendicular to the x-axis and passing through the point (-5,4,-4 is:

attempt at solution:
a plane that hits the x-axis at a 90 degree angel would mean that it always has the same value for x if I'm picturing this correctly? so the direction number for x would be zero i believe.

a(x-xi) + b(y-yi) + c(z-zi) = 0
b(y-(-5)) + c(z-(-4)) = 0
by + 5b + cz + 4z = 0
so i could make b = 4 and z = -5?
4y + 20 - 5z - 20 = 0
4y - 5z = 0...

program says I'm doing it wrong X_x
.

(C) An equation for the plane parallel to the xz-plane and passing through the point (-5,4,-4) is:

attempt at solution:
If it's parallel to the xz plane that means the direction on it's y component must be zero..
a(x-xi) + b(y-yi) + c(z-zi) = 0
a(x-(-5)) + b(y-4) + c(z-(-4)) = 0
ax + 5a + by - 4b + cz + 4c = 0
well b has to be zero so it's parallel to the xz plane...

so xa + 5a -cz + 4c = 0
c could be -5 and a could be 4... ?
4x + 20 +5z -20 = 0
4x + 5z = 0

something went wrong here...?

 

[Mark44]

It's better to post one problem per thread.

Homework Statement

(1 pt)

First of all I'm a bit confused.. it wants the equation for a plane in all three questions.. the linear equation of a plane?

(A) If the positive z-axis points upward, an equation for a horizontal plane through the point (-5,4,-4) is

So it's a horizontal plane which means that the z direction number will be zero..

Any vector along the z-axis will be normal to the plane. One that comes to mind is <0, 0, 1>.

a(x-xi) + b(y-yi) + c(z-zi) = 0
ax - a5 + by - b4 = -
so i'll make a equal 4 and b = -5?
4x - 20 -5y +20 = 0
4x - 5y = 0

program says this is wrong..

Correct - this plane isn't horizontal.

 

[Mark44]

Homework Statement

(B) An equation for the plane perpendicular to the x-axis and passing through the point (-5,4,-4 is:

attempt at solution:
a plane that hits the x-axis at a 90 degree angel would mean that it always has the same value for x if I'm picturing this correctly?

Yes. And what would that value be? This problem is much simpler than you're making it.

so the direction number for x would be zero i believe.

a(x-xi) + b(y-yi) + c(z-zi) = 0
b(y-(-5)) + c(z-(-4)) = 0
by + 5b + cz + 4z = 0
so i could make b = 4 and z = -5?
4y + 20 - 5z - 20 = 0
4y - 5z = 0...

program says I'm doing it wrong X_x

I don't think the concept of a normal to a plane has gelled in your mind just yet. If the plane is perpendicular to the x-axis, give me one normal to this plane.

Do you understand what the variables in this equation represent?
A(x - x₀) + B(y - y₀) + C(z - z₀) = 0

In case you don't, n = <A, B, C> is a normal to the plane, and P₀(x₀, y₀, z₀) is a particular point on the plane.

The equation comes from the fact that if P(x, y, z) is an arbitrary point on the plane, then the vector PP₀ = <x - x₀, y - y₀, z - z₀> is perpendicular to n. Therefore, n $\cdot$ PP₀ = 0. The equation above comes directly from this dot product.