Thx for help tonight yall 3 more questions X_x

  • #1
784
11
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...?
 

Answers and Replies

  • #2
35,019
6,770
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.
 
  • #3
35,019
6,770

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.
 

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