Find coordinates for two points that belong to the plane 2x+3y+5z=15. Show that the vector [2,3,5] is perpendicular to the segment that joins your two points. Explain why [2,3,5] is perpendicular to the plane.

how about the two points

(0,0,3) and (3,3,0) ?

a line containing those two point has direction
[3,3,-3] or [1,1,-1] if reduced

let's take the dot product of [1,1,-1] and [2,3,5]
= 2 + 3 - 5
= 0
so if the dot product is 0, then the vectors are perpendicular.

By now you should have learned that for the plane
ax + by + cz = d
[a,b,c] is a normal to the plane.

what does (a,b,c) is normal to the plane mean? specifically, what do u mean by "normal"?

Sure, let me find the coordinates for two points on the plane first. To do that, we can set up two equations to solve for the x, y, and z coordinates.

Let's choose an arbitrary value for z, say z = 0:

2x + 3y + 5(0) = 15

Simplifying:

2x + 3y = 15

We can choose x = 5, which gives us:

2(5) + 3y = 15

10 + 3y = 15

3y = 5

y = 5/3

So, the first point is (5, 5/3, 0).

Now let's choose another arbitrary value for z, say z = 1:

2x + 3y + 5(1) = 15

Simplifying:

2x + 3y + 5 = 15

2x + 3y = 10

We can choose x = 0, which gives us:

0 + 3y = 10

3y = 10

y = 10/3

So, the second point is (0, 10/3, 1).

To show that the vector [2, 3, 5] is perpendicular to the segment joining these two points, we can calculate the vector from the first point to the second point:

Vector AB = (0 - 5, 10/3 - 5/3, 1 - 0)
= (-5, 5, 1)

Now, to check if [2, 3, 5] is perpendicular to Vector AB, we can calculate their dot product:

[2, 3, 5] · (-5, 5, 1) = (2 * -5) + (3 * 5) + (5 * 1) = -10 + 15 + 5 = 0

Since the dot product is zero, we can conclude that [2, 3, 5] is indeed perpendicular to the segment joining these two points.

Now, let's explain why [2, 3, 5] is perpendicular to the plane 2x + 3y + 5z = 15. The equation of the plane can be rewritten as:

2x + 3y + 5z - 15 = 0

If we take the coefficients of x, y, and z in this equation, we get the normal vector to the plane, which is [2, 3, 5]. Since the dot product of the normal vector and any vector lying in the plane is always zero, we can say that [2, 3, 5] is perpendicular to the plane.

To find coordinates for two points that belong to the plane 2x + 3y + 5z = 15, we can arbitrarily assign values to two variables and solve for the third variable.

Let's assume x = 0 and y = 0 to find the coordinates of the first point:
2(0) + 3(0) + 5z = 15
5z = 15
z = 3

Therefore, the first point is (0, 0, 3).

Now, let's assume x = 1 and y = 0 to find the coordinates of the second point:
2(1) + 3(0) + 5z = 15
2 + 5z = 15
5z = 13
z = 13/5

Therefore, the second point is (1, 0, 13/5).

To show that the vector [2, 3, 5] is perpendicular to the segment joining these two points, we can compute the dot product of the segment vector (the vector joining the two points) and [2, 3, 5]. If the dot product is zero, it indicates that the two vectors are perpendicular.

The segment vector can be found by subtracting the coordinates of the first point from the second point:
Segment vector = [(1 - 0), (0 - 0), ((13/5) - 3)]
= [1, 0, (13/5 - 3)]
= [1, 0, (-2/5)]

Now, calculate the dot product of the segment vector and [2, 3, 5]:
[1 * 2] + [0 * 3] + [(-2/5) * 5]
= 2 + 0 - 2
= 0

Since the dot product of the segment vector and [2, 3, 5] is zero, this means that the two vectors are perpendicular.

Additionally, [2, 3, 5] is perpendicular to the plane 2x + 3y + 5z = 15 because it is the normal vector of the plane. The normal vector of a plane is perpendicular to every vector that lies entirely in the plane. Since [2, 3, 5] lies entirely in the plane 2x + 3y + 5z = 15, it is perpendicular to the plane.

To find coordinates for two points that belong to the plane 2x + 3y + 5z = 15, we can first set one variable as a free variable and solve for the remaining two variables.

Let's solve for x with y and z as free variables:

2x + 3y + 5z = 15

Let's assume y = 0 and z = 0:

2x + 3(0) + 5(0) = 15
2x = 15
x = 15/2

Here, one point on the plane can be (15/2, 0, 0).

Now, let's assume x = 0 and z = 0:

2(0) + 3y + 5(0) = 15
3y = 15
y = 5

Another point on the plane can be (0, 5, 0).

Therefore, two points that belong to the plane 2x + 3y + 5z = 15 are (15/2, 0, 0) and (0, 5, 0).

To show that the vector [2, 3, 5] is perpendicular to the segment that joins these two points, we need to calculate the vector formed by these two points and take the dot product with the given vector.

Vector joining the two points:
(0, 5, 0) - (15/2, 0, 0) = (-15/2, 5, 0)

Now, calculate the dot product of the vector (-15/2, 5, 0) and [2, 3, 5]:

(-15/2 * 2) + (5 * 3) + (0 * 5) = -15 + 15 + 0 = 0

The dot product is equal to 0, which means the vector [2, 3, 5] is perpendicular to the segment that joins the two points.

To explain why [2, 3, 5] is perpendicular to the plane 2x + 3y + 5z = 15, we can observe that the coefficients of x, y, and z in the plane equation (2, 3, 5) are the same as the components of the given vector [2, 3, 5].

The equation 2x + 3y + 5z = 15 can be rewritten as (2, 3, 5) · (x, y, z) = 15, where · represents the dot product.

For a vector to be perpendicular to a plane, the dot product of that vector with any vector on the plane should be zero. Here, the given vector [2, 3, 5] forms a dot product of zero when multiplied with any vector (x, y, z) that satisfies the plane equation.

Therefore, [2, 3, 5] is perpendicular to the plane 2x + 3y + 5z = 15.