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) ยท (x, y, z) = 15, where ยท 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.