verify that the two planes are parallel, and find the distance between the planes. (round your answer to three decimal places)
2x-5z=5
2x-5z=6
To determine if two planes are parallel, we need to compare their normal vectors.
Let's rewrite the equations in the form Ax + By + Cz = D, where A, B, and C are the coefficients of x, y, and z, respectively.
Equation 1: 2x - 5z = 5 => 2x + 0y - 5z = 5 => 2x + 0y - 5z - 5 = 0
Equation 2: 2x - 5z = 6 => 2x + 0y - 5z = 6 => 2x + 0y - 5z - 6 = 0
Comparing the coefficients, we can see that the normal vectors of the two planes are [2, 0, -5] and [2, 0, -5], respectively. Since the normal vectors are the same, the two planes are parallel.
Now, let's find the distance between the parallel planes. We can use the formula d = |D2 - D1| / √(A^2 + B^2 + C^2), where (A, B, C) are the coefficients of the normal vectors of the planes, and (D1, D2) are the constants on the right side of the plane equations.
In this case, the normal vector coefficients are (2, 0, -5), and the constants are (-5) and (-6) for the first and second planes, respectively.
Substituting these values into the formula:
d = |(-6) - (-5)| / √(2^2 + 0^2 + (-5)^2)
= 1 / √(4 + 0 + 25)
= 1 / √29
≈ 0.183 (rounded to three decimal places)
Therefore, the distance between the two planes is approximately 0.183 units.
To determine if the two planes are parallel, we need to compare their normal vectors.
Step 1: Write the equations of the planes in the general form Ax + By + Cz + D = 0.
The first plane equation, 2x - 5z = 5, can be rewritten as 2x + 0y - 5z - 5 = 0.
The second plane equation, 2x - 5z = 6, can be rewritten as 2x + 0y - 5z - 6 = 0.
Step 2: Identify the coefficients of x, y, and z in the general form.
For both planes, the coefficients of x and z are 2 and -5, respectively. The coefficient of y is 0 in both cases.
Step 3: Compare the normal vectors.
Since the coefficients of x, y, and z for both planes are the same, their normal vectors are the same as well.
Step 4: Determine if the normal vectors are parallel.
Since the normal vectors are the same, they are parallel.
Step 5: Find the distance between the planes.
To find the distance between the planes, we can use the formula d = |D2 - D1| / √(A^2 + B^2 + C^2), where (A, B, C) is the normal vector and (D1, D2) are the constant terms in the plane equations.
From the first plane equation, D1 = -5, and from the second plane equation, D2 = -6. The normal vector is (2, 0, -5).
Plugging in these values into the formula, we get:
d = |-6 - (-5)| / √(2^2 + 0^2 + (-5)^2)
= 1 / √29
Rounding the result to three decimal places, the distance between the planes is approximately 0.183.
To check whether the two planes are parallel, we need to examine their normal vectors. The normal vector of the plane 2x-5z=5 is <2, 0, -5>, whereas the normal vector of the plane 2x-5z=6 is also <2, 0, -5>. Since the normal vectors are the same, the two planes are parallel.
To find the distance between the planes, we can use the formula:
distance = |d|/sqrt(a^2 + b^2 + c^2)
where a, b, and c are the coefficients of x, y, and z in the equation of the normal vector, and d is the difference in the constant terms.
In this case, a=2, b=0, c=-5, and d=6-5=1
So, the distance between the planes is:
distance = |1|/sqrt(2^2 + 0^2 + (-5)^2) = 0.111 (rounded to three decimal places)