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)
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
1 answer
1 answer