In this problem you will prove the formula for the distance between two parallel

Question

In this problem you will prove the formula for the distance between two parallel planes through an explicit calculation. Consider two parallel planes 2x + 3y + z = 1 and 2x + 3y + z = 2. Choose a point that lies on any one of the planes. Call this point A Find the equation of a straight line that passes through this point. A. and is perpendicular to the plane on which you chose the point. Find the intersection point of the line you constructed in the last part with the other plane. Call this point B Compute the length AB. This is the distance between parallel planes. Discuss why the choice of the plane and the point A will not change the result of your calculation.

Explanation / Answer

a)I will find a point on the plane 2x+3y+z = 1
The obvious choice is A = (0, 0, 1)
2(0)+3(0)+1 = 1 shows that this point is on the line.
The vector in the direction (2, 3, 1) is perpendicular to this point.
Then, a line perpendicular to the plane at (0, 0, 1) is
(0, 0, 1) + (2, 3, 1)t or (2t, 3t, 1t+1)

b) Let's find where this line (2t, 3t, 1t+1) meets the other plane 2x+3y+z=2
2(2t)+3(3t)+1t+1 = 2
14t +1 = 2
t = 1/14
The point on the other line is
(2(1/14),3(1/14),1/14+1) = (1/7, 3/14, 1 1/14) = B
Substituting into
2x+3y+z = 2(1/7)+3(3/14)+1 1/14 = 2/7 + 9/14 + 1 1/14 = 2
Yes, the point is on the plane.

c) B - A = (1/7, 3/14, 1 1/14) - (0, 0, 1) = (1/7, 3/14, 1/14)
The distance is ((1/7)2 + (3/14)2 + (1/14)2 )= 1/14 0.267261241912424

d) Let's say we move to a point C on the same plane as A.

Then, it is easy to see that B + C - A will be the point on the other plane (after all, C-A creates no change in the sum for the plane equation, since C and A generate the same sum) that is on the line orthogonal to the plane at C and thus, its distance will be the same.

Then, we clearly see that point A is on the line orthogonal to the other plane at B (coefficients of the other plane are the same, and thus it is also normal to the line) and is therefore the closest point on the first plane to the second plane at B. Then, we can move similarly to any other point on the other plane and find the point on the first plane that is closest and it will be the same distance from the selected point on the second plane.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.