Distance between planes help

Question

Find the perpendicular distance between 3x - y + 9z = 15 and 6x - 2y + 18z - 3 = 0. Give your answer to 3 significant figures:

Plane 1: r . (3i - j + 9k) = 15

Plane 2: r . (6i - 2j + 18k) = 3

The direction vector of Plane 2 = 2 x the direction vector of plane 1

Therefore, the planes are parallel.

Plane 1 contains the point A, (5,0,0)

The line that passes through (5,0,0) the plane 2 is

r = 5i + s(6i - 2j + 18k)

= (5 + 6s)i - 2sj + 18sk = 3

At the point where the line and the plane intersect, Q:

((5 + 6s)i - 2sj + 18sk) . (6i - 2j + 18k) = 3

30 + 36s + 4s + 324s = 3

364s = -27

s = -27/364

Therefore, vector OQ = (829/182)i + (27/182)j - (243/182)k

Therefore, vector AQ = -(81/182)i + (27/182)j - (243/182)k

Length AQ = root ((-81/182)^2 + (27/182)^2 + (-243/182)^2)

= 1.42

Textbook says 1.49. Where've I gone wrong? (If anywhere)

Answer 1

Take any point on one plane, say $(x,y,z)=(2,0,1)$ on the first one. Then calculate $$d= {|6x - 2y + 18z - 3|\over \sqrt{36+4+18^2}}={27\over \sqrt{364}}=1.4151845296...$$

Answer 2

We have that the vector $3\mathbf i - \mathbf j + 9\mathbf k$ is perpendicular to both planes.

A unit normal vector perpendicular to both planes is therefore $$\frac{1}{\sqrt{3^2 + 1^1 + 9^2}}(3\mathbf i - \mathbf j + 9\mathbf k) = \frac{1}{\sqrt{91}}(3\mathbf i - \mathbf j + 9\mathbf k).$$

Taking the intersection of the line $y=0,$ $z=0$ with each of the planes we find the position vectors of the two intersections are $5\mathbf i$ and $\frac12\mathbf i.$

Taking the inner product (aka dot product) of each of these position vectors with the unit normal vector gives us the distance of each plane from the origin in the direction of the normal vector. More simply, if we take the difference of the two position vectors, $5\mathbf i - \frac12\mathbf i = \frac92\mathbf i,$ we have a vector between the planes, and the inner product of that vector with the unit normal vector is the distance between the planes in the direction of the normal vector. (But since we don't care about direction for the final answer, we can take the absolute value in the end.)

So we have $$ \left(\frac{1}{\sqrt{91}}(3\mathbf i - \mathbf j + 9\mathbf k)\right) \cdot \frac92\mathbf i = \frac{3}{\sqrt{91}} \cdot \frac92 \approx 1.41518. $$

Again this matches your answer and not the textbook's answer. So either you've copied something incorrectly in the problem statement or there is an error in the book--either is possible, though the more carefully you checked your transcription the more likely the error is to be in the book.

Answer 3

A somewhat simpler method than yours: The constant terms in these equations are proportional to the signed distances of the planes from the origin. If you write the equations in the vector form $\vec n\cdot\vec r=d$ and then divide both sides by $\|\vec n\|$ to make it a unit vector, adjusting the sign if necessary so that both unit vector point in the same direction, the right-hand side is exactly the signed perpendicular distance of the plane from the origin. The difference between these values is the distance between the parallel planes: $$\left|{15\over\|(3,-1,9)\|} - {3\over\|(6,-2,18)\|}\right| = {27\over2\sqrt{91}}\approx1.42,$$ which agrees with your answer.

As to why it differs from the answer in the book, perhaps you misread the problem, or perhaps the textbook has an error. It’s been known to happen (far too often, as a matter of fact).