Question

Find the distance of the planes 3x+4y-z=2 and 6x+8y-2z=3?

Answer 1

can I divide 6x+8y-2z=3 by 2, and then take the difference of the distance over the magnitude?

Answer 2

Hmm interesting proposition. I'm not 100% sure that would work, so I'm going to do it the way that seems most reliable to me, and then we can check. A normal vector \(\vec n\) of both of the planes is given by \(\vec n = \langle 3,4,-1\rangle\). The normalized version of this vector is \(\hat n = \frac{1}{\sqrt{26}}\langle3,4,-1 \rangle\). This will land on the first plane when we scale \(\hat n\) by some constant \(t_1\) and the other for some other constant \(t_2\). Since the direction of this vector is normal to both planes, the value \(|t_1-t_2|\) will represent the separation of the two planes. Substituting \(t_1\hat n\) into the first equation gives us the following.\[3\left(\frac{3t_1}{\sqrt{26}}\right)+4\left(\frac{4t_1}{\sqrt{26}}\right)-\left(\frac{t_1}{\sqrt{26}}\right)=2\]\[t_1=\frac{2\sqrt{26}}{6}=\frac{\sqrt{26}}{3}\]Similarly, substituting \(t_2\hat n\) into the second equation gives us the following.\[6\left(\frac{3t_2}{\sqrt{26}}\right)+8\left(\frac{4t_2}{\sqrt{26}}\right)-2\left(\frac{t_2}{\sqrt{26}}\right)=3\]\[t_2=\frac{3\sqrt{26}}{12}=\frac{\sqrt{26}}{4}\]Thus, we obtain the following.\[|t_1-t_2|=\frac{\sqrt{26}}{3}-\frac{\sqrt{26}}{4}=\boxed{\displaystyle \frac{\sqrt{26}}{12}}\]