Find the distance between the planes.
$$5 x - 2 y + z - 4 = 0,5 x - 2 y + z + 6 = 0$$
A) $\frac { 1 } { 6 }$
B) $\frac { 1 } { 3 } \sqrt { 30 }$
C) $\frac { 5 \sqrt { 30 } } { 6 }$
D) $\frac { 5 } { 6 }$
E) $\sqrt { 30 }$

Question

Quizplus · Accepted Answer

We can find the normal vectors of the planes by looking at their coefficients. The normal vectors will be $\begin{pmatrix} 5 \ -2 \ 1 \end{pmatrix}$ and $\begin{pmatrix} 5 \ -2 \ 1 \end{pmatrix}$. To find the distance between the planes, we take the dot product of the vector between any point on one plane to another with the normal vector of one of the planes, and then divide by the length of the normal vector.

Let's use the point $(0,0,4)$ on the first plane, and the normal vector $\begin{pmatrix} 5 \ -2 \ 1 \end{pmatrix}$ of the first plane. The vector between this point and the second plane is $\begin{pmatrix} x \ y \ z \end{pmatrix} - \begin{pmatrix} 0 \ 0 \ 4 \end{pmatrix} = \begin{pmatrix} x \ y \ z-4 \end{pmatrix}$.

Taking the dot product, we get:

$$\begin{pmatrix} 5 \ -2 \ 1 \end{pmatrix} \cdot \begin{pmatrix} x \ y \ z-4 \end{pmatrix} = 5x - 2y + z -20$$

To make this easier, we can find the distance between the point $(0,0,4)$ and the line of intersection between the two planes. We already have a normal vector for each plane, and we can find a direction vector for the line of intersection by taking the cross product of the normal vectors:

$$\begin{pmatrix} 5 \ -2 \ 1 \end{pmatrix} 	imes \begin{pmatrix} 5 \ -2 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 6 \ 20 \end{pmatrix}$$

This gives a direction vector of $\begin{pmatrix} 0 \ 6 \ 20 \end{pmatrix}$. We can use this vector and the known point $(0,0,4)$ to write a vector equation for the line of intersection:

$$\begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 4 \end{pmatrix} + t\begin{pmatrix} 0 \ 6 \ 20 \end{pmatrix}$$

Now, we need to find the value of $t$ that gives the point of intersection of this line and the second plane. Substituting into the equation for the second plane, we get:

$$5(0) - 2(0) + (4 + 20t) + 6 = 0 \Rightarrow t = -\frac{1}{6}$$

Substituting this value of $t$ back into the equation for the line of intersection, we get:

$$\begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 4 \end{pmatrix} - \frac{1}{6}\begin{pmatrix} 0 \ 6 \ 20 \end{pmatrix} = \begin{pmatrix} 0 \ -1 \ \frac{2}{3} \end{pmatrix}$$

Now we can find the distance between this point and the first plane (which we already know has a normal vector $\begin{pmatrix} 5 \ -2 \ 1 \end{pmatrix}$). We take the dot product and divide by the length:

$$	ext{Distance} = \frac{\begin{pmatrix} 5 \ -2 \ 1 \end{pmatrix} \cdot \begin{pmatrix} 0 \ -1 \ \frac{2}{3} \end{pmatrix}}{\left\| \begin{pmatrix} 5 \ -2 \ 1 \end{pmatrix} ight\|} = \frac{-2}{\sqrt{30}} = \boxed{\frac{1}{3} \sqrt{30}}$$

Find the Distance Between the Planes 5x−2y+z−4=0,5x−2y+z+6=05 x - 2 y + z - 4 = 0,5 x - 2 y + z + 6 = 05x−2y+z−4=0,5x−2y+z+6=0

Find the Distance Between the Planes $5 x - 2 y + z - 4 = 0,5 x - 2 y + z + 6 = 0$