Question

Show that the planes given by 2x − 3y + 5z = 7 and −6x + 9y − 15z = 8 are parallel, and find the distance between the planes

Short answer

$$\begin{gathered} \text{Therefore, the distance from the point}P(1,2,-3)\text{to the line}\mathbf{r}\left(t\right)=⟨2+5t,-3+t,-4t⟩\text{is} \\ 6.10units \end{gathered}$$

Step-by-step solution

Step 1:Given information

the planes given by 2x − 3y + 5z = 7 and −6x + 9y − 15z = 8

Step 2:Calculation

Consider the point$P(1,2,-3)$and the line with the equation $\mathbf{r}\left(t\right)=⟨2+5t,-3+t,-4t⟩$and the plane with the equation $x-3y+4z=5$.

$\text{The point}{P}_0(2,-3,0)\text{corresponds to}\mathbf{r}\left(0\right)\text{and the direction vector of the line is}\mathbf{d}=⟨5,1,-4⟩\text{.}$

$\text{The norm of the direction vector}\mathbf{d}=⟨5,1,-4⟩\text{is:}$

$\Vert \mathbf{d}\Vert =\sqrt{5^2+1^2+(-4{)}^2}$

$=\sqrt{25+1+16}=\sqrt{42}$

$\text{The vector}\overrightarrow{P_{0}P}\text{is given by:}$

$\overrightarrow{P_{0}P}=⟨1-2,2-(-3),-3-0⟩$

$=⟨-1,5,-3⟩$

$\text{The value of}\mathbf{d}\times \overrightarrow{P_{0}P}\text{is:}$

$\mathbf{d}\times {\overrightarrow{P}}_{0}P=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 5& 1& -4\\ -1& 5& -3\end{array}\right|$

$=-23i-19j+26k$

$=⟨-23,-19,26⟩$

$\text{The distance from point}P(1,2,-3)\text{to the line}\mathbf{r}\left(t\right)=⟨2+5t,-3+t,-4t⟩\text{is:}$

$\frac{||\mathbf{d}\times \overrightarrow{P_{0}P}||}{\Vert \mathbf{d}\Vert }=\frac{\Vert ⟨-23,-19,26⟩\Vert }{\sqrt{42}}$

$=\frac{\sqrt{1566}}{\sqrt{42}}$

$\approx 6.10\text{units}$

$$\begin{gathered} \text{Therefore, the distance from the point}P(1,2,-3)\text{to the line}\mathbf{r}\left(t\right)=⟨2+5t,-3+t,-4t⟩\text{is} \\ 6.10units \end{gathered}$$