Question

Find the distance between the planes $\overrightarrow{r}·\left(\hat{i}+2\hat{j}+3\hat{k}\right)+7=0\mathrm{and}\overrightarrow{r}·\left(2\hat{i}+4\hat{j}+6\hat{k}\right)+7=0.$

Answer 1

$$\begin{gathered} \text{The given planes are} \\ \overrightarrow{r}.\left(\hat{i}+2\hat{j}+3\hat{k}\right)=-7 \\ \Rightarrow x+2y+3z=-7 \\ \text{Multiplying this equation of the plane by 2, we get} \\ 2x+4y+6z=-14...\left(1\right) \\ \text{and} \\ \overrightarrow{r}.\left(2\hat{i}+4\hat{j}+6\hat{k}\right)=-7 \\ \Rightarrow 2x+4y+6z=-7...\left(2\right) \\ \text{We know that the distance between two planes}ax+by+cz=d_1\text{and}ax+by+cz=d_2\text{is}\frac{\left|d_2-d_1\right|}{\sqrt{a^2+b^2+c^2}} \\ \text{So, the required distance} \\ \text{=}\frac{\left|-7-\left(-14\right)\right|}{\sqrt{2^2+4^2+6^2}} \\ \text{=}\frac{\left|7\right|}{\sqrt{4+16+36}} \\ =\frac{7}{\sqrt{56}}\text{units} \end{gathered}$$