Question

The distance between the planes $\overrightarrow{r}.(2i-j+3k)=4$ and $\overrightarrow{r}.(6i-3j+9k)+13=0$ is:

• A. $\frac{5}{3\sqrt{14}}$
• B. $\frac{10}{3\sqrt{14}}$
• C. $\frac{25}{3\sqrt{14}}$
• D. None of these

Answer 1

The correct option is C $\frac{25}{3\sqrt{14}}$

The distance between the parallel planes $r.(2i-j+3k)=4$ and $r.(2i-j+3k)=-\frac{13}{3}$

$[\because r.(6i-3j+9k)+13=0\Rightarrow r.(2i-j+3k)=-\frac{13}{3}]$

is $\frac{|4-(-\frac{13}{3})|}{\sqrt{2^2+{(-1)}^2+3^2}}$ $(\because$ required distance $=\frac{|d-k|}{n})$

$=\frac{|4+\frac{13}{3}|}{\sqrt{4+1+9}}=\frac{25}{3\sqrt{14}}$