Question

The distance between the planes \(\vec{r}\cdot(2\hat{i}+\hat{j}-2\hat{k})+5=0\) and \(\vec{r}\cdot(6\hat{i}+3\hat{j}-6\hat{k})+2=0\) is :

• A. \(\frac{9}{13}\)
• B. \(\frac{15}{4}\)
• C. \(\frac{13}{9}\)
• D. \(\frac{1}{13}\)

Hint:

Make the normal vectors identical, then use the formula for the distance between parallel planes.

Answer 1

The Correct Option is C

The second plane is \(\vec{r}\cdot(6\hat{i}+3\hat{j}-6\hat{k})+2=0\). Divide it by 3 to get \(\vec{r}\cdot(2\hat{i}+\hat{j}-2\hat{k})+\frac{2}{3}=0\).
Now both planes have the same normal \(\vec{n}=2\hat{i}+\hat{j}-2\hat{k}\), so they are parallel, with \(|\vec{n}|=\sqrt{4+1+4}=3\).
For parallel planes \(\vec{r}\cdot\vec{n}+d_1=0\) and \(\vec{r}\cdot\vec{n}+d_2=0\), the distance is \(\frac{|d_1-d_2|}{|\vec{n}|}\).
Distance \(=\frac{|5-\frac{2}{3}|}{3}=\frac{13/3}{3}=\frac{13}{9}\).
Answer: (3) \(\frac{13}{9}\)