Question

Find the direction cosines of perpendicular from the origin to the plane $\overline{r}(2\hat{i}+3\hat{j}+6\hat{k})+7=0$

Answer 1

$\overline{r}(2i+3j+6k)+7=0$.

$\overline{r}(-2i-3j-6k)=7$

now $n=-2i-3j-6k$

$|n|=\sqrt{4+9+36}=\sqrt{49}=7$

$\overline{r}(\frac{-2}{7}i-\frac{3}{7}j-\frac{6}{7}k)=\frac{7}{7}$

Now

dc's are $(\frac{-2}{7},\frac{-3}{7},\frac{-6}{7})$.