The distance of a point vec{a} from the plane | vedclass

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(D) The distance of a point $\vec{a}$ from the plane $\vec{r} \cdot \vec{m} = d$ is given by $\frac{|\vec{a} \cdot \vec{m} - d|}{|\vec{m}|}$.Given the

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(D) The distance of a point $\vec{a}$ from the plane $\vec{r} \cdot \vec{m} = d$ is given by $\frac{|\vec{a} \cdot \vec{m} - d|}{|\vec{m}|}$.Given the plane $\vec{r} \cdot (2\hat{i} + 6\hat{j} - 9\hat{k}) = -1$,we have $\vec{m} = 2\hat{i} + 6\hat{j} - 9\hat{k}$ and $d = -1$.The magnitude $|\vec{m}| = \sqrt{2^2 + 6^2 + (-9)^2} = \sqrt{4 + 36 + 81} = \sqrt{121} = 11$.For the point $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$,the distance $p$ is:$p = \frac{|(1)(2) + (2)(6) + (3)(-9) - (-1)|}{11} = \frac{|2 + 12 - 27 + 1|}{11} = \frac{|-12|}{11} = \frac{12}{11}$.The distance $q$ of the origin $(0, 0, 0)$ from the plane is:$q = \frac{|0 - (-1)|}{11} = \frac{1}{11}$.Therefore,$p - q = \frac{12}{11} - \frac{1}{11} = \frac{11}{11} = 1$.

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