The distance of a point $\overrightarrow{\mathrm{a}}$ from the plane $\overrightarrow{\mathrm{r}} \cdot…
The distance of a point $\overrightarrow{\mathrm{a}}$ from the plane $\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{m}}=\mathrm{q}$ is given by $\frac{|\vec{a} \cdot \vec{m}-q|}{|m|}$. If the distance of the point $\hat{i}+2 \hat{j}+3 \hat{k}$ from the plane $\vec{r} \cdot(2 \hat{i}+6 \hat{j}-9 \hat{k})=-1$ is $p$ and the distance of the origin from this plane is $\mathrm{q}$, then $\mathrm{p}-\mathrm{q}=$
$6$
$5$
$2$
$1$
Solution
Distance from point $\vec{i}+2 \vec{j}+3 \vec{k}$ from the plane
$\begin{aligned}
& \vec{r} \cdot(2 \vec{i}+b \vec{j}-9 \vec{k})=-1 \text { is qp } \\
& \Rightarrow \frac{|2+12-27+1|}{\sqrt{4+36+81}}=\frac{12}{11}=p
\end{aligned}$
and distance from origin to the given plane is
$\frac{|0+1|}{11}=\frac{1}{11}=q$
Now $p-q=\frac{12}{11}-\frac{1}{11}=1$