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Q.
The distance of the plane $\vec{r}\cdot\left(\frac{2}{7}\hat{i}+\frac{3}{7}\hat{j}-\frac{6}{7}\hat{k}\right)=1$ from the origin is
Three Dimensional Geometry
Solution:
$\vec{r}\cdot\left(\frac{2}{7}\hat{i}+\frac{3}{7}\hat{j}-\frac{6}{7}\hat{k}\right)=1$
Here $\vec{n}=\frac{2}{7}\hat{i}+\frac{3}{7}\hat{j}-\frac{6}{7}\hat{k}$
$\left|\vec{n}\right|=\sqrt{\left(\frac{2}{7}\right)^{2}+\left(\frac{3}{7}\right)^{2}+\left(\frac{-6}{7}\right)^{2}}$
$=\sqrt{\frac{49}{49}}=1$
So, $\vec{n}$ is a unit vector. Hence given eq. of plane is in normal from.
$\therefore $ Distance from origin $= 1$