Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

The distance of the point ' $a$ ' from the plane $\vec{r}.\vec{n}=q$ measured in the direction of the unit vector $\hat{b}$ is $=\frac{q-\vec{a}\cdot \vec{n}}{\hat{b}\cdot \vec{n}}$ Here $\vec{a}=\hat{i}+2\hat{j}+3\hat{k},\vec{n}=\hat{i}+\hat{j}+\hat{k}$ and $q=5$ Also $\hat{b}=\frac{2\hat{i}+3\hat{j}-6\hat{k}}{\sqrt{(2{)}^2+(3{)}^2+(-6{)}^2}}=\frac{2\hat{i}+3\hat{j}-6\hat{k}}{7}$ $\therefore$ The required distance

$$\begin{array}{rlr}& =\frac{5-(\hat{i}+2\hat{j}+3\hat{k})\cdot (\hat{i}+\hat{j}+\hat{k})}{\frac{1}{7}(2\hat{i}+3\hat{j}-6\hat{k})\cdot (\hat{i}+\hat{j}+\hat{k})}& \frac{5-(1+2+3)}{\frac{1}{7}(2+3-6)}=7\end{array}$$