Question

If O is the origin and the coordinates of A are (a, b, c). Find the direction cosines of OA and the equation of the plane through A at right angles to OA. [NCERT EXEMPLAR]

Answer 1

It is given that O is the origin and the coordinates of A are (a, b, c).

The direction ratios of OA are proportional to

$a-0,b-0,c-0$ or a, b, c

∴ Direction cosines of OA are

$\frac{a}{\sqrt{a^2+b^2+c^2}},\frac{b}{\sqrt{a^2+b^2+c^2}},\frac{c}{\sqrt{a^2+b^2+c^2}}$

The normal vector to the required plane is $a\hat{i}+b\hat{j}+c\hat{k}$.

The vector equation of the plane through A(a, b, c) and perpendicular to OA is

$$\begin{gathered} \left[\overrightarrow{r}-\left(a\hat{i}+b\hat{j}+c\hat{k}\right)\right].\left(a\hat{i}+b\hat{j}+c\hat{k}\right)=0\left[\left(\overrightarrow{r}-\overrightarrow{a}\right).\overrightarrow{n}=0\right] \\ \Rightarrow \overrightarrow{r}.\left(a\hat{i}+b\hat{j}+c\hat{k}\right)=\left(a\hat{i}+b\hat{j}+c\hat{k}\right).\left(a\hat{i}+b\hat{j}+c\hat{k}\right) \\ \Rightarrow \overrightarrow{r}.\left(a\hat{i}+b\hat{j}+c\hat{k}\right)=a^2+b^2+c^2 \end{gathered}$$

The Cartesian equation of this plane is

$$\begin{gathered} \left(x\hat{i}+y\hat{j}+z\hat{k}\right).\left(a\hat{i}+b\hat{j}+c\hat{k}\right)=a^2+b^2+c^2 \\ \mathrm{Or}ax+by+cz=a^2+b^2+c^2 \end{gathered}$$