Line and Plane — Maths STD 12 Science — Question

is $\bar{r} \cdot \bar{n}=\bar{a} \cdot \bar{n} \ldots(1)$

The position vectors $\bar{a}$ and $\bar{b}$ of the given points $A$ and $B$ are $\bar{a}=2 \hat{i}+3 \hat{j}+6 \hat{k}$ and

$\bar{b}=4 \hat{i}+3 \hat{j}-2 \hat{k}$

If $M$ is the midpoint of segment $A B$, the position vector $\bar{m}$ of $M$ is given by

$\begin{aligned} \bar{m}=\frac{\bar{a}+\bar{b}}{2} & =\frac{(2 \hat{i}+3 \hat{j}+6 \hat{k})+(4 \hat{i}+3 \hat{j}-2 \hat{k})}{2} \\ & =\frac{6 \hat{i}+6 \hat{j}+4 \hat{k}}{2}=3 \hat{i}+3 \hat{j}+2 \hat{k}\end{aligned}$

The plane passes through $\mathrm{M}(\bar{m})$.

$\mathrm{AB}$ is perpendicular to the plane

If $\bar{n}$ is normal to the plane, then $\bar{n}=\overline{\mathrm{AB}}$

$\begin{aligned} & \therefore \bar{n}=\bar{b}-\bar{a}=(4 \hat{i}+3 \hat{j}-2 \hat{k})-(2 \hat{i}+3 \hat{j}+6 \hat{k}) \\ & =2 \hat{i}-8 \hat{k} \\ & \therefore \bar{m} \cdot \bar{n}=(3 \hat{i}+3 \hat{j}+2 \hat{k}) \cdot(2 \hat{i}-8 \hat{k}) \\ & =(3)(2)+(3)(0)+(2)(-8) \\ & =6+0-16=-10 \text {. } \\ & \end{aligned}$

$\therefore$ from (1), the vector equation of the required plane is

$\begin{aligned} & \bar{r} \cdot \bar{n}=\bar{m} \cdot \bar{n} \\ & \text { i.e. } \bar{r} \cdot(2 \hat{i}-8 \hat{k})=-10 \\ & \text { i.e. } \bar{r} \cdot(\hat{i}-4 \hat{k})=-5\end{aligned}$