Question

Find the equation of the plane through the point $(4,-3,2)$ and perpendicular to the line of intersection of the planes $x-y+2z-3=0$ and $2x-y-3z=0$. Find the point of intersection of the line $\overrightarrow{r}=\hat{i}+2\hat{j}-\hat{k}+\lambda (\hat{i}+3\hat{j}-9\hat{k})$ and the plane obtained above.

Answer 1

given,

plane passing through the pt $P(4,-3,2)$ and perpendicular to line of intersection of planes $x-y+2z-3=0$ and $2x-y-3z=0$

Let equation of plane be $\overrightarrow{r}-\overrightarrow{p}).\overline{n}=0$

Let equation of line passing through line of intersection of $2$ plane is $L_1=\overrightarrow{a}+\lambda \overrightarrow{b}$

$\overrightarrow{n}\parallel \overrightarrow{b}$

$P_1:x-y+2z-3=0\to \overrightarrow{n_1}(1,-1,2)$

$P_2:2x-y-3z=0\to \overrightarrow{n_2}(2,-1,-3)$

$L_1\perp n_1,L_1\perp n_2$ $\overrightarrow{L_1}\to \overrightarrow{n_1}\times \overrightarrow{n_2}=\left|\begin{array}{ccc}i& j& k\\ 1& -1& 2\\ 2& -1& -3\end{array}\right|$

$\overrightarrow{b}=5\hat{i}+7\hat{j}+\hat{k}$

$[\overrightarrow{r}-\overrightarrow{p}].\overrightarrow{n}=0$

$(\overrightarrow{r}-(4\hat{i}-3\hat{j}+2\hat{k}\left)\right).(5\hat{i}+7\hat{j}+\hat{k})=0$

$\left[\right(x-4)\hat{i}+(y+3)\hat{j}+(2-z\left)\hat{k}\right].(5\hat{i}+7\hat{j}+\hat{k})=0$

$5(x-4)+7(y+3)+(z-2)=0$

$5x+7y+z-20+21-2=0$

$5x+7y+z-1=0\to$ equ of plane

$\overrightarrow{v}=\hat{i}+2\hat{j}-\hat{k}+\lambda (\hat{i}+3\hat{j}-9\hat{k})$

$\overrightarrow{r}=(1+\lambda )\hat{i}+(2+3\lambda )\hat{j}+(-1-9\lambda )\hat{k}$

Any point on the line is $\left[\right(1+\lambda ),(2+3\lambda ),(-1-9\lambda \left)\right]$

satisfies the plane equation

$5(1+\lambda )+7(2+3\lambda )-1(-1-9\lambda )=0$

$20+35\lambda =0\Rightarrow \lambda =-4/7$

point $\left[\right(1-4/7),(2-\frac{12}{7}),(-1+\frac{36}{7})]$

$(3/7,2/7,29/7)$

$\therefore$ point of intersection of line & plane, is $=(3/7,2/7,29/7)$