Question

Find the vector equation of the plane passing through the points $(2,1,-1)$ and $(-1,3,4)$ and perpendicular to the plane $x-2y+4z=10.$ Also show that the plane thus obtained contains the line $\overrightarrow{r}=-\hat{i}+3\hat{j}+4\hat{k}+\lambda (3\hat{i}-2\hat{j}-5\hat{k}).$

Answer 1

Let the equation of plane through $(2,1,-1)$ be
$a(x-2)+b(y-1)+c(z+1)=0....\left(i\right)$
$\because \left(i\right)$ passes through $(-1,3,4)$
$\therefore a(-1-2)+b(3-1)+c(4+1)=0$

$\Rightarrow -3a+2b+5c=0...\left(ii\right)$

Also plane (i) is perpendicular to plane $x-2y+4z=10$
$\Rightarrow \overrightarrow{n_1}\perp \overrightarrow{n_2}\Rightarrow \overrightarrow{n_1}.\overrightarrow{n_2}=0$

$\therefore 1a-2b+4c=0...\left(iii\right)$

From (ii) and (iii), we get
$\frac{a}{8+10}=\frac{b}{5+12}=\frac{c}{6-2}\Rightarrow \frac{a}{18}=\frac{b}{17}=\frac{c}{4}=\lambda \left(say\right)$

$\Rightarrow a=18\lambda ,b=17\lambda ,c=4\lambda$

Putting the value of a, b, c in (i), we get
$18\lambda (x-2)+17\lambda (y-1)+4\lambda (z+1)=0$
$\Rightarrow 18x-36+17y-17+4z+4=0$
$\Rightarrow 18x+17y+4z=49$
$\therefore$ Required vector equation of plane is
$\overrightarrow{r}.(18\hat{i}+17\hat{j}+4\hat{k})=49...\left(iv\right)$
Obviously plane (iv) contains the line
$\overrightarrow{r}=(-\hat{i}+3\hat{j}+4\hat{k})+\lambda (3\hat{i}-2\hat{j}-5\hat{k})....\left(v\right)$
Since, point $(-\hat{i}+3\hat{j}+4\hat{k})$ satisfy equation (iv) and vector $(18\hat{i}+17\hat{j}+4\hat{k})$ is perpendicular to, $(3\hat{i}-2\hat{j}+5\hat{k}),$ as $(-\hat{i}+3\hat{j}+4\hat{k}).(18\hat{i}+17\hat{j}+4\hat{k})=-18+51+16=49$ and $(18\hat{i}+17\hat{j}+4\hat{k}).(3\hat{i}-2\hat{j}-5\hat{k})=54-34-20=0$
Therefore, (iv) contains line (v).