Question

Find the vector equation of the plane passing through the intersection of the planes $\overrightarrow{r,}(2\hat{i}+2\hat{j}-3\hat{k})=7,\overrightarrow{r}(2\hat{i}+5\hat{j}+3\hat{k})=9$ and through the point (2,1,3)

Answer 1

We have,

Given that equation of planes are

$\overrightarrow{r}.(2\hat{i}+2\hat{j}-3\hat{k})=7......\left(1\right)$

$\overrightarrow{r}.(2\hat{i}+5\hat{j}+3\hat{k})=9......\left(2\right)$

And through the point $(2,1,3)$.

So, we know that,

On comparing that, equation (1) and (2) to, we get,

$\overrightarrow{r}.\overrightarrow{n_1}=d_1$ and$\overrightarrow{r}.\overrightarrow{n_2}=d_2$

$n_1=2\hat{i}+2\hat{j}-3\hat{k}and{d}_1=7$

$n_2=2\hat{i}+5\hat{j}+3\hat{k}and{d}_2=9$

So, equation of plane is

$\overrightarrow{r}.(\overrightarrow{n_1}+\lambda \overrightarrow{n_2})=d_1+\lambda d_2$

$\Rightarrow \overrightarrow{r}.[(2\hat{i}+2\hat{j}-3\hat{k})+\lambda (2\hat{i}+5\hat{j}+3\hat{k})]=7+9\lambda$

$\Rightarrow \overrightarrow{r}.[2\hat{i}+2\hat{j}-3\hat{k}+2\lambda \hat{i}+5\lambda \hat{j}+3\lambda \hat{k}]=9\lambda +7$

$\Rightarrow \overrightarrow{r}.[(2+2\lambda )\hat{i}+(2+5\lambda )\hat{j}+(-3+3\lambda )\hat{k}]=9\lambda +7......\left(3\right)$

Now, to find $$\lambda$$, put $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$

So,

$(x\hat{i}+y\hat{j}+z\hat{k}).[(2+2\lambda )\hat{i}+(2+5\lambda )\hat{j}+(-3+3\lambda )\hat{k}]=9\lambda +7$

$x(2+2\lambda )\hat{i}+y(2+5\lambda )\hat{j}+z(-3+3\lambda )\hat{k}=9\lambda +7$

But the plane passes through the point $(2,1,3)$.

Now,

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

$\Rightarrow 4+4\lambda +2+5\lambda -9+9\lambda =9\lambda +7$

$\Rightarrow 18\lambda -9\lambda =7+3$

$\Rightarrow 9\lambda =10$

$\Rightarrow \lambda =\frac{10}{9}$

Putting value of $$\lambda$$ in equation (3) and we get,

$\overrightarrow{r}.[(2+2\lambda )\hat{i}+(2+5\lambda )\hat{j}+(-3+3\lambda )\hat{k}]=9\lambda +7$

$\Rightarrow \overrightarrow{r}.[(2+\frac{20}{9})\hat{i}+(2+\frac{50}{9})\hat{j}+(-3+\frac{30}{9})\hat{k}]=10+7$

$\Rightarrow \overrightarrow{r}.[\frac{38}{9}\hat{i}+\frac{68}{9}\hat{j}+\frac{3}{9}\hat{k}]=17$

$\Rightarrow \frac{1}{9}\overrightarrow{r}.[38\hat{i}+68\hat{j}+3\hat{k}]=17$

$\Rightarrow \overrightarrow{r}.[38\hat{i}+68\hat{j}+3\hat{k}]=17\times 9$

$\Rightarrow \overrightarrow{r}.[38\hat{i}+68\hat{j}+3\hat{k}]=153$

Hence, this is the answer.