Question

Find the equation of the plane through the point (1,2,3) and normal to the planes x-y-z=5& 2x-5y-32=7.

Solution ,, Here, equation of the plane passing through the point (1,2,3) is, a(x-1) + b(x-2) + ((z-3) = 0 -(i). ( This Is the very first step of the solution.plz explain be how that came with derivation in Cartesian coordinate form only )

Answer 1

Step-by-step explanation:

The given planes are$\rm\:x-y-z=5$ and $\rm\: 2x-5y-3z=7$

The required plane is perpendicular to both planes, so, the normal to the required plane will be parallel to $\vec{n}_{1}\times\vec{n}_{2}$,

where $\vec{n}_{1}\:\:\&\:\:\vec{n}_{2}$ are normals to the given plane,

So,

$\vec{n}_{1} = \hat{i} - \hat{j} - \hat{k} \: \: \: \& \: \: \: \vec{n}_{2} = 2 \hat{i} - 5 \hat{j} - 3 \hat{k}$

$\implies \vec{n}_{1} \times \vec{n}_{2} = - 2\hat{i} + \hat{j} - 3 \hat{k}$

So, the required plane is passing through $\vec{a}=\hat{i}+2\hat{j}+3\hat{k}$ and its normal is $\vec{n}=-2\hat{i}+\hat{j}-3\hat{k}$

Now, required equation of plane:

$( \vec{r} - \vec{a}). \vec{n} = 0$

Here, $\vec{r}$ is position vector of general point

$\implies \: \vec{r}. \vec{n} - \vec{a}. \vec{n} = 0$

$\implies \: \vec{r}. \vec{n} = \vec{a}. \vec{n}$

$\implies \: \vec{r}. ( - 2\hat{i}+\hat{j} - 3\hat{k}) = ( \hat{i}+2\hat{j}+3\hat{k}).( - 2 \hat{i}+\hat{j} - 3\hat{k})$

$\implies \:( x\hat{i}+y\hat{j}+z\hat{k}). ( - 2\hat{i}+\hat{j} - 3\hat{k}) = ( \hat{i}+2\hat{j}+3\hat{k}).( - 2 \hat{i}+\hat{j} - 3\hat{k})$

$\implies \: - 2x+y - 3z =( - 2 + 2 - 9)$

$\implies \: - 2x+y - 3z = - 9$

$\implies \: 2x - y + 3z = 9$