Question

Find the equation of the plane passing through (a,b,c)and parallel to the plane $\overrightarrow{r}$.($\hat i+\hat j+\hat k$)=2.

Answer 1

Any plane parallel to the plane $\overrightarrow{r} . \hat i+\hat j+\hat k$ =2, is of the form

$\overrightarrow{r}$.($\hat i+\hat j+\hat k$) = λ...(1)

The plane passes through the point (a,b,c).

Therefore, the position vector r→ of this point is $\overrightarrow{r}$=$a\hat i+b\hat j+c\hat k$

Therefore, equation(1) becomes

($a\hat i+b\hat j+c\hat k$).($\hat i+\hat j+\hat k$)=λ

⇒a+b+c=λ

Substituting λ=a+b+c in equation(1), we obtain

$\overrightarrow{r}$.($\hat i+\hat j+\hat k$)=a+b+c...(2)

This is the vector equation of the required plane.

Substituting$\overrightarrow{r}$=$x\hat i+y\hat j+z\hat k$ in equation(2), we obtain

($x\hat i+y\hat j+z\hat k$).($\hat i+\hat j+\hat k$)=a+b+c

⇒x+y+z=a+b+c.