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.