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

Answers (1)

Given that the plane is passing through (a,b,c) and is parallel to the plane \overrightarrow{r}.(\widehat{i}+\widehat{j}+\widehat{k})=2

So, we have

The position vector of the point A(a,b,c) is, \vec{r_{A}} = a\widehat{i}+b\widehat{j}+c\widehat{k}

and any plane which is parallel to the plane, \overrightarrow{r}.(\widehat{i}+\widehat{j}+\widehat{k})=2 is of the form,

\overrightarrow{r}.(\widehat{i}+\widehat{j}+\widehat{k})=\lambda.                      .......................(1)

Therefore the equation we get,

( a\widehat{i}+b\widehat{j}+c\widehat{k}).(\widehat{i}+\widehat{j}+\widehat{k})=\lambda

Or, a+b+c = \lambda

So, now substituting the value of \lambda = a+b+c in equation (1), we get

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

So, this is the required equation of the plane .

Now, substituting \vec{r} = x\widehat{i}+y\widehat{j}+z\widehat{k} in equation (2), we get

(x\widehat{i}+y\widehat{j}+z\widehat{k}).(\widehat{i}+\widehat{j}+\widehat{k})=a+b+c

Or, x+y+z = a+b+c

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