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

Answers (1)

Here \vec{n_{1}} =2 \widehat{i}+2\widehat{j}-3\widehat{k} and \vec{n_{2}} = 2\widehat{i}+5\widehat{j}+3\widehat{k} 

and d_{1} = 7  and  d_{2} = 9

Hence, using the relation \vec{r}.(\vec{n_{1}}+\lambda\vec{n_{2}}) = d_{1}+\lambda d_{2}, we get

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

or     \vec{r}.[(2+2\lambda)\widehat{i}+(2+5\lambda)\widehat{j}+(3\lambda-3)\widehat{k}] = 7+9\lambda             ..............(1)

where, \lambda is some real number.

Taking \vec{r} = x\widehat{i}+y\widehat{j}+z\widehat{k}, we get

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

or   x(2+2\lambda) + y(2+5\lambda) +z(3\lambda-3) = 7+9\lambda

or   2x+2y-3z-7 + \lambda(2x+5y+3z-9) = 0                       .............(2)

Given that the plane passes through the point (2,1,3), it must satisfy (2), i.e.,

                 (4+2-9-7) + \lambda(4+5+9-9) = 0

or                \lambda = \frac{10}{9}

Putting the values of \lambda in (1), we get

\vec{r}\left [\left ( 2+\frac{20}{9} \right )\widehat{i}+\left ( 2+\frac{50}{9} \right )\widehat{j}+\left ( \frac{10}{3}-3 \right )\widehat{k} \right ] = 7+10

or       \vec{r}\left ( \frac{38}{9}\widehat{i}+\frac{68}{9}\widehat{j}+\frac{1}{3}\widehat{k} \right ) = 17

or      \vec{r}.\left ( 38\widehat{i}+68\widehat{j}+3\widehat{k} \right ) = 153

which is the required vector equation of the plane.

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