15.  Find the equation of the plane passing through the line of intersection of the planes \overrightarrow{r}.\left ( \widehat{i}+\widehat{j}+\widehat{k} \right )=1 and \overrightarrow{r}.\left (2 \widehat{i}+3\widehat{j}-\widehat{k} \right )+4=0   and parallel to x-axis.

Answers (1)

So, the given planes are:

\overrightarrow{r}.\left ( \widehat{i}+\widehat{j}+\widehat{k} \right )=1  and  \overrightarrow{r}.\left (2 \widehat{i}+3\widehat{j}-\widehat{k} \right )+4=0

The equation of any plane passing through the line of intersection of these planes is 

[\overrightarrow{r}.\left ( \widehat{i}+\widehat{j}+\widehat{k} \right )-1] + \lambda \left [ \overrightarrow{r}.\left (2 \widehat{i}+3\widehat{j}-\widehat{k} \right )+4\right ] = 0

\vec{r}.[(2\lambda+1)\widehat{i}+(3\lambda+1)\widehat{j}+(1-\lambda)\widehat{k}]+(4\lambda+1) = 0           ..............(1)

Its direction ratios are (2\lambda+1) , (3\lambda+1),  and (1-\lambda)  = 0 

The required plane is parallel to the x-axis.

Therefore, its normal is perpendicular to the x-axis.

The direction ratios of the x-axis are 1,0, and 0.

\therefore \1.(2\lambda+1) + 0(\3\lambda+1)+0(1-\lambda) = 0

\implies 2\lambda+1 = 0

\implies \lambda = -\frac{1}{2}

Substituting \lambda = -\frac{1}{2} in equation (1), we obtain

\implies \vec{r}.\left [ -\frac{1}{2}\widehat{j}+\frac{3}{2}\widehat{k} \right ]+(-3)= 0

\implies \vec{r}(\widehat{j}-3\widehat{k})+6= 0

So, the Cartesian equation is y -3z+6 = 0.

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