Find the equation of the plane through the line of intersection of the planes ` vec rdot(( hat i+3 hat j)-6=0\ a n d\ vec rdot((3 hat i- hat j-4 hat k)=0,` which is at a unit distance from the origin.

Explanation

Given planes are

`vec{r} cdot(hat{i}+3 hat{j})-6=0` and `vec{r} cdot(3 hat{i}-hat{j}-4 hat{k})=0`

Equation of family of planes through intersection of these planes is


`vec{r} cdot(hat{i}+3 hat{j})-6+lambda[vec{r} cdot(3 hat{i}-hat{j}-4 hat{k})]=0`

`Rightarrow vec{r} cdot[(1+3 lambda) hat{i}+(3-lambda) hat{j}+hat{k}(-4 lambda)]=6`

`Rightarrow frac{vec{r} cdot[(1+lambda) hat{i}+(3-lambda) hat{j}+hat{k}(-4 lambda)]}{sqrt{(1+3 lambda)^{2}+(3-lambda)^{2}+(-4 lambda)^{2}}}=frac{6}{sqrt{(1+3 lambda)^{2}+(3-lambda)^{2}+(-4 lambda)^{2}}}`

Since, the perpendicular distance from origin is unity.

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