The vector equation of the plane through the line of interectionb of the planes x+y+z=1 and 2x+3y+4z=5 which is perpendicular to the plane x-y+z=0 is :

  • Option 1)

    \vec{r}\times \left ( \vec{i}-\vec{k} \right )+2=0

  • Option 2)

    \vec{r}\cdot \left ( \vec{i}-\vec{k} \right )-2=0

  • Option 3)

    \vec{r}\times \left ( \vec{i}+\vec{k} \right )+2=0

     

  • Option 4)

    \vec{r}\cdot \left ( \vec{i}-\vec{k} \right )+2=0

 

Answers (1)

Equation of plane passing through line of interaction of plane P_1 and P_2 is 

\left ( x+y+z-1 \right )+\lambda \left ( 2x+3y+4z-5 \right )=0\cdots (i)

\Rightarrow \left ( 1+2\lambda \right )x+\left ( 1+3\lambda \right )y+\left ( 1+4\lambda \right )z-1-5\lambda =0\;\;\;\dots(ii)

plane (ii) and x-y+z=0 is perpendicular.

so, 

\left ( 1+2\lambda \right )\times 1+\left ( 1+3\lambda \right )\left ( -1 \right )+\left ( 1+4\lambda \right )\left ( 1 \right )=0

1+2\lambda -1-3\lambda +1+4\lambda =0

3\lambda +1=0

\lambda =-\frac{1}{3}

Equation of plane is 

\left ( 1-\frac{2}{3} \right )x+\left ( 1-\frac{3}{3} \right )y+\left ( 1+4\left ( -\frac{1}{3} \right ) \right )z-1+\frac{5}{3}=0

\frac{1}{3}x+0y-\frac{1}{3}z+\frac{2}{3}=0

\Rightarrow x-z+2=0

vector equation.              \vec{r}\cdot \left ( \hat{i}-\hat{k}\right )+2=0

 


Option 1)

\vec{r}\times \left ( \vec{i}-\vec{k} \right )+2=0

Option 2)

\vec{r}\cdot \left ( \vec{i}-\vec{k} \right )-2=0

Option 3)

\vec{r}\times \left ( \vec{i}+\vec{k} \right )+2=0

 

Option 4)

\vec{r}\cdot \left ( \vec{i}-\vec{k} \right )+2=0

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