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The vector equation of the plane through the line of intersection of the planes

Question:

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

  1. $\overrightarrow{\mathrm{r}} \times(\hat{\mathrm{i}}+\hat{\mathrm{k}})+2=0$

  2. $\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}-\hat{\mathrm{k}})-2=0$

  3. $\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}-\hat{\mathrm{k}})+2=0$

  4. $\overrightarrow{\mathrm{r}} \times(\hat{\mathrm{i}}-\hat{\mathrm{k}})+2=0$


Correct Option: , 3

Solution:

Let the plane be

$(x+y+z-1)+\lambda(2 x+3 y+4 z-5)=0$

$\Rightarrow(2 \lambda+1) x+(3 \lambda+1) y+(4 \lambda+1) z-(5 \lambda+1)=0$

$\perp$ to the plane $x-y+z=0$

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

$\Rightarrow$ the required plane is $x-z+2=0$

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