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  • Tell me? Equation of line of intersection of planes 2x-y+2z-1=0 and 2x+y-2z+1=0

Equation of line of intersection of planes 2x-y+2z-1=0 and 2x+y-2z+1=0

  • Option 1) x/0= (y+1)/8=z/4

     

  • Option 2) x=y-1=z+1

     

  • Option 3) x/2= (y-1)/4=(z-1)/3

     

  • Option 4) none of these 

     

 

Answers (1)

best_answer

As we have learned

Equation of line as intersection of two planes -

Let the two intersecting planes be

ax+by+cz+d= 0 and 

a_{1}x+b_{1}y+c_{1}z+d_{1}= 0

then the parallel vector of line formed their intersection can be obtained by

\begin{vmatrix} \hat{i} &\hat{j} & \hat{k}\\ a&b &c \\ a_{1} & b_{1} & c_{1} \end{vmatrix}= A\hat{i}+B\hat{j}+C\hat{k}(assumed)

and points can be obtained by putting z= 0 and solving

ax+by+d= 0 and 

a_{1}x+b_{1}y+d_{1}= 0 say \alpha ,\beta

Now the equation will be

\frac{x-\alpha }{A}=\frac{y-\beta }{B}=\frac{z-0 }{C}

 

-

 

 Vector parallel to line required, will be along cross product of normals to both planes \therefore \vec{n_1}\times\vec{n_2}=\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ 2 & -1&2\\ 2& -1& -2 \end{vmatrix}=8\hat{i}+4\hat{k}

Now we need one common point of both planes, which will also lie on line of intersectio, so let us put z=0 an both planes so equation become 2x-y=1 and 2x+y=-1 on solving we get x=0, y=-1 

\thereforeone common point is (0,-1,0)

\therefore line of intersection is 

\frac{x-0}{0}=\frac{y+1}{8}=\frac{z-0}{4}


Option 1)

\frac{x}{0}=\frac{y+1}{8}=\frac{z}{4}

Option 2)

x=y-1=z+1

Option 3)

\frac{x}{2}=\frac{y-1}{4}=\frac{z-1}{3}

Option 4)

none of these

Posted by

Himanshu

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