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Answer please! - The equation of the plane containing the straight line and the perpendicular to the plane containin - Three Dimensional Geometry - JEE Main

The equation of the plane containing the straight line x/2=y/3=z/4 and the perpendicular to the plane containing the straight lines x/3=y/4=z/2 and x/4=y/2=z/3 is:

  • Option 1)

     

    5x+2y-4z=0

  • Option 2)

    x+2y-2z=0

     

  • Option 3)

     

    x-2y+z=0

  • Option 4)

     

    3x+2y-3z=0

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Equation of any plane passing through the line of intersection of two planes (Cartesian form ) -

  The equation of any plane passing through the line of intersection of two planes

ax+by+cz+d= 0 and

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

\left ( ax+by+cz+d \right )+\lambda \left ( a_{1}x+b_{1}y+c_{1}z+d _{1}\right )= 0

 

 

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From the concept we have learned vector along the normal to the plane containing the lines

\frac{x}{3} = \frac{y}{4} = \frac{z}{2} and \frac{x}{4} = \frac{y}{2} = \frac{z}{3} is (8\hat{i} - \hat{j} - 10\hat{k})

vector perpendicular to the vectors 2\hat{i} + 3\hat{j} + 4\hat{k} and 8\hat{i} -\hat{j} -10\hat{k}  is  26\hat{i} -52\hat{j} +26\hat{k} 

So, the required plane is 

26x -52y +26z = 0 or  x -2y +z = 0

 


Option 1)

 

5x+2y-4z=0

Option 2)

x+2y-2z=0

 

Option 3)

 

x-2y+z=0

Option 4)

 

3x+2y-3z=0

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