Question

A plane passes through the point (1, -2, 5) and is perpendicular to the line joining the origin to the point 3i + j - k. . Find the vector and Cartesian forms of the equation of the plane.

Answer 1

As per the given criteria the required plane is passing through Q (1, -2, 5) and is perpendicular to OP, where point O is the origin and position vector of point P is 3i + j - k. 

Let the position vector of this point Q be 

And it is also given the plane is normal to the line joining the points O(0,0,0) and position vector of point P is  \(3\hat i + \hat j - \hat k\)

Then  \(\vec n=\vec {OP}\)

⇒ \(\vec n\) = Position vector of \(\vec P\) - Position vector of \(\vec O\)

We know that vector equation of a plane passing through point  \(\vec a\) and perpendicular/normal to the vector  \(\vec n\)  is given by

\((\vec r-\vec a).\vec n=0\)

Substituting the values from eqn(i) and eqn(ii) in the above equation, we get

 (by multiplying the two vectors using the formula \(\vec A.\vec B\) = A_xB_x+ A_yB_y + A_zB_z)

⇒ \(\vec r.(3\hat i+\hat j-\hat k)+4=0\) is the vector equation of a required plane.

Let  \(\vec r=(\text x\hat i+y\hat j+z\hat k)\)

Then, the above vector equation of the plane becomes,

⇒ 3x + y - z - 4 = 0

This is the Cartesian form of equation of the required plane.