A plane P passes through the line of intersection of the planes 2x - y + 3z = 2, x + y - z = 1 and the point (1 ,0 ,1 ).Q. What is the equation of the plane P?a)2x + 5y - 2 = 0b)5x + 2y - 5 = 0c)x + z -2 = 0d)2x - y - 2z = 0Correct answer is option 'B'. Can you explain this answer?

Question

Answer

Explanation:

Finding the Equation of Plane P:
- The equation of plane P passing through the line of intersection of the given planes can be found by finding the cross product of the normals of the two planes passing through the intersection line.
- The normal vectors of the planes 2x - y + 3z = 2 and x + y - z = 1 are <2, -1,='' 3=''> and <1, 1,='' -1=''> respectively.
- The cross product of these normal vectors is calculated as: <2, -1,='' 3=''> x <1, 1,='' -1=''> = i(3 - 1) - j(2 + 3) + k(-1 - 2) = 2i - 5j - 3k.
- Since the plane P passes through the point (1, 0, 1), the equation of plane P is given by the dot product of the normal vector and the vector from the point (1, 0, 1) to any point (x, y, z) on the plane P equal to 0.
- Substituting the values, we get 2(x-1) - 5(y-0) - 3(z-1) = 0 which simplifies to 2x - 5y - 3z + 2 + 3 = 0.
- Therefore, the equation of plane P passing through the line of intersection of the given planes is 2x - 5y - 3z - 5 = 0, which is equivalent to option 'B' (5x + 2y - 5 = 0).

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