Q23 of 26 Page 1

Find the distance of the point from the plane 3 x + y – z + 2 = 0 measured parallel to the line . Also, find the foot of the perpendicular from the given point upon the given plane.

OR


Find the equation of the plane passing through the line of intersection of the planes and and passing through the point (3, –2, –1). Also, find the angle between the two given planes.

Given Data:




Plane 3x + y – z + 2 = 0


Equation of line passing through (3,–2, 1) and parallel to the line




Any general point on this line is (2α+3,–3α–2, α+1)


As line will intersect plane 3x + y – z + 2 = 0 at some point,


3(2α+3)–3α–2 – α–1 + 2 = 0


α = –4


Therefore this point is (–5, 10,–3)


Distance between the points,



d = 4√14


Now the line passing through (3,–2, 1) and perpendicular to the plane 3x + y – z + 2 = 0 is



Any general point on this line is (3β+3, β–2, –β+1)


If it pass through the plane then it should suffice the plane equation


3(3β+3) + β–2 + β –1 + 2 = 0



Putting this value in (3β+3, β–2,–β+1) we get



OR


Given Data:


Point: (3, –2, –1)


Planes:




The intersection line of two planes is




If this line passes through the point (3, –2, –1)


(2+α)3 – 2(3+α) – 1(–1–2α) + 1=0



Therefore equation of plane is




The angle between the two given planes is






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