Q. 235.0( 1 Vote )

Find the distance

Answer :

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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