Q. 245.0( 2 Votes )

Find the equation

Answer :

Given: Point P(1, 1, 1) and the line:


To find the equation of the plane passing through the given point and given line.


To prove: plane contains line,


Formula used:


Equation of line passing through point P(x1, y1, z1):


a(x – x1) + b(y – y1) + c(z – z1) = 0


Therefore,


Equation of line passing through point P(1, 1, 1):


a(x – 1) + b(y – 1) + c(z – 1) = 0


Since a plane contains the line


So, the plane passes through the point(-3, 1, 5) and parallel to the line with direction ratios(3, -1, -5)


a(-3 – 1) + b(1 – 1) + c(5 – 1) = 0


-4a + 4c = 0


4c = 4a


a = c……………(1)


Plane parallel to the line with direction ratios(3, -1, -5)


3a – b – 5c = 0


From (1):


3c – b – 5c = 0


- b – 2c = 0


b = -2c


Now,


a(x – 1) + b(y – 1) + c(z – 1) = 0


c(x – 1) + (-2c)(y – 1) + c(z – 1) = 0


(x – 1) – 2(y – 1) + (z – 1) = 0


x – 1 – 2y + 2 + z – 1 = 0


x – 2y + z = 0


Vector equation of plane:



If this plane contains line:


Then,



(1)(1) + (-2)(-2) + (-5)(1) = 0


1 + 4 – 5 = 0


0 = 0


which is true


So, plane, x – 2y + z = 0 contain line,


Hence Proved


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