find the direction ratios of the normal to the plane,which passes through points (1,0,0) and (0,1,0) and makes an angle (pi/4) with the plane x+y=3 .aslo find the equation of the plane.
Answers
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Let the equation of the plane passing through
(1 ,0,0) be,
a ( x-1) + b( y-0) + c (z -0) _______(1)
Here a, b and c are direction ratios of normal to the plane
if this plane passing through (1,0,0)
then,
a(0-1) + b(1-0) + c ( 0-0) = 0
=> -a + b = 0
=> a = b _________(2)
It is given that plane (1) makes an angle of π/4 with plane x+y=3
cos (π/4) = a×1 + b ×1 + c × 0
√a^2 +b^2 +c^2 √1^2 +1^1+0^2
=> 1 = a + b
√2 √a^2 + b^2 + c^2 √2
=> √a^2 + b^2 + c^2 = a + b
Squaring on both the sides, we get
=> a^2 + b^2 + c^2 =a^2 + b^2 + 2ab
=> c^2 = 2ab ________(3)
From (2) and (3), we get
c + 2a
So, the equation of the plane is
a ( x-1) + a (y-0) + √2 a(z-0) = 0
=> x-1 +y + √2 z =0
=> x + y + √2 z
Hence the direction ratios of the normal to the plane are proportional to 1, 1 ,+ √2
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