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Find the equation of plane passing through (3,2,0)and contain the line (x-3/1,y-6/5,z-4/4)


3 years ago

Arun
24742 Points
							Let the equation of a plane passing through (3,2,0) bea(x−3)+b(y−2)+c(z−0)=0   ..(1)The line x−31=y−65=z−44passes through the point (3,2,0) and has direction ratios 1,5,4.If (1) contains the lines, it must pass through (3,6,4) and must be parallelto the line.Therefore,a(3−3)+b(6−2)+c(4−0)=0⇒4b+4c=0⇒b+c=0     ...(2)and a+5b+4c=0     ...(3)On solving (2) and (3), we geta=c and b=−cSo, From eq (1), we getc(x−3)−c(y−2)+c(z−0)=0⇒c[(x−3)−(y−2)+z]=0⇒(x−3)−(y−2)+z=0⇒x−y+z−1=0⇒x−y+z=1 is the required equation of plane.

3 years ago
Arun
24742 Points
							Let the equation of plane passing through (3,2,0) isa(x-3) + b(y-2) + c(z-0) = 0 x-3/1 = y-6/5 = z-4/4Pass through (3,2,0) and has direction ratios as 1,5,4 If the equation of plane contains this, it should pass from(3,6,4) and parallel to (1,5,4)Hencea(3-3) + b(6-2) + c *4 = 04b + 4c = 0b +c = 0And a + 5b +4 c = 0 On solving these two equationsa = c & b= -cHence equation of plane will bec(x-3) - c(y-2) +cz = 0x-3 -y +2 +z= 0x - y +z -1 = 0x -y +z = 1

3 years ago
Rishi Sharma
614 Points
							Dear Student,Please find below the solution to your problem.Let the equation of plane passing through (3,2,0) isa(x-3) + b(y-2) + c(z-0) = 0x-3/1 = y-6/5 = z-4/4Pass through (3,2,0) and has direction ratios as 1,5,4If the equation of plane contains this, it should pass from (3,6,4) and parallel to (1,5,4)Hencea(3-3) + b(6-2) + c *4 = 04b + 4c = 0b +c = 0Anda + 5b +4 c = 0On solving these two equationsa = c & b= -cHence equation of plane will bec(x-3) - c(y-2) +cz = 0x-3 -y +2 +z= 0x - y +z -1 = 0x -y +z = 1Thanks and Regards

3 months ago
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