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

Arun
25757 Points
5 years ago
Let the equation of a plane passing through (3,2,0) bea(x3)+b(y2)+c(z0)=0   ..(1)The line x31=y65=z44passes 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(33)+b(62)+c(40)=04b+4c=0b+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(x3)c(y2)+c(z0)=0c[(x3)(y2)+z]=0(x3)(y2)+z=0xy+z1=0xy+z=1 is the required equation of plane.
Arun
25757 Points
5 years ago
Let the equation of plane passing through (3,2,0) is
a(x-3) + b(y-2) + c(z-0) = 0

x-3/1 = y-6/5 = z-4/4
Pass 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)
Hence
a(3-3) + b(6-2) + c *4 = 0
4b + 4c = 0
b +c = 0
And
a + 5b +4 c = 0

On solving these two equations
a = c & b= -c
Hence equation of plane will be
c(x-3) - c(y-2) +cz = 0
x-3 -y +2 +z= 0
x - y +z -1 = 0
x -y +z = 1
Rishi Sharma
3 years ago
Dear Student,

Let the equation of plane passing through (3,2,0) is
a(x-3) + b(y-2) + c(z-0) = 0
x-3/1 = y-6/5 = z-4/4
Pass 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)
Hence
a(3-3) + b(6-2) + c *4 = 0
4b + 4c = 0
b +c = 0
And
a + 5b +4 c = 0
On solving these two equations
a = c & b= -c
Hence equation of plane will be
c(x-3) - c(y-2) +cz = 0
x-3 -y +2 +z= 0
x - y +z -1 = 0
x -y +z = 1

Thanks and Regards